biol 204 Flashcards

midterm

1
Q

components of the cell theory

A

1) All organisms are composed of one or more cells (some organisms like prokaryotes are composed of a single cell)
2) The cell is the smallest unit that has the properties of life. If cells are broken open, the property of life is lost. They are unable to grow, reproduce, or respond to stimuli in a coordinated, potentially independent fashion
3) Cells arise only from growth and division of pre-existing cells. Although DNA and RNA contain the information required to manufacture an array of biological molecules, they cannot orchestrate the formation of the entire cell, new cells can only arise from pre-existing cells

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2
Q

components of primordial earth

A

h2s, co2, methane and ammonia

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3
Q

DNA

A

a large double stranded helix molecule that contains a unique alphabet that provides the instructions for assembling many of the important components of a cell organism from simpler molecules

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4
Q

RNA

A

single strand of nucleotides which is used for protein synthesis and carries genetic information for many viruses.

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5
Q

Protein

A

Molecules that carry out most of the activities of life, including the synthesis of all other biological molecules.

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6
Q

Difference between prokaryotic and eukaryotic cells

A

Eukaryote cells are 10x bigger, have nucleus, and are multicellular. They are characterized by an endomembrane system, which consists of the nuclear envelope, the endoplasmic reticulum, and the Golgi complex. They also have specialized motor (contractile) proteins that move cells and internal parts. Prokaryotes are more complex, have no organelles and lack a true membrane bound nucleus, instead their DNA floats freely as a nucleoid, much less internal membrane organization, are biochemically more versatile.

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7
Q

cyanobacteria

A

The earliest form of photosynthesis relied on compounds such as h2s and ferrous iron (Fe2+), which could be easily oxidized b energy trapped from sunlight (which would reduce CO2 into sugars). However, aprox 3 billion years ago, a group of prokaryotes called cyanobacteria appeared and could be used for something more common than h2s or Fe2+ as an electron donor for photosynthesis. Cyanobacteria were able to harness electrons from water and could thrive virtually anywhere there was sunlight.

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8
Q

endoplasmic reticulum

A

extensive interconnected network of membranous channels and vesicles. Each vesicle is formed by a single membrane that surrounds an enclosed space called the lumen of the ER. The ER occurs in two forms…
rough&smooth

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9
Q

Golgi complex

A

consists of a stack of flattened membranous sacs, usually located between rough er&plasma membrane

  • receives proteins made in the ER and transports to the complex in vesicles
  • more chemical modifications of proteins occur, those proteins are then sorted into other vesicles, then regulated movement of several types of proteins
  • proteins secreted from the cell are transported to the plasma membrane by secretory vesicles (release contents to the exterior by exocytosis). Vesicles may also form by the reverse process, calledendocytosis, which brings molecules into the cell from exterior
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10
Q

theory of endosymbiosis

A

prokaryotic ancestors of modern mitochondria and chloroplasts were engulfed by larger prokaryotic cells, forming a mutually advantageous relationship called a symbiosis and over time, the host cell and the endosymbionts became inseparable parts of the same organism.

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11
Q

factors in endosymbiosis

A

rise in atmospheric O2 is thought to be a key factor in endosymbiosis
Morphology – the form or shape of the mitochondria and chloroplasts is similar to a prokaryotic cell. Mitochondria resemble aerobic prokaryotes, chloroplasts resemble cyanobacteria
Reproduction – a cell cannot make a mitochondrion from a chloroplast, they are derived only from pre-existing mitochondria or chloroplasts. Both divide by binary fission, which is how prokaryotic cells divide
Genetic Information- mitochondria and chloroplasts contain DNA (as they should if they were free living cells). Where free-living bacterium contain a few thousand protein-coding genes, DNA in energy-transducing organelles contain less than 100 (because many of the genes have been relocated to the nucleus).
Transcription and Translation –chloroplasts and mitochondria contain a complete transcription and translational machinery, including the enzymes and ribosomes necessary to synthesize the proteins encoded by their DNA. Ribosomes of mitochondria and chloroplasts are similar to the type found in prokaryotes.
Electron Transport – Similar to free-living prokaryotes, mitochondria and chloroplasts can generate energy in the form of ATP through the presence of their own electron transport chain

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12
Q

exocytosis

A

in eukaryotes, the process by which a secretory vesicle fuses with the plasma membrane and releases the vesicle contents to the exterior

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13
Q

9+2 structure

A

A bundle of microtubules extends from the base to the tip of a eukaryotic flagellum or cilium. In the bundle, a circle of nine double microtubules surrounds a central pair of single micro- tubules, forming what is known as the 9 􏰑 2 complex. Dynein motor proteins slide the microtubules of the 9 􏰑 2 complex over each other to produce the flagellar or ciliar movements

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14
Q

endocytosis

A

brings molecules into the cell from the exterior

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15
Q

ribosomes

A

the organelle in contemporary organisms required for protein synthesis. It is interesting to note that the modern ribosome, which plays a key role as an intermediate between RNA and protein, is composed of about two-thirds RNA and one-third protein. Interestingly, it has recently been shown that the RNA of the ribosome, not the protein, actually catalyzes the incorporation of amino acids onto a growing peptide chain. Thus, the ribosome may be considered a type of ribozyme

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16
Q

protobionts

A

term given to a group of abiotically produced organic molecules that are surrounded by a membrane or membrane like structure. Development of protobionts was important because it allowed for an internal environment to develop that was distinctly different from the external environment: the concentration of key molecule could be higher and attain more order in a closed space. May have formed spontaneously. Ex- liposomes.

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17
Q

ribozymes

A

Group of RNA molecules that themselves could act as catalysts. They catalyze reactions on the precursor RNA molecules that lead to their own synthesis, as well on unrelated RNA molecules. Discovery of ribozymes revolutionized thinking about the origin of life, instead of the contemporary system that requires all 3 molecules (DNA, RNA, protein) early life may have existed in an RNA world – where a single type of molecule could serve as both a carrier of information and a catalyst.

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18
Q

panspermia hypothesis

A

life on earth could have had an extraterrestrial origin:

1) Although life seems very complex, it arose relatively quickly after the formation of Earth. The Earth formed 4.6 billion years ago, and we have clear fossil evidence of life dated to about 3.5 billion years ago and chemical evidence to about 3.9 billion years ago. Given that primordial Earth had to cool after being formed, many scientists argue that this window for the development of life is very narrow.
2) Research in the past decade has shown that life is far more resilient than previously thought and could possibly survive for years in space. Extremophiles, which are mostly prokaryotes, can thrive under very harsh conditions of temperature, pressure, and nutrients and might be able to survive in a dormant state in interstellar space. Prolonged dormancy is a property of the spores of a range of organisms, including a number of prokaryotes and simple eukaryotes. Spores are highly resistant to changes in the external environment and can be restored to active growth after exposure to high levels of radiation, water deficiency, and/or exposure to extreme temperatures. Given this, one cannot discount the possibility that simple life forms came to Earth about 4 billion years ago and initiated the evolution of life as we know it

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19
Q

autotroph

A

self nourishment, mostly plants, synthesize organic carbon molecules from using inorganic co2

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20
Q

heterotrophs

A

mostly animals, obtain carbon from organic molecules (either living hosts or from organic molecules in products, wastes, remains of dead organisms)

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21
Q

chemoautotroph

A

obtain energy by oxidizing inorganic or organic substances

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22
Q

phototrophs

A

obtain energy from light

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23
Q

photoautotroph

A

energy from sunlight, carbon from environmental co2

*some photosynthetic bacteria, some proteins, some plants

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24
Q

photoheterotroph

A

every from sunlight, carbon from organic substances

*some photosynthetic bacteria

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25
Q

chemoautotroph

A

energy from inorganic chemicals, carbon from environmental co2
*some bacteria and archness, not found in eukaryotes

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26
Q

chemoheterotroph

A

energy from inorganic chemicals, carbon from organic sources

*some bacteria and archaens, proteins, fungi, animals, plants

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27
Q

natural selection

A

changes in environment favour particular genotypes that survive

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28
Q

evolution

A
  • Evolution is a gradual change in the characteristics of a population of organisms over time, and can be the result of selection
  • evolution is a central key to understanding the diversity of life on earth
  • theory of evolutionexplains both the unity and diversity of all life; tells us all organisms alive today descended from a common ancestor, which explains why all organisms share features such as the use for ATP as an energy source, DNA as genetic material, and plasma membranes composed of lipid bilayers
  • evolution also tells us species change over time as a result of natural selection
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29
Q

adaptive radiation

A

adaptive radiations, therefore, occur when an evolutionary breakthrough allows diversification of life. Adaptive radiations are a recurring theme in the development of biodiversity. At one level, we can see it in the evolution of the fauna and flora of the Hawaiian Islands, arguably the most isolated landmasses in the world. Adaptive radiation on islands reminds us that species can be adapting to new ways of life at the same time in different parts of the world. The result is a mosaic of life, with many examples of parallel and convergent evolution

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30
Q

challenges from organisms moving from water to land

A

matters of support, conservation of water, reproduction, and disposal of wastes. Other facts of life are also different for organisms living on land as opposed to in water. Some of the differences between water and air include density and viscosity, which, in turn, affect rates of diffusion and availability of oxygen.

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31
Q

opportunities from organisms moving to land

A

easy access and less energy expended to get oxygen. terrestrial plants had developed specialized sexual organs, stems with mechanisms for fluid transport, structural elements such as wood to provide mechanical support, roots for anchorage, leaves as sites of photosynthesis, stomata in the leaves to allow passage of CO2 and O2, and seed transport. Terrestrial animals of this period had skeletons for support and anchoring muscles (allowing locomotion), organs for gaseous exchange (breathing atmospheric oxygen), and systems for circulating materials within the body. Terrestrial animals and plants also had waterproof coverings to minimize the chances of desiccation. Terrestrial animals used nontoxic excretory products (urea and uric acid), whereas aquatic ones still relied heavily on ammonia

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32
Q

seasonal variation -northern hemisphere

A

receives its maximum illumination, and the Southern Hemisphere its minimum, on the June solstice (around June 21), when the sun shines directly over the Tropic of Cancer. The reverse is true on the December solstice (around December 21), when the sun shines directly over the Tropic of Capricorn (23.5􏱯 S latitude). Twice each year, on the vernal and autumnal equinoxes (around March 21 and September 21, respectively), the sun shines directly over the equator

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33
Q

topography

A

In the Northern Hemisphere, south-facing slopes are warmer and drier than north-facing slopes because they receive more solar radiation. In addition, adiabatic cooling causes air temperature to decline also establish regional and local rainfall patterns.

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34
Q

parasitism

A

a symbiotic relationship in which one organism benefits at the expense of another

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35
Q

forms of energy

A

heat, chemical, electrical and mechanical energy

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36
Q

first law of thermodynamics

A

Energy can be transformed from one form into another or transferred from one to another, but it cannot be created or destroyed. Example is Niagara falls – water at the top of the falls has high potential energy because of its location within earths gravitational field. As water moves over the waterfall, its potential energy is converted into kinetic energy. When it reaches the bottom of the waterfall, the kinetic energy of the water is dissipated into various forms of potential and kinetic energy. At Niagara falls, the kinetic energy of the moving water is converted into electricity through hydroelectric turbines. A living example is – 40% of glucose is harnessed in cellular respiration

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37
Q

second law of thermodynamics

A

the total disorder (entropy) of a system and its surroundings always increases. The physical disintegration of an organized system is the second law in action. Life does not apply as it is highly organized.

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38
Q

entropy

A

the unusable energy that is produced during energy transformations results in an increase in the disorder or randomness of the universe. In thermodynamics, this randomness or disorder is a quantity called entropy. Ex: cup of coffee gets cold, new car doesn’t stay new and loses its ‘‘new car smell’’, etc

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39
Q

what determines a spontaneous reaction

A
  1. Reactions tend to be spontaneous if the products have less potential energy than the reactants. The reaction is exothermic (process that releases energy), producing a large amount of heat, as the products have less potential energy than the reactants. Alternatively, reactions that absorb energy are endothermic, if the products have more potential energy than the reactants.
  2. Reactions tend to be spontaneous when the products are less ordered than the reactants. Reactions tend to occur spontaneously if the entropy of the products is greater than the entropy of the reactants, that is, if the products are more random than the reactants. (ex- melting of ice)
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40
Q

equilibrium

A

maximum stability. It is represented by ΔG=0, a state in which the reaction does not stop but rather a state in which the rate of the forward reaction equals the rate of the backward reaction. As a system moves toward equilibrium, the free energy of the system becomes progressively lower and reaches its lowest point and maximum stability when the system is at equilibrium.

