BIOLOGY EXAM Flashcards

Question

What is the purpose of cellulose?

Answer 1

major component of cell walls and provides rich supply of energy to organisms. Humans cannot digest cellulose because they lack the enzyme.

Answer 2

forms exoskeleton of mushrooms, insects, and crustaceans; used for contact lenses and biodegradable stitches.

Answer 3

most common energy-storing molecules, provide twice as much energy as carbohydrates. major component of cell membranes in living organisms.

Answer 4

non-polar molecules and hydrophobic

Answer 5

fatty acids, fats, phospholipids, steriods, waxes

Answer 6

consists of a carboxyl group and a hydrocarbon chain

Answer 7

saturated and unsaturated

Answer 8

Contain single bonds between carbons; easily stack

Answer 9

Has one or more double bonds between carbons, which causes kinks

Answer 10

consists of a glycerol molecule and fatty acids. Glycerol is a 3 carbon carbon alcohol with a hydroxyl group on each carbon. Glycerol forms the backbone of fats, to which 3 fatty acids are attached by a dehydration reaction via an ester linkage.

Answer 11

primary lipids of cell membranes in living organisms. | Ex. in the cell membrane

Answer 12

It is made of a glycerol backbone, 2 fatty acids, and a charged phosphate group. The fatty acid tail is hydrophobic and the phosphate + polar unit head is hydrophilic.

Answer 13

extremely essential for structural support, movement, storage, immune system, transport materials, signalling, enzymes | Ex. immunoglobulins, hemoglobins, keratin, fibrin, collagen

Answer 14

This is a structural change in proteins due to an increase in temperature or changes in pH, that results in an altered function/dysfunction.

Answer 15

1. A central carbon 2. An amino group 3. A carboxyl group 4. An H atom 5. A side chain (R group)

Answer 16

There are a total of 20 amino acids; 12 can be synthesized by our cells, 8 are considered essential and must be obtained from our diet.

Answer 17

Amino acids connected by peptide bonds. Each polypeptide has a C-terminus (carboxyl end) and a N-terminus (amino end)

Answer 18

primary, secondary, tertiary, quartnernary

Answer 19

specific linear sequence of amino acids; if one amino acid is changed in the sequence, it could render the protein dysfunctional.

Answer 20

folds and coils at various locations of the polypeptide due to hydrogen bonding between electronegative N and O atoms and partially positive H-atoms. There are 2 different orientations: alpha helix and beta pleated sheet.

Answer 21

coiled shape, H bonds form between the carboxyl group and the amino groups of amino acids along the same chain (every 4 amino acids)

Answer 22

2 separate polypeptide strands that run parallel to each other interact due to H bonds, an accordion shape appears.

Answer 23

3D structure is determined by intermolecular reactions between R groups in the polypeptide chain. Intermolecular reactions include ionic bonds, hydrophobic interactions, hydrogen bonds, or disulfide bridges.

Answer 24

Some proteins consist of 2 or more polypeptide chains aggregated into one functional macromolecule. The final structure is critical as its orientation and shape is directly related to its function.

Answer 25

Informational macromolecules that stores hereditary information

Answer 26

nucleic acids are polymers of monomers called nucleotides linked together by dehydration synthesis reactions via phosphodiester bonds.

Answer 27

1. nitrogenous bases 2. pentose 3. phosphate groups

Answer 28

1. Pyrimidine: 6 membered, single ring of C and N (cytosine, thymine, and uracil) 2. Purine: 5 membered ring fused to a 6 membered ring (adenine and guanine)

Answer 29

1. Ribose: sugar in RNA with oxygen 2. Deoxyribose: sugar in DNA without an onxygen

Answer 30

5 carbon of the pentose

Answer 31

movement of molecules in biological systems that doesn’t require energy. Particles move from high to low concentration (with the concentration gradient). Passive transport includes simple diffusion, faciltated diffusion, and osmosis.

Answer 32

- movement of molecules from an area of high to low concentration unassisted - can occur across cellular membranes if the solutes are small enough - influenced by temperature and pressure - rate of diffusion steadily increases

Answer 33

- occurs if the rate of diffusion is too slow or if the cell needs ions and large polar molecules - diffusion of these molecules across a membrane occurs with the help of transmembrane proteins called transport proteins - the rate of diffusion is influenced by the concentration gradient, the efficiency of transport molecules, or the number of transport proteins

Answer 34

movement of water across semi permeable membrane; occurs when the membrane is impermeable to the solute in question. The 3 types of solutions include hypotonic, hypertonic, and isotonic.