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41
Q

Enzymes

A

protein based catalysts which speed up a spontaneous reaction by lowering the activation energy of spontaneous (exergonic) reactions. They are crucial since you cannot use heat due to the fact that heat would cause more problems (like destroying cell structure) and it could speed up all other reactions that should not be going faster than needed.

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42
Q

how enzymes catalyze reactions

A

1: bring reactants close together
2: exposing reactants to an electrical charge
3: changing conformation of the substrate.

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43
Q

conditions that affect enzyme activity

A

changes in the concentration of substrate and other molecules that bind to enzymes. In addition, a number of control mechanisms modify enzyme activity, thereby adjusting reaction rates to meet a cell’s requirements for chemical products. As well, changes in temperature and pH can have a significant impact on enzyme activity.

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44
Q

competitive inhibition

A

where there is an inhibitor molecule that competes with the substrate (same shape) for the active site and can slow down the reaction or stop altogether if there is too many competitive inhibitors.

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45
Q

non competitive inhibition

A

inhibitor binds elsewhere on the enzyme which distort the activation site (allosteric inhibition) (sometimes the inhibitor in any inhibition case may for covalent bonds and completely disable the enzyme)

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46
Q

feed back inhibition

A

occurs in regulations where the product of a reaction also acts as its regulator. It acts in a similar fashion to allosteric inhibition by minimizing wastage of cellular energy and resources during molecular synthesis in the pathway. Both processes are similar in controlling enzyme activity

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47
Q

allosteric inhibition

A

a reversible way of regulating enzyme activity. It uses the binding of regulatory molecule and the allosteric site, causing an increase or decrease in enzyme activity. An allosteric inhibitor binds to an allosteric site on the enzyme in order to release the substrate, and the enzyme will go from high-affinity state to low-affinity state because it is catalyzing too many molecules. Allosteric inhibitors are often a product of their metabolic pathway. If accumulates in excess, the inhibitor will slows or stops the reaction producing it. If product is scarce, it will increase production and decrease inhibition. The product of the pathway regulates the reaction by acting on the primary enzyme.

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48
Q

fluid mosaic model

A

Membranes are fluid and the membrane proteins move across the membrane (not rigid). Experiment was conducted with rat cell and human cell (each got membrane proteins dyed a different color) when they fused together the proteins moved from the original starting position and moved across the cell. Lots of different protein that scatter the membrane and different membranes have different proteins

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49
Q

factors in simple diffusion

A

size and charge, determines the ease with which a molecule can move across a membrane. Very small nonpolar molecules such as O2 and CO2 are readily soluble in the hydrophobic interior of a membrane and move rapidly from one side to the other. In contrast, the membrane is practically impermeable to charged molecules, including ions such as C1-, Na+, and phosphate (PO4-). The presence of a charge and a hydration shell of water surrounding the ion contribute to ions being prevented from entering the hydrophobic core of the membrane.

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50
Q

how cells move different materials across a cell membrane

A

Facilitated diffusion allows transport to be helped or facilitated by protein complexes that span the membrane (although facilitated diffusion uses specific transporters, the movement of molecules is still initiated by diffusion based on a concentration gradient against the membrane).
Channel proteins form hydrophilic pathways in the membrane through which water and ions can pas, which aids the diffusion of molecules through the membrane (gives them an avenue where they don’t have to interact with the hydrophobic portions of the membrane). Carrier proteins also form passageways through the lipid bilayer, binging a specific single solute (such as sugar molecule or amino acid) and transports it across the lipid bilayer (uniport transport). The carrier protein undergoes conformational changes which progressively move the solute binding site from one side of the membrane to the other, transporting the solute.

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51
Q

how ATP is synthesized by chemiosmosis

A

ATP is formed by hydrogen protons being pumped into the mitochondrial intermembrane from complex 1,3 and 4. More hydrogen in intermembrane makes a proton-motive force which the hydrogen protons want to return into the matrix and can do so through the ATP synthase which pump the protons back into the matrix and in turn changes the conformation of ATP synthase which makes a turning motion that synthesized ADP +P into ATP.

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52
Q

how cellular respiration is regulated in the cell

A

Most metabolic pathways are regulated by supply and demand through the process of feedback inhibition: the end products of the pathway inhibit an enzyme early in the pathway

Not surprisingly, the rate of cellular respiration is controlled by key metabolic intermediates. The rate of sugar oxidation by glycolysis is closely regulated by several mechanisms to match the cell’s need for ATP. For example, if excess ATP is present in the cytosol, it binds to phosphofructokinase, inhibiting its action. The resulting decrease in the concentration of fructose- 1,6-bisphosphate slows or stops the subsequent reactions of glycolysis and, as a consequence, the remainder of cellular respiration. Thus, glycolysis does not oxidize fuel substances needlessly when ATP is in adequate supply.

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53
Q

describe how chlorophyll in photosystems traps light, and explain how this results in an electron transport chain.

A

1) Absorption of photons by antenna complex and funneling of energy to the reaction centre results in an electron with P680 being raised from ground to excited state (P680)
2) P680
can be easily oxidized to P680+ by the primary electron acceptor of photosystem II, a molecule called pheophytin, which initiates electron transport by donation to plastoquinone (PQ) which is analogous to ubiquinone of respiratory electron transport

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54
Q

explain how pigment molecules absorb light energy.

A

The absorption of a photon by a pigment molecule excites a single electron, moving it from the ground state to an excited state. The difference in energy level between the ground state and the excited state is equivalent to the energy of the photon of light that was absorbed. Can either return to ground state, returns to ground state, or donated to electron accepting molecule

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55
Q

main pigments in photosynthetic apparatus

A

P680 → p680* in PSII and p700 → p700* in PSI and they are arranged at the reaction centre where the photon in funnelled to get to that excited state and electron transport chain

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56
Q

phases of the Calvin cycle

A

1 carbon fixation
2 reduction
3regeneration

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57
Q

rubisco

A

Rubisco is slow and inefficient enzyme. As an oxygenase, rubisco catalyzes the combination of RuBP with O2 rather than with CO2, forming toxic products that cannot be used in photosynthesis. The toxic products are eliminated by reactions that release carbon in inorganic form as CO2, greatly reducing the efficiency of photosynthesis. The entire process is called photorespiration because it uses oxygen and releases CO2

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58
Q

CAM plants

A

CAM plants typically live in regions that are hot and dry during the day and cool at night. Their fleshy leaves or stems have a low surface-to-volume ratio, and their stomata are reduced in number. Further, the stomata open only at night, when they release O2 that accumulates from photosynthesis during the day and allow CO2 to enter the leaves. The entering CO2 is fixed by the C4 pathway into malate, which accumulates throughout the night and is stored in large cell vacuoles.
Daylight initiates the second phase of the strategy. As the sun comes up and the temperature rises, the stomata close, reducing water loss and cutting off the exchange of gases with the atmosphere. Malate diffuses from cell vacuoles into the cytosol, where it is oxidized to pyruvate, and CO2 is released in high concentration. The high CO2 concentration favours the carboxylase activity of rubisco, allowing the Calvin cycle to proceed at maximum efficiency with little loss of organic carbon from photorespiration. The pyruvate produced by malate breakdown accumulates during the day; as night falls, it enters the C4 reactions, con- verting it back to malate. During the night, oxygen is released by the plants, and more CO2 enters.

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59
Q

c4 plants

A

In c4 plants, the C4 cycle occurs in mesophyll cells, which lie lose to the surface of leaves and stems, where O2 is abundant. The malate intermediate of the C4 cycle diffuses from the mesophyll cells to bundle sheath cells, located in deeper tissues, where O2 is less abundant. In these cells, in which the Calvin cycle operates, the malate enters chloroplasts and is converted to pyruvate and CO2. Because O2 concentration is low and CO2 concentration is high because of its release by malate breakdown, the oxygenase activity of rubisco is inhibited, and the carboxylation reaction runs highly efficiently. The pyruvate produced by malate oxidation returns to the mesophyll cells to enter another turn of the C4 cycle.

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60
Q

importance of the thylakoid membrane

A

It houses the molecules that carry out the light reactions of photosynthesis, including the pigments, electron transfer carriers, and ATP synthase enzymes for ATP production.

The enzymes that catalyze the reactions of the Calvin Cycle are found in the chloroplast.
-But cells lacking a chloroplast can still be photosynthetic. Many photosynthetic prokaryotes also have thylakoid membranes tthat are formed from infoldings of the plasma membrane.

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61
Q

what is CO2 converted to in the Calvin cycle

A

Three carbon sugars, which can be readily combined to form six-carbon monosaccharides, including glucose.

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62
Q

function of a photosystem

A

The function of a photosystem is to trap photons of light and use the energy to oxidize a reaction centre chlorophyll, with the electron being transferred to the primary electron acceptor.

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63
Q

if eukaryotes require their daughter cells to be exact genetic copies of the parental cell - which process is used?

A

mitosis

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64
Q

when cells are needed that are different from the parent cells - which process is used?

A

meiosis

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65
Q

diploid

A

2 copies of each chromosome, also referred to as 2n

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66
Q

haploid

A

1 copy of each chromosome, also referred to as n

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67
Q

number of chromosome sets in a cell is called a ___

A

ploidy of the cell or species

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68
Q

replication of the DNA of each individual chromosome creates ___

A

two identical molecules called sister chromatids

held together until mitosis separates them

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69
Q

equal distribution of daughter chromosomes to each of the two cells the result from cell division is called___

A

chromosome segregation

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70
Q

The precision of chromosome replication and segregation in the mitotic cell cycle creates a group of cells called a__

A

clone

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71
Q

true or false: all cells of a clone are genetically identical

A

yes except in the case of genetic mutation

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72
Q

interphase

A

-cell grows and replicates DNA in preparation for mitosis (M phase) and cytokinesis
-g1 phase: cell makes various RNAs, proteins, other types of cellular molecules, but NOT nuclear DNA (period of cell growth before DNA replicates)
-if the cell is going to divide, DNA replication begins, initiating the S (synthesis) phase
-S phase is the period when DNA replicates and chromosomal proteins are duplicated, cell enters G2 phase when complete
-The g2 phase: cell continues to synthesize RNAs, proteins, continues to grow. Period after DNA replicates, cell prepares for division
During these steps chromosomes are loose, but organized in nucleas. After this, mitosis begins

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73
Q

which phase of the cell cycle varies in length between species?

A

G1 phase

thus, whether cells divide rapidly or slowly depends on length of g1

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74
Q

what is the g0 phase?

A

g1 is the stage in which many cells stop dividing. this state of division arrest is designated as the g- phase.
ex: in humans, cells in nervous system reach g0 phase when fully mature

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75
Q

what are the stages of mitosis? (list the names)

A

prophase, metaphase, anaphase telophase

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76
Q

prophase

A
  • greatly extended chromosomes that were replaced in interphase condense and compact into rod like structures
  • while condensation is in progress, nucleus becomes smaller and eventually disappears in most species: reflects a shut down of all types of RNA synthesis
  • in the cytoplasm, mitotic spindle begins to form between the 2 centrosomes as they migrate to opposite ends to form spindle poles
  • spindle developer as bundles of microtubules that radiate from the spindle poles
  • nuclear envelope breaks down, heralding the beginning of pro metaphase
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77
Q

prometaphase

A
  • bundles of spindle microtubules grow from centrosomes at the opposing spindle poles toward the centre of the cell
  • some of the developing spindle enters the former nuclear area and attaches to the chromosomes
  • remember* each chromosome is made up of 2 identical sister chromatids held together only at their centromeres
  • complex of several proteins, kintechore, has formed on each chromatid at centromere
  • kintechore microtubules bind to kintechores, which determine the outcome of mitosis (bc they attach the sister chromatids of each chromosome to microtubules leading to the opposite spindle poles) , microtubules that do not attach to kintechores overlap those from the opposite spindle pole
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78
Q

metaphase

A
  • spindle reaches its final form and spindle microtubules move chromosomes into alignment at spindle midpoint (metaphase plate)
  • chromosomes complete condensation and assume characteristic shape as determined by location of the centromere and length and thickness of chromatid arms
  • once chromosomes are assembled at spindle midpoint with 2 sister chromatids of each one attached to microtubules leading to opposite spindle poles, can metaphase give way to actual separation of chromatids
  • complete collection of metaphase chromosomes arranged according to size and shape, forms the KARYOTYPE of given species
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79
Q

anaphase

A
  • sister chromatids separate and move to opposite spindle poles
  • first signs of chromosome movement can be seen at the centromeres as the kintechores are the first sections to move toward opposite poles
  • movement continues until separated chromatids, now called daughter chromosomes, have reached the 2 poles, and chromosome segregation has now completed
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80
Q

telophase

A
  • spindle disassembles and chromosomes at each spindle pole decondense and return to the extended state typical of interphase
  • nucleolus disappears, RNA transcription resumes, new nuclear envelope forms around the chromosomes at each pole, producing 2 daughter nuclei.
  • nuclear division is completely, cell has 2 nuclei
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81
Q

cytokinesis

A

division of the cytoplasm
usually follows nuclear division stage of mitosis and produces 2 daughter cells, each one containing 2 daughter nuclei
-in most cells, cytokinesis begins during telophase or late anaphase
-when cytokinesis is completed, daughter nuclei have progressed the interphase state and entered the g1 phase of the next cell cycle

proceeds….
in animals: the furrow girdles the cell and gradually deepens until it cuts the cytoplasm in 2
in plants: cell plate forms between daughter nuclei and grows laterally until it divides the cytoplasm

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82
Q

furrowing

A
  • the layer of microtubules that remains at the former spindle midpoint expands laterally until it stretches entirely across the dividing cell
  • as layer develops, band of microfilaments forms just inside plasma membrane, forming a belt that follows the inside boundary of the cell plane in the microtubule layer
  • powered by motor proteins, microfilaments slide together, tightening the band and constricting the cell - forming a groove or furrow in plasma membrane that gradually deepens until daughter cells are separated
  • cytoplasmic division isolates the daughter nuclei in the 2 cells at distributes organelles/ other structures approximately equally
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83
Q

cell plate formation

A
  • layer of microtubules that persists at the former spindle midpoint serves as organizing site for vesicles produced by the ER and Golgi complex
  • as vesicles collect, layer expands until it spreads entirely across the dividing cell, during this, vesicles fuse together and contents assemble into new cell wall- the cell plate- stretching completely across the former spindle midpoint
  • this separates cytoplasm and its organelles into 2 parts and isolates the daughter nuclei in separate cells
  • plasma membrane that lines the two surfaces of the cell plate are derived from vesicle membranes
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84
Q

centrosome

A

site near the nucleus from which microtubules radiate outward in all directions

  • centrosome is the main MTOC (microtubule organizing centre) of the cell, anchoring the microtubule cytoskeleton during interphase and positioning many of the cytoplasmic organelles
  • contains a pair of controls, arranged at right angles to each other
  • as prophase begins in the M phase, centrosome separates into 2 parts (step3)
  • duplicated centrosomes, with centrioles inside, continue to separate until they reach opposite ends of the nucleas
  • as centrosomes move apart, microtubules between them lengthen and increase in number
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85
Q

cytoplasmic division in prokaryotes

A
  • occurs through an inward growth of the plasma membrane, along which new cell wall material is assembled to cut the cell into two parts
  • new wall divides the two replicated DNA molecules and cytoplasmic structures and molecules equally between the daughter cells
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86
Q

discuss binary fission(3 steps) of prokaryotes in a little more detail

A

Replication of the bacterial chromosome begins at a site called the “origin” through reactions catalyzed by enzymes located in the middle of the cell. Once the origin of replication is duplicated, the two origins migrate to the two ends of the cells. Division of the cytoplasm then occurs through a partition of cell wall material that grows inward until the cell is separated into two parts

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87
Q
  1. describe how the genetic material of eukaryotes is organized (i.e., in chromosomes)
A

• Human DNA is organized into chromosomes which are diploid (is when there are 2 copies which is shown as n). Human chromosomes contain 23 diploid chromosome pairs, since human chromosomes are 2n humans have 2*23 = 46 chromosomes. Sometimes it is organized into haploids (1n which would mostly be microorganisms). Also, they can be organized as ploidy (mostly plants and have multiple chromosomes). When DNA is separated (chromosome segregation) it would be split into 2 sister chromatids

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88
Q
  1. describe the function of mitosis as equal distribution of chromosomes and DNA from the dividing parent cell to two daughter cells.
A

• Chromosomes consist of 2 genetically identical pairs of chromatids which are bound together at the centromere. When they are pulled to each spindle pole the identical pairs are separated and each spindle pole contains 23 chromatids from the 46 that were aligned in metaphase. Replication of the DNA of each individual chromosome creates two identical molecules called sister chromatids. Newly formed sister chromatids are held together until mitosis separates them, placing one in each of the two daughter nuclei. As a result of this precise division, each daughter nucleus receives exactly the same number and types of chromosomes, and contains the same genetic information, as the parent cell entering the division. The equal distribution of daughter chromosomes to each of the two cells that result from cell division is called chromosome segregation.

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89
Q
  1. contrast how binary fission (the cell division in prokaryotes) differs from that in eukaryotes.
A

• Most prokaryotes have a single circular DNA (bacterial Chromosome) vs the multiple chromatids in eukaryotes. DNA is not separated by microtubuli but rather from the origin of replication (ori) in a zipper like fashion where DNA is separated actively on opposite ends of the cell. When replication is complete the cell division begins as the plasma membrane grows inward, and a new cell wall is synthesized. Prokaryotic cells only have one circular DNA and if each daughter cell gets one copy the genetic information was separated evenly where eukaryotic cells have much more DNA to split and more chance of error.

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90
Q

what is genetic recombination

A

Since mutations are relatively rare, diversity is amplified through various mechanisms that shuffle existing mutations into dif- ferent combinations.

  • This process, of literally cutting and pasting DNA backbones into new combinations, is called genetic recombination and is very widespread in nature.
  • Genetic recombination allows “jumping genes” to move, inserts some viruses into the chro- mosome of their hosts, underlies the spread of anti- biotic resistance among bacteria, and is at the heart of meiosis in eukaryotes. Genetic recombination puts the “sexual” in sexual reproduction; without genetic recombination, reproduction is “asexual,” and off- spring are simply identical clones of their parent
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91
Q

what things does genetic recombination require?

A

two DNA molecules that differ from one another in at least two places, a mechanism for bringing the DNA molecules into close proximity, and a collection of enzymes to “cut,” “exchange,” and “paste” the DNA back together.

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92
Q

what holds DNA together?

A

The sugar–phosphate backbone is held together by strong covalent bonds, whereas the bases pair with their partners through relatively weak hydrogen bonds

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93
Q

What would happen if two circular DNA molecules were involved in a single recombination event?

A

TWO CIRCULAR MOLECULES ‘FUSE’ TOGETHER AS A RESULT OF A SINGLE RECOMBINATION EVENT

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94
Q

____ allows DNA on different molecules to line up and recombine precisely.

A

homology

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95
Q

phototroph

A

Strains that are able to synthesize the necessary amino acids for protein synthesis

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96
Q

auxotrophs

A

mutant strains that are unable to synthesize amino acids

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97
Q

a strain that cannot manufacture its own arginine is represented by

A

arg-

normal form is arg+

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98
Q

conjugation

A

The basis of a kind of sexual reproduction in bacteria
- Instead of fusing, bacterial cells conjugate
-Contact each other by a long tubular
structure called a sex pilus and then form a
cytoplasmic bridge
-Part or all of the DNA of one cell moves into
the other through the bridge. The donated DNA can then recombine with homologous sequences of the recipient cell’s DNA
-Conjugation is initiated by the bacterial cell that has the fertility plasmid, or the ‘F Factor’ (It carries about 20 or so genes, several encoding proteins for the sex pilus). One strand of the F factor breaks at a specific point and begins to move from F+ to F- cell as the F factor replicates. No genetic recombination occurs between the DNA of two different cells in such a mating
• Hfr Cells and Recombination- Occasionally F factor can integrate into the chromosome in a single crossover event. Due to this phenomenon, some of the actual chromosome is transferred into F- cell, which results in recombination of the two cell’s DNA

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99
Q

transformation

A

Living cells of some species absorb pieces of DNA released from cells that have disintegrated. The entering DNA fragments can recombine with the recipient cell’s DNA.

(Griffith and pneumonia in mice -> . In 1928 Dr Griffith was studying how pneumonia was cause Griffith found that a mixture of heat-killed virulent cells plus living nonvirulent cells still caused pneumonia. The living nonvirulent cells had been transformed to virulence by something released from the dead cells. The substance derived from the killed virulent cells, the substance capable of transforming nonvirulent bacteria to the virulent form, was DNA. virulent cells recombine with the chromosomal DNA of the nonvirulent cells in much the same way as genetic recombination takes place in conjugation. Only some bacteria can use this form of genetic recombination.)

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100
Q

transduction

A

DNA is transferred from one cell to another by mistake inside the head of an infecting virus.
-During infection, the host bacterial chromosome is degraded to provide raw material for synthesis of new bacteriophage chromosomes. Sometimes, a fragment of host chromosome avoids degradation and is packed into the head of a new phage by mistake

In general, transduction begins when new phages assemble in an infected bacterial cell; they sometimes incorporate fragments of the host cell DNA along with, or instead of, the viral DNA. After the host cell is killed, the new phages that are released may then attach to another cell and inject the bacterial DNA (and the viral DNA if it is present) into that recipient cell. The recipient cell becomes partial diploid.
Generalized transduction: in which all donor genes are equally likely to be transferred, is associated with some virulent bacteriophages, which kill their host cells during each cycle of infection (the lytic cycle). the host bacterial chromosome is degraded to provide raw material for synthesis of new phage chromosomes. However, sometimes a fragment of host chromosome avoids degradation and is packed into the head of a new phage by mistake. This particular phage now contains a small random sample of bacterial genes instead of phage genes.
Specialized transduction: Lambda is a temperate bacteriophage. That is, when it first infects a new host, it determines whether the host is likely to be a good one. Is it starving? Is it suffering from DNA damage? If the host cell passes this molecular health checkup, then the lambda chromosome lines up with a small region of homology on the bacterial chromosome and a phage-coded enzyme catalyzes a single recombination event. The phage is thus integrated into the host chromosomal DNA and, in this state, is called a prophage. The prophage is then replicated to daughter cells as long as conditions are favorable (it stays in the lysogenic cycle). If, however, the host cell becomes inhospitable (perhaps as a result of ultraviolet-induced DNA damage), the prophage activates several genes, releases itself from the chromosome by a recombination event, and proceeds to manufacture new phage, which are released as the cell bursts as a result of lytic growth. the “mistake” occurs when the prophage is excised from the chromosome. Sometimes this recombination event is imprecise; bacterial DNA is removed from the host chromosome, and some prophage DNA is left behind. As a result, this bacterial DNA is packaged into new phage and carried to recipient cells. Since the transducing phage is defective, having left some of its genes behind in the host, it does not kill its new host

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101
Q

CONTRAST THE CHARACTERISTICS OF F-, F+, AND Hfr CELLS

A

F- IS THE RECIPIENT CELL WHILE F+ IS THE DONOR CELL. BETWEEN THESE TWO CELLS, NO GENETIC RECOMBINATION OCCURS. THE Hfr CELL IS A SPECIAL DONOR CELL THAT CAN TRANSFER GENES ON A BACTERIAL CHROMOSOME TO A RECIPIENT BACTERIUM (BECAUSE THE F FACTOR IS INCORPORATED INTO THE SINGLE BACTERIAL CHROMOSOME)

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102
Q

Q-EXPLAIN WHY ALL GENES HAVE AN EQUAL LIKELIHOOD OF TRANSFER BY GENERALIZED TRANSDUCTION BUT NOT BY SPECIALIZED TRANSDUCTION

A

SINCE GENERALIZED TRANSDUCTION TRANSFERS RANDOM FRAGMENTS OF THE HOST CHROMOSOME, ALL HOST GENES ARE TRANSFERRED AT EQUAL FREQUENCY. SPECIALIZED TRANSDUCTION ONLY TRANSFERS GENES LYING CLOSE TO THE POINT OF INSERTION OF THE PROPHAGE

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103
Q

T/F meiosis makes gametes in plants

A

F- only in animals

In house plants/fungi - the haploid products are spores

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104
Q

Outline the life cycle in animals

A

Follow the pattern in which the diploid phase dominates the life cycle , the haploid phase is reduced, and meiosis is followed directly by gamete formation. In male animals, each of the four nuclei produced by meiosis is enclosed in a separate cell by cytoplasmic divisions, and each of the four cells differentiates into a functional sperm cell. In female animals, only one of the four nuclei becomes functional as an egg cell nucleus. Fertilization restores the diploid phase of the life cycle. Thus, animals are haploids only as sperm or eggs, and no mitotic divisions occur during the haploid phase of the life cycle.

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105
Q

outline the life cycle in most plants/fungi

A

In all plants (except bryophytes), the diploid sporophyte generation is the most visible part of the plant. The gametophyte haploid generation is reduced to an almost microscopic stage that develops in the reproductive parts of the sporophytes—in flowering plants, in the structures of the flower. The female gametophyte remains in the flower; the male gametophyte is released from flowers as microscopic pollen grains. When pollen contacts a flower of the same species, it releases a haploid nucleus that fertilizes a haploid egg cell of a female gametophyte in the flower. The resulting cell reproduces by mitosis to form a sporophyte. Sphagnum moss (commonly known as “peat moss”) is a good example of a plant in which the gametophyte is the most visible and familiar stage of the life cycle. In this case, the sporophyte is reduced and develops from a zygote within the body of the gametophyte. Vast peatlands of Sphagnum gametophytes are industrially harvested in many parts of the world for fuel and horticultural use.