Answer 35

if the extracellular solution is lower in solute concentration than the inside of the cell, water will move INTO the cell, causing it to expand; bursting is possible as well.

Answer 36

if the extracellular solution is higher in concentration than the inside of the cell, water will move OUT of the cell, causing it to shrivel.

Answer 37

solute concentration both inside and outside the cell is equal so water moves in and out of the cell at an equal rate.

Answer 38

This process requires cell energy to move materials in and out of the cell; particles move from high concentration to low concentration (against the concentration gradient) There are 4 types of active transport that inlcude, primary active transport, secondary active transport, endocytosis, and exocytosis.

Answer 39

identical to facilitated diffusion except for the fact that the molecules move against the concentration gradient, which requires ATP

Answer 40

uses the concentration gradient of an ion, created by an electrochemical gradient. It is facilitated by symport and antiport.

Answer 41

a solute moves through the membrane channel in the same direction as the driving ion.

Answer 42

driving ions move through membranes in one direction, providing energy for active transport of another solute in the opposite direction.

Answer 43

the process of importing large molecules or even whole cells from the exterior of the cell into the cytosol. There are 3 types of endocytic pathways: pinocytosis, receptor-mediated, and phagocytosis.

Answer 44

cells engluf extracellular water and any other molecules in the solution. **(cell drinking)**

Answer 45

outer cell contains surface receptor proteins that recognize and bind to specific molecules, bringing them into the cell.

Answer 46

cells engulf solid particles **(cell eating)**

Answer 47

large molecules held within the cell are transpotyed to the external environment

Answer 48

an atom LOSES an electron. The reducing agent is the molecule being oxidized and losing electrons.

Answer 49

an atom GAINS an electron. The oxidizing agent is the molecule being reduced and gaining electrons.

Answer 50

a chemical reaction in which energy is ABSORBED, giving the products MORE potential energy than the reactants.

Answer 51

chemical reaction in which energy is RELEASED; the product has LESS potential energy than the reactants.

Answer 52

6C glucose is split into 3C | glucose + 2 ADP + 2 Pi +2 NAD+ -> 2 pyruvate + 2 ATP +2 NADH

Answer 53

2 ATP net 2 NADH 2 pyruvate

Answer 54

The carboxyl group from pyruvate is removed to form CO₂.

Answer 55

2 pyruvates (from glycolysis) become 2 molecules of Acetyl-CoA. | 2 pyruvate + 2 NAD+ + 2 CoA -> 2 Acetyl-CoA + 2 NADH + 2 H+ + 2 CO₂

Answer 56

2e- and 2H+ are transferred to NAD+, reducing it to 2 NADH + H+

Answer 57

A sulfur-containing compound called coenzyme A (CoA) bonds to the acetyl group, producing Acetyl-CoA which will then enter the Krebs cycle.

Answer 58

mitochondrial matrix

Answer 59

all of the C atoms in glucose have been completely consumed. ## Footnote oxaloacetate + Acetyl-CoA + ADP + Pi + 3 NAD+ +FAD -> CoA + ATP + 3 NADH

Answer 60

4 CO₂ 2 ATP 6 NADH 2 FADH

Answer 61

to build a proton gradient (high concentration of H+) and a proton-motive force (called an electrochemical gradient) in the intermembrane space of the mitochondria for ATP synthesis.

Answer 62

inner mitochondrial membrane

Answer 63

Oxygen begins the movement of electrons by interacting with the last protein complex; 2 electrons are removed. Together the 2 electrons, oxygen, and 2H+ from the matrix forms water.

Answer 64

The oxidation of NADH, reduces the first protein complex. The oxidation of FADH2, reduces the second protein complex.

Answer 65

As the electrons are shuttled down from one protein complex to the next, protons (H+ ions) are pumped out into intermembrane space and create an electrochemical gradient. Electrons shuttle through the ETC until they reach the final electron acceptor, oxygen.