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106
Q

outline the life cycle specifically in some fungi

A

During fertilization, two haploid gametes, usually designated simply as positive (+) or negative (-) because they are similar in structure, fuse to form a diploid nucleus. This nucleus immediately enters meiosis, producing four haploid cells. These cells develop directly or after one or more mitotic divisions into haploid spores. These spores germinate to produce haploid individuals, the gametophytes, which grow or increase in number by mitotic divisions. Eventually, positive and negative gametes are formed in these individuals by differentiation of some of the cells produced by the mitotic divisions. Because the gametes are produced by mitosis, all the gametes of an individual are genetically identical.

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107
Q

explain why meiosis results in a reduction of genetic material.

A

Genetic material is cut in half through the process of meiosis since once the male gamete meets with the female gamete they can form a diploid cell and share each half of the chromosome.

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108
Q

essence of ‘‘difference’’ in meiosis

A

two kinds of difference: halved chromosome number and recombined chro- mosomal DNA sequence.

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109
Q

essence of ‘‘sameness’’ in meiosis

A

chromosomes are replicated and partitioned to ensure that cells produced by the process have the same number of chromosomes, with the same DNA sequence, as the cell that began the process. In this way, somatic cells are produced for most of the requirements of multicellular bodies

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110
Q

paternal chromosome

A

derived from the male parent of the organism

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111
Q

maternal chromosome

A

derived from the female parent of the organism

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112
Q

in dogs (for example), what gives each individual offspring a unique Como of traits, such as size, coat, colour, susceptibility to certain diseases and disorders, aspects of behaviour/intelligence?

A

alleles

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113
Q

meiocytes

A

cells that are destined to divide by meiosis

114
Q

Meosis I

A

During meiosis I, chromosomes behave dramatically differently than they do during mitosis. That is, early in meiosis I, homologous chromo- somes find their partners and pair lengthwise, gene for gene, in a process called synapsis. During this intimate pairing, recombination occurs, and chro- mosomal segments are exchanged. As the meiocyte continues through to the end of the first division, the members of each homologous pair are sepa- rated into one or the other of the two daughter cells. These daughter cells still contain replicated chro- mosomes with two chromatids each; however, the number of such chromosomes is only half that of the original meiocyte

115
Q

meiosis II

A

the sister chromatids are sepa- rated into different cells. A total of four cells, each with the haploid number of chromosomes and a novel collection of alleles, is the final result of the two meiotic divisions

116
Q

T/F- DNA replication occurs during interkinesis

A

F - no DNA replication occurs during interkinesis

117
Q

Identify (list only) the main phases of meiosis

A
  • Prophase 1
  • Prometaphase 1
  • Metaphase and anaphase 1
  • telophase 1 and interkinesis
  • prophase 2, pro metaphase, and metaphase 2
  • anaphase 2 and telophase 2
118
Q

explain why meiosis results in genetic recombination

A

since the chromosomes that were homologous were brought into close proximity so that the DNA is able to be exchanged between the similar regions on the chromosome.

119
Q

explain principle of transposition with examples

A
  • Transposable elements are done in 2 ways (1) cut and paste where the DNA is cut from the sequence and pasted in a different region. Does Not have to be homologous. (2) copy and paste where the DNA is replicated then it is moved to another region. TE’s are never free flowing and are in contact with target site.
  • Two main types of transposition in bacteria (1) insertion sequence: each end of the IS there is identical DNA running in opposite directions (ex: 1234 - IS - 4321) (2) transposon: has the same inverted DNA as IS with one or more genes in the middle mostly insertion sequences. These non-IS genes included in transposons are carried along as the TEs move from place to place. Many antibiotics, such as penicillin…, that were once successful in curing bacterial infections have lost much of their effectiveness because of resistance genes carried in transposons.
  • Eukaryotic cells have 2 main types of transpositions: transposons: which are similar to bacterial transposons cut and paste and retrotransposons: copy and paste method. DNA element integrated into the chromosomal DNA, is transcribed into a complementary RNA copy. Next, an enzyme called reverse transcriptase, which is encoded by one of the genes of the retrotransposon, uses the RNA as a template to make a DNA copy of the retrotransposon. The DNA copy is then inserted into the DNA at a new location, leaving the original in place
120
Q

prometaphase 1

A

the nuclear envelope breaks down and the spindle enters the former nuclear area
- two chromosomes of each pair attach to kinetochore microtubules leading to opposite spindle poles. That is, both sister chromatids of one homologue attach to microtubules leading to one spindle pole, whereas both sister chromatids of the other homologue attach to microtubules leading to the opposite pole.

121
Q

prophase 1

A

two sister chromatids, begin to fold and condense into thread like structures in the nucleus. The 2 homologous chromosomes line up side by side forming a tetrad (DNA is exchanged at his point) followed by spindle forming in the cytoplasm.

122
Q

metaphase and anaphase 1

A

movements of the spindle microtubules have aligned the recombined tetrads on the equatorial plane—the metaphase plate—between the two spindle poles. Then the two chromosomes of each homologous pair separate and move to opposite spindle poles during anaphase I.The movement segregates homologous pairs, delivering a haploid set of chromosomes to each pole of the spindle. However, all the chromosomes at the poles are still double structures composed of two sister chromatids

123
Q

Telophase 1 and interkinesis

A

Telophase I is a brief, transitory stage in which there is little or no change in the chromosomes. New nuclear envelopes form in some species but not in others. Telophase I is followed by an interkinesis in which the single spindle of the first meiotic division disassembles and the microtubules reassemble into two new spindles for the second division. There is no DNA replication between the first and the second division.

124
Q

prophase II, prometaphase II, metaphase II

A
  • During prophase of meiosis II, the chromosomes condense and the spindle begins to form.
  • During prometaphase II, the nuclear envelope breaks down, the spindle enters the former nuclear area, and spindle microtubules leading to opposite spindle poles attach to the two kinetochores of each chromosome.
  • At meta-phase II, move- ments of the chromosomes within the spindle bring them to rest on the metaphase plate
125
Q

Anaphase and Telophase II

A

Anaphase II begins as the sister chromatids of each chromosome separate from each other and move to opposite spindle poles. At the completion of ana- phase II, the separated chromatids—now called chromosomes—have been segregated to the two poles.
-During telophase II, the chromatids decondense to the extended interphase state, the spindles disas- semble, and new nuclear envelopes form around the masses of chromatin. The result is four haploid cells, each with a nucleus containing half the number of chromosomes present in a somatic cell of the same species. These chromosomes all carry various new combinations of maternal and paternal alleles.

126
Q

nondisjunction

A

failed chromosome segregation at meiosis I or II
-ex during meiosis I, both chromosomes of a homologous pair may connect to the same spindle pole in anaphase I. in the resulting nondisjunction, the spindle fails to separate the homologues chromosomes of the tetrad, and one pole receives both chromosomes of the homologous pair, where the other pole has no copies of the pair. meiosis II will proceed to separate the chromatids of the extra chromosome as usual, with the result that gametes will have two copes of this chromosome.

127
Q

during meiosis II, what would happen if a homologous pair connected to the same spindle pole in anaphase I?

A

the spindle fails to separate the homologous chromosomes of the tetrad. As a result, one pole receives both chromo- somes of the homologous pair, whereas the other pole has no copies of that chromosome. Meiosis II will proceed to separate the chromatids of the extra chromosome as usual, with the result that gametes will have two copies of this chromosome (instead of one). Zygotes that receive an extra chromosome because of nondisjunction therefore have three copies of a given chromosome instead of two. In humans, most zygotes of this kind do not result in live births. One exception is Down syndrome, which results from three copies of chromosome 21. Down syndrome involves characteristic alterations in body and facial structure, mental retardation, and significantly reduced fertility

128
Q

During meiosis and fertilization (sexually reproducing organisms), genetic variability arises from which four sources

A

(1) genetic recombination, (2) the differing combinations of maternal and paternal chromosomes segregated to the poles during anaphase I,
(3) the differing combinations of recombinant chromatids segregated to the poles during anaphase II, and
(4) the particular sets of male and female gametes that unite in fertilization

129
Q

synaptonemal complex

A

a protein framework that holds together pairing homologous chromosomes,, which disassembles and disappears when the exchange of segments (producing new combos of alleles) is complete

130
Q

crossovers/ chiasmata

A

regions in which non-sister chromatids cross one another

131
Q

4 sources of genetic variability produced by meiosis

A

1- recombination (chromatids generate new combinations of alleles by physically exchanging segments- exchange process involves precise breakage and joining of DNA molecules. catalyzed by enzymes and occurs while the homologous chromosomes are held tightly together by S-complex. crossovers visible between chromosomes at late prophase I reflect the exchange of chromatid segments that occurred during the molecular steps of genetic recombination)
2- random segregation of homologous chromosomes (homologous pairs separate at anaphase I of meiosis, segregating random combinations of maternal and paternal chromosomes to the spindle poles.
3- random segregation of the chromatids of replicated chromosomes at meiosis II
4- random joining of male and female gametes in fertilization

132
Q

transposable/mobile elements

A

DNA sequences that can move from place to place in the DNA. The TEs may move from one location in the DNA to another or generate duplicated copies that insert in new locations while leaving the “parent” copy in its original location

133
Q

transposition

A

mechanism of movement of TEs.
occurs at a low frequency in either of two ways, depending on the type of element: (1) a cut-and-paste process, in which the TE leaves its original location and transposes to a new location, and (2) a copy-and-paste process, in which a copy of a TE transposes to a new location, leaving the original TE behind

134
Q

why are TEs important?

A

For example, they produce mutations by transposing into genes and knocking out their functions, and they increase or decrease gene expression by transposing into regulatory sequences of genes. as such, TEs are biological mutagens that increase genetic variability

135
Q

types of TEs (bacterial)

A

2 major types: insertion sequences (IS) and transposons.

  • Insertion sequences are the simplest TEs. They are relatively small and contain only genes for their transposition, notably the gene for transposase, an enzyme that catalyzes some of the recombination reactions for inserting or removing the TE from the DNA at each of the two ends IS a short inverted repeat sequence- the same DNA sequence running in opposite directions (shown by directional arrows in the figure). The inverted repeat sequences enable the transposase enzyme to identify the ends of the TE when it catalyzes transposition. The inverted repeat sequence is an IS element on both the F factor and the bacterial chromosome that provides the homology needed for the creation of the Hfr strains described in Section 10.2.
    transposon: has an inverted repeat sequence at each end enclosing a central region with one or more genes. In a number of bacterial transposons, the inverted repeat sequences are insertion sequences, which provide the transposase for movement of the element (see Figure 10.18). Additional genes in the central region typically code for antibiotic resistance; they originated from the main bacterial DNA circle or from plasmids. These non-IS genes included in transposons are carried along as the TEs move from place to place.
136
Q

describe the connection between antibiotic resistance and transposons?

A

Many antibiotics, such as penicillin, erythromycin, tetracycline, ampicillin, and streptomycin, that were once successful in curing bacterial infections have lost much of their effectiveness because of resistance genes carried in transposons. Movements of the trans- posons, particularly to plasmids that can be transferred by conjugation within and between bacterial species, greatly increase the spread of genes, providing anti- biotic resistance to infecting cells. Resistance genes have made many bacterial diseases difficult or impossible to treat with standard antibiotics.