Answer 66

process where ATP is created using an electrochemical gradient

Answer 67

The electrochemical gradient causes H+ ions to flow back into the matrix (with concentration gradient) through the ATP Synthase Complex As H+ pass through it powers enzyme which allows ATP to form from ADP and Pi (reserve in the matrix)

Answer 68

The purpose is to capture light energy to produce ATP and NADPH (energy carriers). Occurs in thylakoid membranes of chloroplasts.

Answer 69

photoexcitation, noncyclic elctron transport, and chemiosmosis

Answer 70

an electron in chlorophyll absorbs a photon (light energy) that excites the electron which allows them to move between the different photosystems.

Answer 71

absorbs wavelengths that are 700 nm. This is the second photosystem in the ETC, but it was the first photosystem discovered. Electrons reaching PSI are re-excited by light energy, which passes to an electron acceptor and then to the transport system, where NADP+ is reduced to NADPH.

Answer 72

absorbs wavelengths that are 680 nm. This is the first photosystem in the ETC. Chlorophyll P680 in PS II absorbs energy. The boosted electron moves through a transport system that releases energy (H+) for the production of ATP. Electrons fill the “hole” left by electron boost of PS I. Electrons from splitting of water fills the “hole” left in PSII and produces oxygen as a by-product. P680+ (PSII) is very electronegative – it can pull electrons from H 2 O molecules and split them! These electrons, along with an enzyme subunit called the water-splitting complex, reduce P680+ to P680 returning it to its neutral state and able to be oxidized again.

Answer 73

- Chlorophyll P680 in PSII absorbs light energy - Boosted electron moves through a transport system that releases energy for the production of ATP - Electron fills the “hole” left by electron boost of PS I - Electron from splitting of water fills the “hole” left in PSII and produces oxygen as a by- product - P680 + (PS II) is very electronegative – it can pull electrons from H2O molecules and split them. These electrons in the water-splitting complex reduce P680 + to P680 returning it to its neutral state and able to be oxidized again. - Chlorophyll P700 in PS I absorbs light energy - Energy “hole” is filled by the electron from PS II - Boosted electrons from PS I pass to an acceptor, then to the transport system. - Finally, the electron, and an H+ ion from the stroma joins with NADP+ to form NADPH. - NADPH is the final electron acceptor and is used in the Calvin Cycle. - A proton gradient is formed as the H + ions accumulate in the thylakoid lumen from: 1) splitting of water 2) the pumping of H + into the thylakoid lumen from the stroma. - This gradient produces electrochemical energy that will be used to make ATP. - Protons move through the ATP synthase complex, forming ATP since light is required for ATP formation, the process is called photophosphorylation. ATP will be used in the Calvin Cycle. - Considered to be non-cyclic because once an electron is lost by a reaction centre it does not return to that system, as it ends up in NADPH

Answer 74

A proton gradient is formed as the H + ions accumulate in the thylakoid lumen from 1) splitting of water and 2) the pumping of H + into the thylakoid lumen from the stroma. This gradient produces electrochemical energy that will be used to make ATP. Protons move through the ATP synthase complex, forming ATP (from ADP + Pi).

Answer 75

- Photosystem II (PSII): Captures light to excite electrons and split water. - Electron Transport Chain (ETC): Transfers electrons and pumps H+ - Photosystem I (PSI): Re-excites electrons and reduces NADP+ to NADPH. - ATP Synthase: Produces ATP from ADP + Pi.

Answer 76

- ATP and NADPH (used in dark reactions). - Oxygen (O2) as a by-product

Answer 77

3 turns; occurs in the stroma of chloroplasts; the 3 stages include carbon fixation, reduction reactions, RuBP regeneration

Answer 78

- CO₂ is added to RuBP (ribulose 1, 5-bisphosphate) which is a 5 carbon molecule to form a 6 carbon intermediate. - This 6 carbon will then split into two 3 carbon molecules of PGA (3-phosphoglycerate). This first reaction occurs 3 times to produce 6 molecules of PGA. - All the reactions in this phase are catalyzed by the enzyme RuBisCO

Answer 79

- Each PGA is phosphorylated from the hydrolysis of 6 ATP to create 6 molecules of 1,3-BPG (1,3-bisphosphoglycerate) - NADPH is oxidized to NADP+. The 2 e- from each of the 6 NADH reduce 1,3-BPG to G3P (3-phosphoglycerate) - 6 Pi molecules are released and can go back to the light reactions to make more ATP - Only 1 of the 6 G3P is used to make glucose and other organic molecules.

Answer 80

- The remaining 5 G3P are combined and rearranged to form 3 molecules of RuBP - 5 molecules of G3P x 3 carbons per G3P = 15 carbons - 15 carbons / 5 carbons per RuBP = 3 RuBP molecules - The production of 1 glucose molecule requires 2 G3P, so it would take 6 turns of the calvin cycle to produce 1 glucose molecule.

Answer 81

an unfavorable process which decreases the production of sugars by photosynthesis. In hot, dry environments, guard cells decrease the size of stomata opening in order to conserve water. This decreases the CO₂ in the leaves, while increasing the O₂. O₂ now competes with CO₂ for rubisco’s active site, and since there is more oxygen, it binds to RuBisCO more often. This results in the oxidation of RuBP, instead of its carboxylation to PGA which decreases the production of carbohydrates. Photorespiration is not ideal because it requires more ATP and resources, slows the Calvin Cycle down due to the increase in substeps, and releases carbon (CO₂) back into the air.

Answer 82

sugar and phosphate group

Answer 83

- ester bond: phosphate and sugar (phosphate group and #5 carbon) - glycosyl bond: sugar and nitrogenous base (base and #1 carbon) - hydrogen bond: nitrogenous bases on both strands - phosphodiester bond: between nucleotides (phosphate group and adjacent #3 carbon)

Answer 84

a triplet of bases that will code for a specific amino acid.

Answer 85

AUG (methionine)

Answer 86

- DNA replication begins at specific sites known as origins of replication. In order to efficiently replicate long strands of eukaryotic DNA, there are multiple replication origins. - A special enzyme called HELICASE recognizes the origins of replication and starts breaking the hydrogen bonds between nitrogenous bases. - The double stranded DNA begins to unwind into single strands. As they separate, they form a Y- shaped structure, called THE REPLICATION FORK - While the DNA is being unwound, tension is created. This is relieved by a class of enzymes called TOPOISOMERASES. These enzymes will relieve the tension by cutting and rejoining strands near the replication fork. - Once separated, the complementary bases have a tendency to rejoin. SINGLE-STRANDED BINDING PROTEINS(SSBs) will prevent this by attaching to the strands, and preventing H-bonds from reforming (annealing) - Separation can occur in both directions from a replication origin creating a replication bubble. This allows for more efficient replication.

Answer 87

1. seperating the DNA strands 2. building the complementary strands 3. proofreading and repairing errors

Answer 88

- DNA polymerases add nucleotides to the 3’ end of a new developing complementary DNA strand. DNA POLYMERASE III reads the parent strand only in the 3’ to 5’ direction. - DNA polymerase builds the new strand using NUCLEOSIDE TRIPHOSPHATE, similar to the nucleotides in finished DNA. (Energy is needed for the new strand to be built, this energy is provided by removing 2 phosphates from the nucleoside triphosphate) - DNA polymerase III cannot begin adding nucleotides without an attachment point. RNA PRIMASE builds a small complementary RNA segment (RNA primer) on the parent template strand at the 3’ end. - The strands are built in the 5’ to 3’ direction in two different ways: - LEADING STRAND – the template parent strand that begins at the 3’ end. Therefore, the complementary strand is continuously made moving toward the replication fork. - LAGGING STRAND– the template parent strand that begins at the 5’ end. Therefore, the complementary strand is made discontinuously in short segments called OKAZAKI FRAGMENTS. - DNA replication is discontinuous on the lagging strand because the replication fork must open first before nucleotides can be added. Therefore, every time the fork is opened further, another RNA primer is required at the 3’ end before DNA polymerase III can add nucleotides. - RNA primers are replaced with DNA nucleotides by DNA POLYMERASE I. - Finally, DNA LIGASE catalyzes the reaction that connects the DNA segments together, including the Okazaki fragments, by a phosphodiester bond.

Answer 89

the template parent strand that begins at the 3’ end. Therefore, the complementary strand is continuously made moving toward the replication fork.