137
Q

2 classes of Eukaryotic TEs

A

transposons and retrotransposons- distinguished by the way the TE sequence moves from place to place the DNA

Eukaryotic transposons are similar to bac- terial transposons in their general structure and in the ways they transpose. However, members of the other class of eukaryotic TEs, the retrotrans- posons, transpose by a copy-and-paste mechanism that is unlike any of the other TEs we have discussed. Retrotransposons have this name because transposition occurs via an intermediate RNA copy of the TE

138
Q

process of a retrotransposon

A

First, the retrotransposon, which is a DNA element integrated into the chromosomal DNA, is transcribed into a complementary RNA copy. Next, an enzyme called reverse transcriptase, which is encoded by one of the genes of the retrotransposon, uses the RNA as a template to make a DNA copy of the retrotransposon.
The DNA copy is then inserted into the DNA at a new location, leaving the original in place. This inser- tion step involves breaking and rejoining DNA back- bones, as we have seen several times in this chapter.
Once TEs are inserted into chromosomes, they become more or less permanent residents, duplicated and passed on during cell division along with the rest of the DNA. TEs inserted into the DNA of reproductive cells may be inherited, thereby becoming a permanent part of the genetic material of a species

139
Q

compare retrotransposons to retroviruses

A

pg 224

140
Q

F factor

A

fertility plasmid. F factor carries several genes as ell as a replication origin that permits a copy tube passed on to each daughter cell during the usual process of bacterial cell division - which is an example of vertical inheritance from one generation to the next. However, during conjugation, the F factor has the ability to be copied and passed directly from one cel (donor) to another (recipient) - which is an example of horizontal inheritance

141
Q

F- cell

A

lacks F factor

142
Q

how many genes does the F factor carry

A

aprox 20, several encode proteins of the sex Pilus (or F Pilus)

143
Q

HFR cell

A

high frequency recombination - can ‘‘export’’ copies of chromosomal genes to another cell.
In hfr cells, the origin of transfer is near the middle of the integrated F factor, as a result, only half of the F factor DNA is transferred at the front of the chromosomal DNA
most likely, recipient will be. partial diploid

144
Q

somatic cells

A

body cells

145
Q

synapsis

A

or pairing - during prophase I, the two chromosomes of each homologous pair coming together in a zipper like way

146
Q

tetrads

A

fully paired homologues

147
Q

Which phase (diploid or haploid) dominates the life cycles of animals, plants, and fungi?

A

animals: diploid dominates
plants: alternate between haploid and diploid generations in which depending on the organism, either generation may dominate the life cycle
fungi: haploid dominates

148
Q

discuss antimicrobial resistance and transposons

A

Many antibiotics, such as penicillin, erythromycin, tetracycline, ampicillin, and streptomycin, that were once successful in curing bacterial infections have lost much of their effectiveness because of resistance genes carried in transposons. Movements of the trans- posons, particularly to plasmids that can be transferred by conjugation within and between bacterial species, greatly increase the spread of genes, providing anti- biotic resistance to infecting cells. Resistance genes have made many bacterial diseases difficult or impossible to treat with standard antibiotics

149
Q

TEs were first discovered in

A

maize (corn) by Barbara McClintock

150
Q

compare bacterial transposons and eukaryotic transposons

A

Eukaryotic transposons are similar to bac- terial transposons in their general structure and in the ways they transpose. However, members of the other class of eukaryotic TEs, the retrotrans- posons, transpose by a copy-and-paste mechanism that is unlike any of the other TEs

151
Q

disease that develops when a person has received 2 copies of a gene (one from each parent) that codes for altered subunit of hemoglobin

A

sickle cell disease

152
Q

blending theory of inheritance

A

suggested that hereditary traits blend evenly in off- spring through mixing of the parents’ blood, much like the effect of mixing coffee and cream. Even today, many people assume that parental characteristics such as skin colour, body size, and facial features blend evenly in their offspring, with the traits of the children appearing about halfway between those of their par- ents. Yet if blending takes place, why don’t extremes, such as very tall and very short individuals, gradually disappear over generations as repeated blending takes place? Also, why do children with blue eyes keep turn- ing up among the offspring of brown-eyed parents?
(Mendels experiment with garden peas provided the first few answers for this)

153
Q

describe mendels pea experiment

A

Mendel studied a variety of heritable characteristics called characters, such as flower colour or seed shape. A variation in a character, such as purple or white flower colour, is called a trait. Mendel established that characters are passed to offspring in the form of discrete hereditary factors, which now are known as genes. Mendel observed that, rather than blending evenly, many parental traits appear unchanged in off- spring, whereas others disappear in one generation to reappear unchanged in the next. Although Mendel did not know it, the inheritance patterns he observed are the result of the segregation of chromosomes, on which the genes are located, to gametes in meiosis (see Chapter 10). Mendel’s methods illustrate, perhaps as well as any experiments in the history of science, how rigorous scientific work is conducted: through observation, making hypotheses, and testing the hypotheses with experiments.

154
Q

explain the significance of Mendel’s experiments with regard to dominance, segregation, and independent assortment.

A

• Mendel’s experiments in regards of dominance showed that breeding 2 true breeding pea plants one with purple flowers and one with white flowers would result in a 3:1 for purple flowers. This was because the dominant gene in pea plants is the purple flower and with the punnett square we can now predict the outcome for certain traits.
• With the principle of segregation the it states that during fertilization, fusion of the haploid maternal and paternal gametes produces a diploid nucleus called the zygote nucleus. The zygote nucleus receives one allele for the character from the male gamete and one allele for the same character from the female gamete, reuniting the pairs. This shows us where some genetic traits come from and the probability of certain genes appearing in offspring.
• Mendel’s Law of Independent Assortment states that allele pairs separate independently during the formation of gametes. This means that traits are transmitted to offspring independently of one another. That is, the allele for seed shape that the gamete receives (R or r) has no influence on which allele for seed colour it receives (Y or y) and vice versa
• Mendel Could Predict Both Classes and Proportions of Offspring from His Hypotheses
-Mendel’s findings anticipated in detail the patterns by which genes and chromosomes determine inheritance

155
Q

whats cross pollination

A

most plants self fertilize, mendel prevented this by cutting the anthers, and fertilizing plants with pollen from another plant. this allowed him to test the effects of mating pea plants of different parental ties.

156
Q

dominant/recessive allele

A

mendels hypothesis: Mendel’s second hypothesis stated that if an individual’s pair of genes consists of different alleles, one allele is dominant over the other. This hypothesis assumes that one allele is dominant and the other allele is recessive. When a dominant allele for a trait is paired with a recessive allele for the same trait, the dominant allele is expressed.
a recessive allele is only expressed when two copies of the allele are present

157
Q

principle of segregation

A

As a third hypothesis, Mendel proposed the following: The pairs of alleles that control a character segregate (separate) as gametes are formed; half the gametes carry one allele, and the other half carry the other allele. This hypothesis is now known as Mendel’s Principle of Segregation. During fertilization, fusion of the haploid maternal and paternal gametes produces a diploid nucleus called the zygote nucleus. The zygote nucleus receives one allele for the character from the male gamete and one allele for the same character from the female gamete, reuniting the pairs.

158
Q

homozygous

A

two copies of the same allele

159
Q

heterozygous

A

two different alleles of a gene

160
Q

monohybrid

A

an F1 heterozygote produced from a cross that involves a single character

161
Q

monohybrid cross

A

generally, a cross between two individuals the are heterozygous for the same pair of alleles

162
Q

genotype

A

genetic constitution of an organism

163
Q

phenotype

A

outward appearance

164
Q

mendels 3 hypotheses (specifically)

A
  1. The genes that govern genetic characters occur in pairs in individuals.
  2. If different alleles are present in an individual’s pair of genes, one allele is dominant over the other.
  3. The two alleles of a gene segregate and enter gametes singly
165
Q

when does the product rule apply

A

when two or more events are independent
Because the outcome of one die has no effect on the other one, the two rolls are independent. When two or more events are independent, the probability that they will both occur is calculated using the product rule—their individual probabilities are multiplied. That is, the probability that events A and B both will occur equals the probability of event A multiplied by the probability of event B. For example, the probability of getting a 4 on the first die is 1/6; the probability of a 4 on the second die is also 1/6. Because the events are independent, the probability of getting a 4 on both dice is 1/6x 􏰒 1/6= 􏰀 1/36. Applying this principle to human families, the sex of one child has no effect on the sex of the next child; therefore, the probability of having four girls in a row is the product of their individual probabilities (very close to 1/2 for each birth): 1/2 x􏰒 1/2 􏰒 1/2 x􏰒 1/2= 􏰀 1/16.

166
Q

when does the sum rule apply

A

applies when several different events all give the same outcome; that is, the probability that either event A or event B or event C will occur equals the prob- ability of event A plus the probability of event B plus the probability of event C. Returning to the two dice example, what is the probability of rolling a 7? Several different events all give the same total. One could make a total of 7 from a 1 on the first die and a 6 on the second , or a 5 on the first and a 2 on the second, or a 4 on the first and a 3 on the second. Each of these three combinations would be expected to occur at a frequency of 1/6 x􏰒 1/6= 􏰀 1/36

167
Q

testcross

A

A cross between an individual with the dominant phenotype and a homozygous recessive individual

168
Q

dihybrid cross

A

A cross between two individuals that are heterozygous for two pairs of allele

169
Q

dyhybrid

A

A zygote produced from a cross that involves two characters

170
Q

parallels between genes/chromosomes according to walker Sutton and his conclusion

A

1 -Chromosomes occur in pairs in sexually reproducing, diploid organisms, as do the alleles of each gene. The chromosomes of each pair are separated

The chromosomes of each pair are separated and delivered singly to gametes, as are the alleles of a gene. (MENDEL:(MENDELS: Which was similar to mendel’s The pairs of alleles that control a character segregate (separate) as gametes are formed; half the gametes carry one allele, and the other half carry the other allele)

2-The separation of any pair of chromosomes in meiosis and gamete formation is independent of the separation of other pairs, as in the independent assortment of the alleles of different genes in Mendel’s dihybrid crosses. (MENDEL:Medel: The alleles of the genes that govern the two characters segregate independently during formation of gametes.)

3-Finally, one member of each chromosome pair is derived in fertilization from the male parent, and the other member is derived from the female parent, in an exact parallel with the two alleles of a gene.

From this total coincidence in behaviour, Sutton correctly concluded that genes and their alleles are carried on the chromosomes, a conclusion known today as the chromosome theory of inheritance.

171
Q

site on a chromosome at which a gene Is located

A

locus (plural, loci) of the gene
-locus is a particular DNA sequence that encodes a protein or RNA product responsible for the phenotype controlled by the gene

172
Q

epistasis

A

activity of one gene can influence the activity of a different gene

Genes interact, with one or more alleles of a gene at one locus inhibiting or masking the effects of one or more alleles of a gene at a different locus ex: dog fur appears black or brown depending of the melanin gene. in a different locus there is an inhibitor gene to the melanin and if the gene is inhibiting the black or brown fur the dog will have yellow fur. so the dog fur depends firstly on the inhibitor gene then the fur color gene and the inhibitor gene is said to be epistatis

173
Q

incomplete dominance

A

occurs when the effects of recessive alleles can be detected to some extent in heterozygotes. Ex: Flower colour in snapdragons If true-breeding, red-flowered and white-flowered snapdragon plants are crossed, all the F1 offspring have pink flowers. It may seem as if the colors have blended until you cross the F1 offspring and you will get the 1 red : 2 pink : 1 white. This outcome can be explained by incomplete dominance between a CR allele for red colour and a CW allele for white colour. When one allele is not completely dominant to the other, we use a superscript to signify the character. In this case, C signifies the character for flower colour and the superscripts indicate the alleles (R for red and W for white). Therefore, the initial cross is CRCR (red) X CWCW (white), which produces CRCW F1 (pink) plants.

174
Q

codominance

A

occurs when alleles have approximately equal effects in individuals, making the alleles equally detectable in heterozygotes. EX: The LM and LN alleles of the human MN blood group gene that control this character encode different forms of a glycoprotein molecule located on the surface of red blood cells. If the genotype is LMLM, only the M form of the glycoprotein is present and the blood type is M; if it is LNLN, only the N form is present and the blood type is N. In heterozygotes with the genotype LMLN, both glycoprotein types are present and can be detected, producing the blood type MN. example of codominance.

175
Q

multiple alleles

A

: although alleles do indeed occur in pairs in individuals, multiple alleles (more than two different alleles of a gene) may be present if all the individuals of a population are taken into account. Multiple alleles present no real difficulty in genetic analysis because each diploid individual still has only two of the alleles, allowing gametes to be predicted and traced through crosses by the usual methods. Ex: ABO Blood types; The four blood types—A, B, AB, and O—are produced by different combinations of multiple (three) alleles of a single gene I. The three alleles, designated IA, IB, and i, produce the following blood types:
IAIA = type A blood IBIB= type B blood
IAi = type A blood IBi = type B blood
IAIB = type AB blood ii = type O blood

• In addition, IA and IB are codominant alleles that are each dominant to the i allele. Inheritance of the blood types of the human ABO blood group. Note that although there are three possible alleles in the population, each individual parent carries only two.

176
Q

pleiotropy

A

Single genes affect more than one character of an organism. For example, sickle cell disease (see earlier discussion) is caused by a recessive allele of a single gene that affects hemoglobin structure and function.

177
Q

Theodore Escherich

A

Discoverer of the bacterium that helped lay the foundations for our understanding of recombination (e coli)

178
Q

Joshua Lederberg and Edward Tatum

A

discovered conjugation in bacteria

179
Q

Francois Jacob and Ellie Wollman

A

Scientists who used interrupted-mating experiments to map genes on the E. coli chromosome

180
Q

Oswald Averey

A

Proved that DNA is the transforming factor

181
Q

Fred Griffith

A

Demonstrated transformation in bacteria

182
Q

Joshua Lederberg and Norton Zinder

A

Demonstrated transduction in Salmonella typhimurium

183
Q

T/F - when bacteria divide, they produce clones

A

true

184
Q

t/f -

A

F - only certain species unless forced in a lab

185
Q

t/f - During conjugation, DNA is transferred from the donor to the F– recipient. Eventually, all bacteria in a culture will become F+.