Answer 90

the template parent strand that begins at the 5’ end. Therefore, the complementary strand is made discontinuously in short segments called OKAZAKI FRAGMENTS.

Answer 91

- DNA polymerase III will proofread each nucleotide as it is being added. If there is a mistake, the enzyme functions as an exonuclease where it backtracks and replaces the wrong nucleotide before continuing on. - DNA polymerase II will work to repair any damage to DNA that occurs between replication events. - DNA polymerase I will fill the gaps with the correct nucleotide bases and DNA ligase connects

Answer 92

DNA is TRANSCRIBED into complementary messenger RNA, and ribosomes TRANSLATE messenger RNA into a specific sequence of amino acids, which are used to build proteins

Answer 93

RNA: A carrier of genetic information, similar to DNA. There are 3 major types of RNA: - mRNA (messenger RNA) varies in length, depending on the gene that has been copied acts as the intermediary between DNA and the ribosomes translated into proteins by ribosomes RNA version of the gene encoded by DNA - tRNA (transfer RNA) functions as the delivery system of amino acids to ribosomes as they synthesize proteins very short, only 70-90 base pairs long - rRNA (ribosomal RNA) binds with proteins to form the ribosomes varies in lengths

Answer 94

1. initiaion 2. elongation 3. termination 4. post-transcriptional modifications

Answer 95

RNA polymerase binds to the DNA and unwinds starting at the promoter. More specifically, RNA polymerase binds to the TATA box because it has a high percentage of adenine and thymine, which only have 2 hydrogen bonds that are easy to break.

Answer 96

RNA polymerase continues transcribing as it begins to build the single strand RNA molecule without a primer. RNA is made in the 5’ to 3’ direction using the 3’ to 5’ strand called the template strand. The opposite DNA strand is referred to as the coding strand, as it contains the exact same sequence as the RNA molecule. While RNA polymerase is transcribing, another RNA polymerase may start transcription again at the promoter.

Answer 97

Transcription of a gene is terminated when the RNA polymerase recognizes a termination sequence. RNA transcript is released and RNA polymerase is free to bind to another promoter region. The termination sequence is a string of adenines, which makes up the Poly-A tail

Answer 98

The transcribed RNA is a precursor to mRNA and is vulnerable to conditions outside of the cell nucleus. It must be modified by a Poly-A tail, a 5’ Cap and splicing. - Poly-A polymerase adds a chain of 50-250 adenines to the 3’ end of the pre-mRNA, producing the Poly-A tail, which allows for it to be translated efficiently and protects it from attacks of enzymes in the cytosol. - The 5’ Cap, a sequence of 7 guanines, are added to the start of the pre-mRNA which functions as the initial attachment site that the ribosome recognizes and will use. - An enzyme-protein complex called a spliceosome excises introns (a non-coding sequence of DNA/RNA) and splices the exons (sequence of DNA/RNA that codes for a gene) together.

Answer 99

1. intiation 2. elongation 3. termination

Answer 100

Small and large ribosomal subunits associate with an mRNA molecule at the 5’ cap and an initiator tRNA. The first tRNA for initiation has the anticodon UAC that binds to the AUG (Methionine) start codon on mRNA Step 1: Initiator Met-tRNA forms a complex with the small ribosomal subunit Step 2: The complex binds to the mRNA 5’ cap. It will move along the mRNA until it reaches the AUG start codon Step 3: The start codon is recognized by the anticodon of the Met-tRNA and the large ribosomal subunit binds to complete the ribosome at the P-site. The ribosome will now read the mRNA nucleotide bases 3 at a time (i.e. codons). To ensure the bases are read correctly (to make the correct protein), a reading frame (a particular system for separating a base pair sequence into readable codons) is established at the tRNA-AUG pairing location on the mRNA.

Answer 101

Step 1: Elongation begins when the initiator tRNA is bound to the P site, the A site is empty Step 2: The second tRNA with the appropriate anticodon and amino acid binds to the mRNA codon in the A site of the ribosome The 1st amino acid (Met) is cleaved from the tRNA in the P site and forms a peptide bond with the amino acid on the tRNA in the A site. The new polypeptide chain is attached to the tRNA in the A site and the empty tRNA is in the P site Step 3: The ribosome moves to the next codon on mRNA. tRNA in the A site moves to the P site, tRNA in the P site moves to the E site. The A site becomes open again to accept the next amino acid attached to its tRNA

Answer 102

- There is no tRNA for the stop codons (UAA, UAG, UGA). When the A site of the ribosome arrives at the stop codon on the mRNA, a protein release factor (RF) binds to this site instead of tRNA. - This causes the polypeptide to be released from the P site and it is detached from the ribosome. The ribosomal subunits separate and detach from the mRNA. - The polypeptide becomes a functional protein once it goes through some processing and forms its 3D shape. Some proteins are composed of two or more polypeptide chains (e.g., hemoglobin). These polypeptides are produced via separate translation events and come together to form a single functioning protein.

Answer 103

Enzyme Induction: The lac operon is a cluster of genes that contain the DNA sequences to regulate the protein needed for the metabolism of lactose. The lac operon is known as an inducible operon because the inducer (signal molecule e.g., lactose) inactivates the repressor and allows the gene to be transcribed and translated to produce a protein.

Answer 104

1. promotos 2. operator 3. coding regions 4. repressor proteins

Answer 105

site where transcription occurs/RNA polymerase binds

Answer 106

sequence of bases that regulatory factors will bind to in order to control transcription

Answer 107

binds to the operator to repress transcription

Answer 108

- lacI protein binds to the operator, covering part of the promoter. - RNA Polymerase cannot bind to the promoter - Transcription is blocked. Therefore, no mRNA made, no translation occurs = no enzymes made.

Answer 109

- Lactose binds to lacI protein, changing its shape (induced-fit) - lacI can no longer bind to the operator - Transcription and translation proceed and the lactose metabolism enzymes are made

Answer 110

The trp operon is an example of enzyme repression: the operon is repressed (turned off) when high levels of tryptophan are present. Therefore no tryptophan will be produced.

Answer 111

- Repressor protein is inactive and does not bind to the operator. - RNA polymerase binds to the promoter region and transcription occurs. - Tryptophan is produced.

Answer 112

- Tryptophan acts as a signal molecule and activates the repressor protein – in this situation, tryptophan is called a corepressor. - Tryptophan will bind to the trp repressor, which activates it. The active trp repressor will bind to the operator and stop transcription. - No tryptophan is produced.

Answer 113

one amino acid is exchanged for another

Answer 114

a codon for an amino acid is replaced by a stop codon.

Answer 115

a mutation that does not change the amino acid coded for, and therefore is no phenotypic change (ex. error in an intron, wobble hypothesis)

Answer 116

mutation that causes the reading frame of codons to change. Results in different amin acids being incorporated into polypeptide.

Answer 117

missense, nonsense, silent, and frameshift

Answer 118

substitution, insertion, deletion, inversion

Answer 119

the replacement of one base with another; results in missense, nonsense, silent

Answer 120

the addition of an extra base pair in a DNA sequence; results in a frameshift mutation

Answer 121

the elimination of a base pair or group of base pairs from DNA sequence; results in frameshift mutation

Answer 122

the reversal of 2 adjoining base pairs in a DNA sequence; results in missense mutation

Answer 123

- Primary mechanism of homeostasis - A stimulus (from change in internal/external environment) triggers a response that will compensate that change to maintain equilibrium - Ex: household thermostat – when the room temperature moves away from the desired set point, the integrator circuit in the thermostat activates an electrical effector (either furnace of a/c), which returns the temperature in the room to the set point.

Answer 124

- a response of a system that acts to increase the effect of any changes made to the system - sets in motion a chain of events that intensify a change from an original condition usually does not result in homeostasis; generally associated with system instability - Ex. childbirth: Uterine contractions stimulate the release of oxytocin; Oxytocin increases and intensifies the contractions, which results in more oxytocin to be released and stronger contractions; This positive feedback cycle continues, and eventually leads to the delivery of a baby

Answer 125

Negative feedback stops when the system returns to equilibrium, whereas positive feedback continues until an external limit or intervention halts the process.