A

F – it depends upon whether the donor is F+, in which case the recipient will become F+, but if the donor is an Hfr then the entire F sequence will not get transferred so the recipient remains F-

186
Q

Temperate viruses follow both the lysogenic and lytic life cycles.

A

true

187
Q

compare and contrast the type of transduction process of bacteriophage P22 with that of bacteriophage λ.

A

P22 encodes an enzyme that degrades the host DNA into short fragments which can become integrated in a random fashion when the phage particles assemble; in bacteriophage λ, the phage chromosome becomes integrated into a specific region of the host’s chromosome; when triggered to excise the phage may accidentally remove a host gene on either side of the prophage ( gal or bio);

188
Q

compare bacteriophage and prophage

A

bacteriophage is the infectious form of a virus whereas a prophage is the form when the phage DNA is integrated into the host’s chromosome.

189
Q

Explain the basis of Down syndrome in humans

A

It is a condition that results from nondisjunction of chromosome 21 so that babies are born with 3 copies of this chromosome instead of 2

190
Q

what is interkinesis

A

It is the brief interlude between meiosis I and II during which no DNA replication occurs

191
Q

What is the relationship between synapsis and tetrads?

A

Tetrads are the homologous pairs of sister chromatids which form through the process of synapsis

192
Q

T/F -Nondisjunction is a common problem in meiosis.

A

F – it is a rare event and generally zygotes that form from cells that have suffered nondisjunction rarely survive;

193
Q

T/F-During gamete formation in humans, the X and Y chromosomes behave as homologues.

A

True

194
Q

linked genes

A

Genes on the same chromosome are known as linked genes, and the phenomenon is called linkage.

195
Q

autosomes

A

Chromosomes other than the sex chromosomes are called autosomes; genes on these chromosomes have the same patterns of inheritance in both sexes. In humans, chromosomes 1 to 22 are the autosomes.

196
Q

SRY gene

A

SRY (for sex-determining region of the Y), appears to be the “master switch” that directs development toward maleness at an early point in embryonic development.
For the first month or so of embryonic development in humans and other mammals, the rudimentary struc- tures that give rise to reproductive organs and tissues are the same in XX or XY embryos. After 6 to 8 weeks, the SRY gene becomes active in XY embryos, producing a protein that regulates the expression of other genes, thereby stimulating part of these structures to develop as testes. As a part of stimulation by hormones secreted in the developing testes and elsewhere, tissues degener- ate that would otherwise develop into female structures such as the vagina and oviducts. The remaining struc- tures develop into the penis and scrotum. In XX embryos, which do not have a copy of the SRY gene, development proceeds toward female reproductive structures. The rudimentary male structures degenerate in XX embryos because the hormones released by the developing testes in XY embryos are not present

197
Q

sex linkage arises from what things

A

2 differences:
between males and females: (1) males have one X chro- mosome and therefore one allele for each gene on this chromosome; females have two copies of the X chromo- some and therefore two alleles for all genes on the X chromosome; (2) males also have one copy of the Y chromosome and one allele for each gene on this chro- mosome; females have no Y chromosome and therefore no Y alleles at all. Y chromosomes are present in males but not females.

198
Q

reciprocal cross

A

phenotypes were switched between the sexes

199
Q

deletion

A

A deletion occurs if a broken segment is lost from a chromosome

200
Q

duplication

A

A duplication occurs if a segment is broken from one chromosome and inserted into its homologue. In the receiving homologue, the alleles in the inserted fragment are added to the ones already there

201
Q

translocation

A

A translocation occurs if a broken segment is attached to a different, nonhomologous chromosome

202
Q

inversion

A

An inversion occurs if a broken segment reattaches to the same chromosome from which it was lost, but in reversed orientation, so that the order of genes is reversed

203
Q

To be inherited, chromosomal alterations_____

A

must occur or be included in cells of the germ line leading to development of eggs or sperm.

204
Q

what Mendel referred to as characters, we now refer to as

A

genes

205
Q

f1 generation

A

the first generation of offspring from the cross

206
Q

plants used in the initial cross are called the ___

mendel

A

P generation

207
Q

f2 generation

A

offspring from the f1 generation

208
Q

masking a gene

A

dominance

209
Q

dihybrid

A

a zygote produced from a cross that involves 2 characters

210
Q

Compare and contrast pleiotropic and polygenic characters

A

In the case of a pleiotropic character, a single gene would give rise to multiple traits whereas in polygenic inheritance and single character (e.g. the height of a human adult) is determined by the combined effects of multiple genes.

211
Q

The phage life cycle begins when___

A

a phage attaches to the surface of a bacterium. For phages such as T2, the infected cell quickly stops producing its own molecules and instead starts making progeny phages. After about 100 to 200 phages are assembled inside the bacterial cell, a viral enzyme breaks down the cell wall, killing the cell and releasing the new phages. The whole life cycle takes approximately 90 mins

212
Q

How did Hershey and Chase exploit the life cycle of a phage to gain evidence for DNA as the hereditary material?

A
  • they infected E. coli growing in the presence of radioactive 32p or 35s with phage T2. The progeny phages were either labelled in their DNA with 32p or in their protein with 35s
  • fresh e coli cells were infected with the radioactively labeled phages
  • After infecting the bacteria, the cells were mixed in a blender to remove the phage coats from the cell surface. The components were analyzed for radioactivity (result = no radioactivity within cell; 35s in phage coat)
  • Progeny phages analyzed for radioactivity (result=no radioactivity in progeny phages)
  • 32p showed within the cell, not page coat
  • conclusion = 32P, the radioisotope used to label DNA, was found within phage-infected cells and in progeny phages, indicating that DNA is the genetic material. 35S, the radioisotope used to label proteins, was found in phage coats after infection, but was not found in the infected cell or in progeny phages, showing that protein is not the genetic material.
213
Q

the research of Griffith, Avery and his coworkers, and Hershey and chase established with term

A

transformation: the conversion of a cells hereditary type of the uptake of DNA released by the breakdown of another cell, as in the Griffith and Avery experiments.

214
Q

DNA is composed of which 4 nucleotides

A

Adenine (purine)
Guanine (purine)
Thymine (pyrimidines)
Cytosine (pyrimidines)

215
Q
  1. outline the structure of DNA, identify its components, and describe how the components form a double helix.
A

• DNA is formed by 2 sugar-phosphate backbones. The sugar is deoxyribose and it is linked to a phosphate group. The phosphate group connects to the 3 carbon in deoxyribose on one end and a 5 carbon on the other end. This creates the polypeptide chain, the polynucleotide chain of DNA has polarity, or directionality which reads 3’ to 5’. Other than the phosphate group which is bound to the deoxyribose a nucleotide base pair is also bonded to deoxyribose. The 4 possible nucleotides A (adenine), T (thymine), C (cytosine), G (guanine), where A and G are purines and T and C are pyrimidines and A bonds with T and C bonds with G to make complemented base pairs. The structure of a double-stranded model for DNA structure in which two polynucleotide chains twist around each other in a right-handed way, like a double-spiral staircase. This double helix is formed by the hydrogen bonds that form in between the base pairs (AT, CG). Each full turn of the double helix takes up 3.4 nm along the length of the molecule; therefore, 10 base pairs are packed into a full turn.

216
Q
  1. identify that the two complementary strands of a DNA molecule are antiparallel.
A

• DNA is configured in a way where it is read 3’ to 5’ and since there are 2 strands in a double helix the only way for the 2 strands to fit together in a stable chemical way only if they are antiparallel, that is, only if they run in opposite directions.

217
Q

Chargaff ’s rules

A

Erwin Chargaff, measured the amounts of nitrogenous bases in DNA and discovered that they occur in definite ratios. He observed that the amount of purines equals the amount of pyrimidines, but more specifically, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine; these relationships are known as Chargaff ’s rules.

218
Q

phosphodiester bond

A

In a polynucleotide chain, the deoxyribose sugars are linked by phosphate groups in an alter- nating sugar–phosphate–sugar–phosphate pattern, forming a sugar–phosphate backbon. Each phosphate group is a “bridge” between the 3􏰕 carbon of one sugar and the 5􏰕 carbon of the next sugar; the entire linkage, including the bridging phosphate group, is called a phosphodiester bond

219
Q

whats the 5’ and 3’ end?

A

the two ends of the DNA polynucleotide chain are not the same: at one end, a phosphate group is bound to the 5􏰕 carbon of a deoxyribose sugar, whereas at the other end, a hydroxyl group is bonded to the 3􏰕 carbon of a deoxyribose sugar. Consequently, the two ends are called the 5’􏰕 end and 3’􏰕 end

220
Q

complementary base pairing

A

Watson and Crick pro- posed that the purine–pyrimidine base pairs in DNA are A-T and G-C pairs. That is, wherever an A occurs in one strand, a T must be opposite it in the other strand; wherever a G occurs in one strand, a C must be opposite it. This feature of DNA is called complementary base pairing, and one strand is said to be complementary to the other. The base pairs, which fit together like pieces of a jigsaw puzzle, are stabilized by hydrogen bonds—two between A and T and three between G and C

221
Q

how much space does each base pair occupy along the axis of the double helix

A

0.34nm

222
Q

antiparallel

A

Watson and Crick also realized that the two strands of a double helix fit together in a stable chemical way only if they are antiparallel, that is, only if they run in opposite directions. In other words, the 3􏰕 end of one strand is opposite the 5􏰕 end of the other strand. This antiparallel arrangement is highly significant for the process of replication

223
Q

how many base pairs per turn of the double helix

A

10

often only 8 are visible in diagrams

224
Q

semiconservative replication

A

model of replication proposed by Watson and crick. two strands of the original molecule serve as templates for the two strands of a new DNA molecule and then rewind into an all “old” molecule. After the two complementary copies separate from their templates, they wind together into an all “new” molecule. In the dispersive replication model, neither parental strand is conserved, and both chains of each replicated molecule contain old and new segments

225
Q

DNA polymerases

A
  • the primary enzymes of DNA replication
  • One of the characteristics of DNA polymerase is that it can add a nucleotide only to the 3􏰕 end of an existing nucleotide chain.
226
Q

The nucleoside triphosphates used in DNA replication differ from ATP by ___

A

having the sugar deoxyribose rather than the sugar ribose. Because four different bases are found in DNA—adenine (A), guanine (G), cytosine (C), and thymine (T)—four different nucleoside triphosphates are used for DNA replication. By analogy with the ATP naming, the nucleoside triphosphates for DNA replica- tion are given the short names dATP, dGTP, dCTP, and dTTP, where the “d” stands for “deoxyribose.”

227
Q

____ of the bond between two phosphates provides the energy for the formation of the new bond

A

hydrolysis

228
Q

key molecular events of DNA replication

A
  1. The two strands of the DNA molecule unwind for replication to occur.
  2. Nucleotides are added only to an existing chain.
  3. The overall direction of new synthesis is in the 5􏰕→ 3􏰕 direction, which is a direction antiparallel
    to that of the template strand.
  4. Nucleotides enter into a newly synthesized chain
    according to the A-T and G-C complementary base-pairing rules.
229
Q

replication fork

A

the Y-shaped structure produced from DNA unwinding

consists of the two unwound template strands transitioning to double-helical DNA.

230
Q

DNA helicase

A

catalyzes the unwinding, which exposes both strands for the next steps in replication. The helicase uses the energy of ATP hydrolysis to unwind the DNA helix. The exposed single-stranded segments of DNA become coated with single-stranded binding proteins, which stabilize the DNA for the replication process. These proteins are displaced as the replication enzymes make the new polynucleotide chain on the template strands

231
Q

topoisomerases

A

overtwisting and strain of DNA ahead of the replication fork during replication are avoided by the action of enzymes known as topoisomerases, which remove the overtwisting as it forms.

232
Q

can a new strand begin since there is no existing strand in place? (DNA replication)

A

a primer, made of RNA instead of DNA. The primer, assembled by the enzyme primase, is laid down as the first series of nucleotides in a new DNA strand. RNA primers are removed and replaced with DNA later in replication

233
Q

leading strand/lagging strand

leading/lagging strand template

A

The new DNA strand assembled in the direction of DNA un- winding is called the leading strand of DNA replication; the strand assembled discontinuously in the opposite direction is called the lagging strand. The template strand for the leading strand is the leading strand template, and the template strand for the lagging strand is the lagging strand template

234
Q

How is the new DNA strand made in the opposite direction to the unwinding?