Answer 126

The maintenance of body temperature within a range that allows cells to function efficiently

Answer 127

1. thermoreceptors in skin detect rise in body temperature 2. hypothalamus turns on cooling system 3. a) sweat glands initiate sweating (evaporation fo sweat causes cooling) b) blood vessels dilate (blood flows to skin and loses heat to skin) 4. body temperature returns to normal 5. hypothalamus turns of cooling system

Answer 128

1. thermoreceptors in skin detect drop in body temperature 2. hypothalamus turns on warming system 3. a) arterioles constrict (limits blood flow and reduces heat loss from skin) b) goosebumps (erect hair traps warm air near skin and reduces heat loss) c) shivering (muscle contractions generate heat by increaing metabolism) 4. body temperature returns to normal 5. hypothalamus turns off warming system

Answer 129

Filtration of water and dissolved solutes occurs as blood is forced through walls of glomerulu into Bowman's capsule by fluid in capillaries. Filtrates include Na+, Cl-, water, H+, glucose, amino acids, vitamins, urea, and uric acid.

Answer 130

1. Filtration – movement of fluids from the blood in the glomerulus to the Bowman’s capsule 2. Reabsorption – transfer of fluids from nephron into peritubular capillaries 3. Secretion – transfer of fluids from peritubular capillaries into nephron

Answer 131

(passage of bodily fluids through selectively permeable membrane) - Occurs in the glomerulus - Occurs when blood under high pressure (>65mmHg) from the afferent arterioles moves into glomerulus - High pressure forces water and dissolved solutes (glucose, amino acids, NaCl, H+, vitamins, minerals, urea) through walls of glomerulus and into bowman’s capsule - Filtration is based on size of molecules; thus, useful molecules also enter the filtrate and must be reabsorbed - Larger particles (proteins, red blood cells, platelets) are too large to pass through and remain in arteriole blood

Answer 132

(return substances to the blood) **Proximal Tubule** - Lined with cells that contain many mitochondria to power selective reabsorption - Some substances are reabsorbed by active transport and some by passive transport - Na+, K+, glucose, and amino acids are actively transported back into the bloodstream (reabsorbed) - Positive ions carry negatively charged ions with them (like HCO3- and Cl-) - The pH balance is maintained by the movement of HCO3 - and H+ - Export of solutes out of nephron creates an osmotic gradient that draws water out of the nephron and into interstitial fluid (reabsorption of H2O) - NH3 and H+ (via ion exchange) are secreted into the tubule to eventually exit as waste **Descending Loop of Henle** - Reabsorption of H2O (filtrate therefore becomes very concentrated) **Ascending Loop of Henle** - Reabsorption of NaCl (made easy by the conc. gradient created in the descending loop) **Distal Tubule** - More selective reabsorption of nutrients - Further pH maintenance - Secretion of K+, H+ & ammonia

Answer 133

(movement of wastes/ toxins from the blood back into nephron) - Remaining filtrate contains some H2O, excess K+ ions, nitrogenous wastes (urea), minerals, salt (NaCl) and other substances (like some drugs) - Enters the collecting duct - Collecting ducts merge into the ureter, which empties into the bladder

Answer 134

1. Hypothalamus: neurohormones 2. Pineal gland 3. Anterior pituitary gland: FSH (follicle-stimulating), LH (luteinizing), ACTH (adrenocorticotropic), TSH (thyroid-stimulating), Prolaction, Endorphin, GH (growth hormone) 4. Posterior pituitary gland: oxytocin, ADH (antidiruetic)

Answer 135

1. Thyroid gland 2. Parathyroid gland

Answer 136

(both in the adrenal glands) 1. cortex: epinephrine and norepinephrine (stress) 2. aldosterone and cortisol

Answer 137

produced by pancreas when blood sugar levels are low, converts glycogen to glucose

Answer 138

produced by pancreas when blood sugar is high and makes cells more permeable to glucose (negative feedback mechanisms control blood glucose levels by increasing or decreasing insulin)

Answer 139

- islets of Langerhans: insulin and glucagon

Answer 140

- regulated by nervous system - adrenal medulla produces epinephrine (adrenaline) and norepinephrine - in times of stress epinephrine and norepinephrine are released into blood; prepares the body for a fight-or- flight response

Answer 141

1. glycogen breakdown into glucose -> increases blood sugar level and results in a sudden burst of energy 2. increases heart rate and respiratory rate and cell metabolism 3. vasodilation -> increases blood supply oxygen and nutrients to tissues (i.e., muscle) 4. retinal dilation -> enhanced visual perception