A

The polymerases make this strand in short lengths that are actually synthesized in the direction opposite to that of DNA unwinding. The short lengths produced by this discontinuous replication are then covalently linked into a continuous polynucleotide chain. The short lengths are called Okazaki fragments, in honour of Reiji Okazaki, the Japanese scientist who first detected them

235
Q
  1. describe the mechanism of DNA replication, and identify the various enzymes involved.
A

DNA helicase, catalyzes the unwinding, which exposes both strands for the next steps in replication. The helicase uses the energy of ATP hydrolysis to unwind the DNA helix. The exposed single-stranded segments of DNA become coated with single-stranded binding proteins, which stabilize the DNA for the replication process. These proteins are displaced as the replication enzymes make the new polynucleotide chain on the template strands. DNA polymerases can add nucleotides only to the 3’ end of an existing strand. in a short nucleotide chain called a primer, made of RNA instead of DNA. The primer, assembled by the enzyme primase, is laid down as the first series of nucleotides in a new DNA strand. RNA primers are removed and replaced with DNA later in replication. DNA polymerases assemble a new DNA strand on a template strand in the 5’ → 3’ direction. Because the two strands of a DNA molecule are antiparallel, only one of the template strands runs in a direction that allows DNA polymerase to make a 5’ → 3’ complementary copy in the direction of unwinding. The polymerases make strands in short lengths that are actually synthesized in the direction opposite to that of DNA unwinding. The short lengths produced by this discontinuous replication are then covalently linked into a continuous polynucleotide chain. The short lengths are called Okazaki fragments. So in the lagging fragment, okazaki fragments are added as the DNA is being pulled apart from helicase and then DNA polymerase fills in between the okazaki fragments. The two newly synthesized fragments are not covalently joined—they have a “nick” between them. Another enzyme, DNA ligase closes the nick.
-replication fork during replication are avoided by the action of enzymes known as topoisomerases, which remove the overtwisting as it forms.

236
Q
  1. explain the function of telomeres at the end of eukaryote chromosomes.
A

• Telomeres are short sequences repeated hundreds to thousands of times, which do not code for proteins. In humans, the repeated sequence, the telomere repeat, is 5’-TTAGGG-3’ on the leading template strand. With each replication, a fraction of the telomere repeats is lost, but the genes are unaffected. The buffering fails only when the entire telomere is lost. The enzyme telomerase can maintain the buffer by adding telomere repeats to the chromosome ends. telomerase adds additional telomere repeats to the end of the template strand before DNA replication begins. The function is to essentially provide a buffer at the ends of chromosomes so that the gene coding section of the DNA is not lost at the end of replication.

237
Q
  1. explain the function of telomeres at the end of eukaryote chromosomes.
A

• Telomeres are short sequences repeated hundreds to thousands of times, which do not code for proteins. In humans, the repeated sequence, the telomere repeat, is 5’-TTAGGG-3’ on the leading template strand. With each replication, a fraction of the telomere repeats is lost, but the genes are unaffected. The buffering fails only when the entire telomere is lost. The enzyme telomerase can maintain the buffer by adding telomere repeats to the chromosome ends. telomerase adds additional telomere repeats to the end of the template strand before DNA replication begins. The function is to essentially provide a buffer at the ends of chromosomes so that the gene coding section of the DNA is not lost at the end of replication.

238
Q
  1. outline the ways in which a cell corrects errors that occur during DNA replication.
A

• 1) The proofreading mechanism, depends on the ability of DNA polymerases to backup and remove mispaired nucleotides from a DNA strand. Only when the most recently added base is correctly paired with its complementary base on the template strand can the DNA polymerases continue to add nucleotides to a growing chain. The correct pairs allow the fully stabilizing hydrogen bonds to form.
2) Any base-pair mismatches that remain after proofreading face still another round of correction by DNA repair mechanisms. These mismatch repair mechanisms increase the accuracy of DNA replication well beyond the one-in-a million errors that persist after proofreading. As noted earlier, the “correct” A-T and G-C base pairs fit together like pieces of a jigsaw puzzle, and their dimensions separate the sugar–phosphate backbone chains by a constant distance. Mispaired bases are too large or small to maintain the correct separation, and they cannot form the hydrogen bonds characteristic of the normal base pairs. As a result, base mismatches distort the structure of the DNA helix. The repair enzymes move along newly replicated DNA molecules, “scanning” the DNA for distortions in the newly synthesized nucleotide chain. If the enzymes encounter a distortion, they remove a portion of the new chain, including the mismatched nucleotides. The gap left by the removal is then filled by a DNA polymerase, using the template strand as a guide. The repair is completed by a DNA ligase, which seals the nucleotide chain into a continuous DNA molecule.

239
Q

base-pair mismatches

A

DNA polymerases make very few errors as they as- semble new nucleotide chains. Most of the mistakes that do occur, called base-pair mismatches, are corrected, either by a proofreading mechanism carried out during replication by the DNA polymerases themselves or by a DNA repair mechanism that corrects mismatched base pairs after replication is complete.

240
Q

mutations

A

differences in DNA sequence that appear and remain in the replicated copies. When a mutation occurs in a gene, it can alter the property of the pro- tein encoded by the gene, which, in turn, may alter how the organism functions. Hence, mutations are highly important to the evolutionary process because they are the ultimate source of the variability in off- spring acted on by natural selection.

241
Q

two major types of proteins associated with DNA structure/regulation in the nucleus of a eukaryote

A

histones and nonhistones

collectively known as chromosomal proteins of eukaryotes

242
Q

chromatin

A

the complex of DNA and its associated proteins, structural building block of a chromosome

243
Q

histones

A
  • small, positively charge basic proteins that are complexed with DNA in the chromosomes of eukaryotes
  • histones link to DNA by an attraction between their positive charges and the negatively charged phosphate groups of DNA
  • 5 types: H1, H2A, H2B, H3, AND H4
  • One function of histones is to pack DNA molecules into the narrow confines of the cell nucleus
  • another function is regulation of DNA activity
244
Q
  1. identify the components of chromatin, and describe the structure of chromatin in eukaryotes.
A

• The histones pack DNA at several levels of chromatin structure. In the most fundamental structure, called a nucleosome, two molecules each of H2A, H2B, H3, and H4 combine to form a bead- like, eight-protein nucleosome core particle around which DNA winds for almost two turns. A short segment of DNA, the linker, extends between one nucleosome and the next. Under the electron microscope, this structure looks like beads on a string. The diameter of the beads (the nucleosomes) gives this structure its name—the 10 nm chromatin fibre. Each nucleosome and linker includes about 200 base pairs of DNA. Nucleosomes compact DNA by a factor of about 7; that is, a length of DNA becomes about 7 times shorter when it is wrapped into nucleosomes. The fifth histone, H1, brings about the next level of chromatin packing. One H1 molecule binds both to the nucleosome at the point where the DNA enters and leaves the core particle and to the linker DNA. This binding causes the nucleosomes to package into a coiled structure 30 nm in diameter, called the 30 nm chromatin fibre or solenoid, with about six nucleosomes per turn. The arrangement of DNA in nucleosomes and solenoids compacts the DNA and probably also protects it from chemical and mechanical damage. In interphase nuclei, chromatin fibres are loosely packed in some regions and densely packed in others. The loosely packed regions are known as euchromatin, and the densely packed regions are called heterochromatin. Chromatin fibres also fold and pack into the thick, rod like chromosomes visible during mitosis and meiosis. heterochromatin represents large blocks of genes that have been turned off and placed in a compact storage form. For example, recall the process of X-chromosome inactivation in mammalian females. As one of the two X chromosomes becomes inactive in cells early in development, it packs down into a block of heterochromatin called the Barr body.

245
Q

euchromatin

A

In interphase nuclei, chromatin fibres are loosely packed in some regions and densely packed in others. The LOOSELY packed regions are known as euchromatin (eu 􏰀 true, regular, or typical)

246
Q

heterochromatin

A

In interphase nuclei, chromatin fibres are loosely packed in some regions and densely packed in others. The DENSELY packed regions are called heterochromatin
-Several experiments indicate that heterochromatin represents large blocks of genes that have been turned off and placed in a compact storage form

247
Q

nonhistone proteins

A

loosely defined as all the proteins associated with DNA that are not histones. Nonhistones vary widely in structure; most are negatively charged or neutral, but some are positively charged. They range in size from polypeptides smaller than histones to some of the largest cellular proteins.
Many nonhistone proteins help control the expression of individual genes. For example, expression of a gene requires that the enzymes and pro- teins for that process be able to access the gene in the chromatin. If a gene is packed into heterochromatin, it is unavailable for activation. If the gene is in the more extended euchromatin, it is more accessible. Many non- histone proteins affect gene accessibility by modifying histones to change how the histones associate with DNA in chromatin, either loosening or tightening the association. Other nonhistone proteins are regulatory proteins that activate or repress the expression of a gene. Yet others are components of the enzyme–protein complexes that are needed for the expression of any gene

248
Q

nuceloid

A

Inside prokaryotic cells, the DNA circle is packed and folded into an irregularly shaped mass called the nucleoid. The DNA of the nucleoid is suspended directly in the cytoplasm with no surrounding membrane.

249
Q

plasmids

A

Many prokaryotic cells also contain other DNA molecules, called plasmids, in addition to the main chromosome of the nucleoid. Most plasmids are circular, although some are linear. Plasmids have replication origins and are duplicated and distributed to daughter ells together with the bacterial chromosome during cell division

250
Q
  1. describe the structure of the nucleoid in bacteria, and explain how it differs from the arrangement of eukaryotic DNA.
A

• DNA. In contrast to the linear DNA in eukaryotes, the primary DNA molecule of most prokaryotic cells is circular, with only one copy per cell. In parallel with eukaryotic terminology, the DNA molecule is called a bacterial chromosome. There are exceptions: some bacteria have two or more different chromosomes in the cell, and some bacterial chromosomes are linear. Inside prokaryotic cells, the DNA circle is packed and folded into an irregularly shaped mass called the nucleoid. The DNA of the nucleoid is suspended directly in the cytoplasm with no surrounding membrane. Many prokaryotic cells also contain other DNA molecules, called plasmids, in addition to the main chromosome of the nucleoid. Most plasmids are circular, although some are linear. The structure of a nucleoid differs from bacteria to eukaryotes in the sense that a nucleoid is a clump or random mass of genetic material DNA and in eukaryotic DNA it is structures a raveld around histones in a organized chromosome.

251
Q

describe how the information contained in a DNA strand is used to synthesize a protein through transcription

A
  • DNA information is in form of A-T-G-C and with this code through translation the code is converted to RNA which is then translated to a codon. The Codon is then made into a protein
  • Transcription: The promoter specifies where on the DNA transcription begins. In prokaryotes, the promoters are immediately upstream of where transcription initiates. With the help of another protein, RNA polymerase recognizes key DNA sequences in the promoter then RNA polymerase binds to the DNA and unwinds it near the beginning of a gene. Unlike DNA polymerases, RNA polymerases can start the complementary copy with no need for a primer already in place. Like DNA, RNA is made in the 5 ́→3 ́ direction using the 3 ́→5 ́ DNA strand as a template. Thus, we refer to the beginning of the RNA strand as the 5 ́end and the other end as the 3 ́end. The RNA polymerase continues adding nucleotides one at a time until the gene is transcribed completely. At this point, the newly synthesized RNA molecule and the enzyme are released from the DNA template. In prokaryotes, there are two types of specific DNA sequences called terminators that signal the end of transcription of the gene.
252
Q

describe how the information contained in a DNA strand is used to synthesize a protein through translation.

A

• Translation: Is the assembly of amino acids into polypeptides. In prokaryotes, translation takes place throughout the cell, whereas in eukaryotes, it occurs mostly in the cytoplasm, although, a few specialized genes are transcribed and translated in mitochondria and chloroplasts.

  • Three main stages: initiation, elongation, termination
  • In translation initiation, a large and small ribosomal subunit associates with an mRNA molecule and the first aminoacyl–tRNA of the new protein chain becomes bound to the AUG start codon. That aminoacyl–tRNA used for initiation is a specialized initiator tRNA with an anticodon to the methionine-specifying AUG start codon. Each step in translation initiation is aided by proteins called initiation factors.

Step 1) Met-tRNA (promoter/initiator) forms a complex with the small ribosomal subunit
Step 2) the complex binds to the 5’ cap of the mRNA and moves along the mRNA (Scanning) until it reaches the first AUG start codon.
Step 3) The large ribosomal subunit binds and GTP is hydrolyzed, the ribosome.