Answer 142

1. has the same effects as epinephrine, also increases blood pressure 2. affects the amygdala – an attention centre in the brain

Answer 143

- estrogen - progesterone - follicle stimulating hormone (FSH) - luteinizing hormones (LH)

Answer 144

ovaries; ovulation

Answer 145

ovaries; luteal phase

Answer 146

pituitary gland; ovulation

Answer 147

pituitary gland; ovulation

Answer 148

- Flow Phase (day 1-5) - Follicular Phase (day 6-13) - Ovulation (day 14) - Luteal Phase (day 15-28)

Answer 149

(Nt): the number of individuals of a specific species that occupies a given area or volume at a given time.

Answer 150

(D): the number of individuals of the same species that occurs per unit area of volume

Answer 151

the rate at which species reproduces with unlimited conditions

Answer 152

1. clumped 2. random 3. uniform

Answer 153

where individuals are grouped more closely together; this can be due to limited seed dispersal or asexual reproduction.

Answer 154

when organisms are distributed independently of each other; this happens when conditions do not vary within a habitat and individuals are neither attracted nor repelled by each other.

Answer 155

when individuals are more widely separated. Individuals are equally spaced throughout a habitat; this happens when individuals compete for food, breeding or nesting grounds.

Answer 156

1. geometric/exponential (J-shaped curve) 2. sigmoid/logarithmic (S-shaped)

Answer 157

J shaped curve; organisms reproduce continuously at a constant rate. Ex: bacteria

Answer 158

S shaped curve; a type of population growth that starts slow, accelerates/grows rapidly, then levels out over time; the population grows slowly when the population size is small bc few individuals are reproducing; it also grows slowly when the population is large bc it approaches the carrying capacity

Answer 159

limit the growth of population and are dependant on the population density 1. competition 2. predation 3. crowding 4. allee effect 5. disease and parasites

Answer 160

- an interaction in which both competing populations lose access to some resources - becomes more intense when more individuals are competing for the same resource 1. interspecific competition: competition between members of different species 2. intraspecific competition: competition between members of the same population

Answer 161

- the interaction between predator and prey - when one organism kills and eats another, is more competitive when the population density of predators is high

Answer 162

- only occurs in high-density population with limited space - individuals living in extremely dense population are less likely to survive, they grow slowly and tend to be small

Answer 163

- a phenomenon that occurs when a population cannot survive or fails to reproduce to offset mortality once the population density is too low - This effect can produce extinction risk.

Answer 164

- Disease is more likely to break out and result in deaths when more individuals are living together in the same place. - Parasites are also more likely to spread under these conditions

Answer 165

limits the growth of a population regardless of the population density - Natural disturbances (fires, earthquakes, storms etc) - Temperature fluctuations

Answer 166

a population's niche is defined by the resources it uses and the environment conditions it requires over its lifetime

Answer 167

the range of conditions and resources that a population can possibly tolerate and use

Answer 168

the range of conditions and resources that a population actually uses in nature

Answer 169

1. camouflage 2. chemical defence 3. behavioural defence 4. mimicry (batesian and mullerian) 5. spines and armour

Answer 170

- an organism mimics the patterns of its environment - ex. rockfish use disruptive coloration to hinder detection and to blend into background

Answer 171

- an organisms is chemical unattractive, either by realsign noxious odious or by concentrating poisonous chemicals in its body - ex. skunks produce a noxious chemical

Answer 172

- this defence can be passive (hiding, freezing or playing dead) or active (fleeing, herding, mobbing, or using distraction displays) - fight and flight - ex. rabbits use both freezing and hiding to avoid becoming prey - ex. squid and octopuses use ink clouds as a form of display distraction

Answer 173

- one species evolves to resemble another - Batesian: a harmless species resembles a poisonous one - Mullerian: two or more poisonous species that share common predators look the same - ex. The viceroy and the monarch butterflies use mullerian mimicry. They are both unpalatable and look very similar

Answer 174

- various animals and plants have hard, thorny, or needle-like structures - ex. porcupine releases hairs that have been modifies ino sharp barbed quills - ex. prickly pear cactus use spines for protection.

BIOLOGY EXAM Flashcards

(200 cards)