  • this is the beginning of translation, aminocytl-tRNA binds the A site.
  • Peptidyl transferase cleaves the amino acid from the P site tRNA and bonds it to the amino acid on the A site tRNA.
    Step 6) The ribosome translocates along the mRNA to the next codon, thereby bringing the tRNA with the growing polypeptide to the P site and moving the empty tRNA to the Esite.
    Step 7) When translocation is complete, The empty tRNA in the E site is released and the cycle is ready to go again. The polypeptide is complete when the RNA ribosome reaches the termination codon. A release factor (RF) binds to the termination codon in the A site. The polypeptide chain is released from the peptidyl-tRNA in the P site. The empty tRNA and release factor are released, and the ribosomal subunits separate
253
Q

genetic code

A

The nucleotide information that specifies the amino acid sequence of a polypeptide

254
Q

codon

A

genetic code is a three-letter code; each three-letter word (triplet) is called a codon.

255
Q
  1. explain the features of the genetic code, including its triplet nature and its redundancy.
A
  • DNA is made from 4 base nucleotides, the nucleotides are arranged in a way that when they are translated they translate to 3 letter codon. These 3 codon which originate from the DNA are what’s used to make protein.
  • Codons is read in the 5 ́→ 3 ́ direction as they appear in mRNAs, substituting U for the T of DNA. Of the 64 codons, 61 specify amino acids. These are known as sense codons. One of these codons, AUG, specifies the amino acid methionine. It is the first codon translated in any mRNA in both prokaryotes and eukaryotes. In that position, AUG is called a start or initiator codon. The three codons that do not specify amino acids—UAA, UAG, and UGA— are stop codons (also called nonsense or termination codons) that act as “periods” indicating the end of a polypeptide-encoding sentence. When a ribosome reaches one of the stop codons, polypeptide synthesis stops and the new polypeptide chain is released from the ribosome.
  • Only two amino acids, methionine and tryptophan, are specified by a single codon. All the rest are represented by at least two, some by as many as six. In other words, there are many synonyms in the nucleic acid code, a feature known as redundancy. For example, UGU and UGC both specify cysteine, whereas CCU, CCC, CCA, and CCG all specify proline.
  • Another feature of the genetic code is that it is commaless; that is, the words of the nucleic acid code are sequential, with no indicators such as commas or spaces to mark the end of one codon and the beginning of the next. Therefore, the code can be read correctly only by starting at the right place—at the first base of the first three-letter codon at the beginning of a coded message (the start codon)— and reading three nucleotides at a time. In other words, there is only one correct reading frame for each mRNA.
  • The code is also universal. With a few exceptions, the same codons specify the same amino acids in all living organisms, and also in viruses.
256
Q
  1. explain the process of transcription from DNA to mRNA.
A

1) RNA polymerase binds to the promoter, unwinds the DNA, and initiates transcription at the start point. RNA polymerase moves along the DNA, unwinding it and adding new RNA nucleotides to the transcript in the 5’ → 3’ direction.
2) Behind the enzyme, the DNA strands reform into a double helix.
3) The complete RNA molecule is released from the template DNA, RNA polymerase leaves the DNA, and the double helix reforms.

257
Q

sense codons

A

Of the 64 codons, 61 specify amino acids. These are known as sense codons

258
Q

start or initiator codon

A

AUG, specifies the amino acid methionine. It is the first codon translated in any mRNA in both prokaryotes and eukaryotes

259
Q

stop codons (also called nonsense or termination codons)

A

The three codons that do not specify amino acids—UAA, UAG, and UGA— are stop codons (also called nonsense or termination codons) that act as “periods” indicating the end of a polypeptide-encoding sentence. When a ribosome reaches one of the stop codons, polypeptide synthesis stops and the new polypeptide chain is released from the ribosome.

260
Q

degeneracy (or redundancy)

A

there are many synonyms in the nucleic acid code, a feature known as degen- eracy (or redundancy). For example, UGU and UGC both specify cysteine, whereas CCU, CCC, CCA, and CCG all specify proline.

261
Q

what is meant by the genetic code is commaless?

A

the words of the nucleic acid code are sequential, with no indicators such as commas or spaces to mark the end of one codon and the beginning of the next. Therefore, the code can be read correctly only by starting at the right place—at the first base of the first three-letter codon at the beginning of a coded message (the start codon)— and reading three nucleotides at a time. In other words, there is only one correct reading frame for each mRNA. For example, if you read the message SADMOMHASMOPCUTOFFBOYTOT three letters at a time, starting with the first letter of the first “codon,” you would find that a mother reluctantly had her small child’s hair cut. However, if you start incorrectly at the second letter of the first codon, you read the gibberish message ADM OMH ASM OPC UTO FFB OYT OT.

262
Q

what is meant by the genetic code being universal?

A

With a few exceptions, the same codons specify the same amino acids in all living organisms, and also in viruses. The universality of the nucleic acid code indicates that it was estab- lished in its present form very early in the evolution of life and has remained virtually unchanged through billions of years of evolutionary history. Minor excep- tions to the universality of the genetic code have been found in a few organisms, including yeast, some pro- tozoans, a prokaryote, and in the genetic systems of mitochondria and chloroplasts

263
Q

promoter

A

control sequence: To initiate transcription, RNA polymerase binds to the promoter, unwinds the DNA in that region, and starts synthesizing an RNA molecule at the transcription start point

264
Q
  1. outline the differences and similarities between transcription and DNA replication.
A

• They are very similar since they are both cutting DNA in half while using the DNA as a template and replicate a copy that is in the opposite order instead of 3’ → 5’ it is now 3’ → 5’. The main differences are that in transcription it converts DNA to RNA which does not include thymine. Also, in transcription only a very specific area is replicated with the use of a promoter and it ends with the terminator.

265
Q

terminators

A

In prokaryotes, there are two types of specific DNA sequences called terminators that signal the end of transcription of the gene. Both types of terminator sequences act after they are transcribed.

266
Q

TATA box

A

A key element of the promoter of most eukary- otic protein-coding genes, the TATA box, is important in transcription initiation. RNA polymerase II itself cannot recognize the promoter sequence. Instead, proteins called transcription factors recognize and bind to the TATA box and then recruit the polymerase. Once the RNA polymerase II–transcription factor complex forms, the polymerase unwinds the DNA and transcription begins.

267
Q

termination of transcription of eukaryotic protein-coding gene vs prokaryotic differences

A

The termination of transcription of a eukaryotic protein-coding gene is different from that of a prokary- otic gene in that there is no terminator sequence at the end of the gene in the DNA. Instead, at the 3 ́end of the gene is a sequence that is to be transcribed into the pre-mRNA. Proteins bind to this polyadenylation signal and cleave the pre-mRNA at that point. This signals the RNA polymerase to stop transcription. Then the enzyme poly(A) polymerase adds a chain of 50 to 250 adenine nucleotides, one nucleotide at a time, to the newly created 3 ́end of the pre-mRNA. No complementary base pairing with a template is needed for this particular type of RNA synthesis. The string of adenine nucleotides, called the poly(A) tail, enables the mRNA produced from the pre-mRNA to be translated efficiently and protects it from attack by RNA-digesting enzymes in the cytoplasm.

268
Q

introns

A

The transcription unit of a protein-coding gene—the RNA-coding sequence—also contains non– protein-coding sequences called introns that interrupt the protein-coding sequence (shown in Figure 14.8). The introns are transcribed into pre-mRNAs but are removed from pre-mRNAs during processing in the nucleus
-The majority of known eukaryotic genes contain at least one intron; some contain more than 60. The original discoverers of introns, Richard Roberts and Phillip Sharp, received a Nobel Prize in 1993 for their findings.

269
Q

exons

A

The amino acid–coding sequences that are retained in finished mRNAs are called exons. The mechanisms by which introns originated in genes remain a mystery. ]

270
Q

mRNA splicing/ spliceosome

A

-occurs in the nucleus, removes introns from pre- mRNAs and joins exons together.

How does this occur? mRNA splicing occurs in a spliceosome, a complex formed between the pre-mRNA and a handful of small ribonucleoprotein particles (snRNPs; pronounced “snurps”) (Generally, a complex of RNA and pro- teins is called a ribonucleoprotein.)

  • Located in the nucleus, each type of snRNP contains a relatively short RNA called a small nuclear RNA (snRNA) bound to a number of proteins The snRNPs bind in a particular order to an intron in the pre-mRNA. The first snRNPs are those with snRNAs that recognize and form complementary base pairs with mRNA sequences at the junctions of the intron and adjacent exons.
  • Other snRNPs are then recruited, leading to looping out of the intron and bringing the two exon ends close together. At this point, the active spliceosome has been formed.
  • The spliceosome cleaves the pre-mRNA at the junc- tion between the 5 ́end of the intron and the adja- cent exon, and the intron loops back to bond with itself near the intron’s 3 ́end. The spliceosome then cleaves the pre-mRNA at the junction between the 3 ́end of the intron and exon 2, releasing the intron and joining together the two exons.
  • Because of the shape of the released intron, it is called a lariat structure. Enzymes degrade the intron, and the snRNPs are used in other mRNA splicing reactions
271
Q

catalytic RNA/ ribozyme

A

An RNA molecule that catalyzes a ratio like a protein enzyme

272
Q

alternative splicing

A

The removal of introns from a given gene is not absolute. That is, in certain tis- sues, or under certain environmental conditions, exons may be joined in different combinations to produce different mRNAs from a single DNA gene sequence. The mechanism, called alternative splicing, greatly increases the number and variety of proteins encoded in the cell nucleus without increasing the size of the genome.
-current data shows that three quarters all human pre-mRNAs are subjected to alternative splicing

273
Q

exon shuffling

A

a process by which existing protein regions or domains, already selected for due to their functions, are mixed into novel combinations to create new proteins. Evolution of new proteins by this mechanism would produce changes much more quickly than by changes in individual amino acids at random points.

274
Q

tRNA

A

tRNAs are small RNAs, about 75 to 90 nucleotides long, with a highly distinctive structure that accomplishes their role in translation. All tRNAs can base-pair with themselves to wind into four double-helical segments, forming a cloverleaf pattern in two dimensions. At the tip of one of the double-helical segments is the anticodon, the three-nucleotide segment that pairs with a codon in mRNAs. At the other end of the cloverleaf is a double-helical segment that links to the amino acid corresponding to the anticodon. For example, a tRNA that is linked to serine (Ser) pairs with the codon 5 ́-AGU-3 ́ in mRNA. The anticodon of the tRNA that pairs with this codon is 3 ́-UCA-5 ́. (The anticodon and codon pair in an antiparallel manner, as do the strands in DNA. We will write anticodons in the 3 ́→ 5 ́ direction to make it easy to see how they pair with codons normally written 5 ́→3 ́.)

275
Q

mRNA

A

The copied RNA derived from the DNA after it has been processed.

276
Q

ribosomes

A

are ribonucleoprotein particles that carry out protein synthesis by translating mRNA into chains of amino acids. Like some automated machines, such as those forming complicated metal parts by a series of machining steps, ribosomes use an information tape—an mRNA molecule—as the directions required to accomplish a task. For ribosomes, the task is joining amino acids in ordered sequences to make a polypeptide chain.

277
Q

aminoacyl–tRNA synthetase(s):

A

one synthetase for each of the 20 amino acids—catalyze aminoacylation. This energy in the aminoacyl–tRNA eventually drives the formation of the peptide bond linking amino acids during translation

278
Q

Peptidyl transferase

A

cleaves the amino acid from the P site tRNA and bonds it to the amino acid on the A site tRNA.

279
Q

wobble hypothesis

A

Recall that 61 of the 64 codons of the genetic code specify an amino acid. Does this mean that 61 different tRNAs read the sense codons? The answer is no. Francis Crick’s wobble hypothesis proposed that the complete set of 61 sense codons can be read by fewer than 61 dis- tinct tRNAs because of the particular pairing properties of the bases in the anticodons. That is, the pairing of the anticodon with the first two nucleotides of the codon is always precise, but the anticodon has more flexibility in pairing with the third nucleotide of the codon. In many cases, the same tRNA’s anticodon can read codons that have either U or C in the third position; for example, a tRNA carrying phenylalanine can read both codons UUU and UUC. Similarly, the same tRNA’s anticodon can read two codons that have A or G in the third position; for example, a tRNA carrying glutamine can pair with both CAA and CAG codons.

280
Q

aminoacylation

A

the process of adding an amino acid to tRNA - or charging (bc the process adds free energy as the acid-tRNA combinations are formed)

281
Q

aminoacyl-tRNA

A

The finished product of charging, a tRNA linked to its “correct” amino acid, is called an aminoacyl–tRNA. Twenty different enzymes called aminoacyl–tRNA synthetases—one synthetase for each of the 20 amino acids—catalyze aminoacylation. This energy in the aminoacyl–tRNA eventually drives the formation of the peptide bond linking amino acids during translation