7. Biology 1 Flashcards
organelles
nucleus, rough er, smooth er, golgi apparatus, mitochondria, centrioles/centrosomes, lysosomes, peroxisomes
nucleus
organelle where DNA is stored (regular DNA, only other place where DNA is found is in the mitochondria, but mitochondrial DNA circular)
nucleolus
site of rRNA transcription and ribosome assembly
Rough ER
covered with ribosomes, transports proteins into ER lumen for translation, almost all proteins are made in the ER, post-translational modifications to proteins and DNA start in RER and continues in golgi
Smooth ER
lipid synthesis/modification (no metabolism, lipid metabolism is done in the mitochondria)
golgi apparatus
post-office, organizes proteins, continues post-translational modification, excrete in vesicles bound for the plasma membrane, back to the ER or to organelles
mitochondria (see diagram)
know mitochondria shape, and function: outer membrane, intermembrane space, inner membrane matrix. contains mitochondrial DNA (from aerobic prokaryotes- endosymbiotic theory)
How do the pH values of the matrix and the intermembrane space compare?
intermembrane space is acidic (around 1.4pH) and matrix is relatively neutral (7.8) due to the proton gradient crossing the inner mitochondrial matrix for ATP production
centrioles/centrosomes
proteins and nucleating factors within which the centrioles are located, it organizes microtubules, flagella, and cilia, and is important to cell division (sets up framework for it)
lysosomes
pH of 5, digests cell pats, fuses with phagocytotic vesicles, participate in cell death, buds off from the golgi
peroxisomes
self-replicate, detoxify chemicals, participate in lipid metabolism
A lab worker must inject a segment of DNA into the nucleus of a living cell. To access the nuclear lumen, the microscopic needle must pierce a minimum of how many layers of lipid membrane?
To enter the nuclear lumen, the needle must pass through the cell membrane (2), plus the outer nuclear membrane (2), plus the inner nuclear membrane (2), for a total of six single layers of lipids.
cytoskeleton
microtubules, intermediate filaments, microfilaments
tubulin
smallest structural unit, forms microtubules, alpha and beta tublulin form a heterodimer called a protofilament, 13 protofilaments make up a hollow 9+2 arrangement that makes up one microtubule
mictotubules
makes up cytoskeleton, with intermediate filaments and mictrofilaments, comprised of protofilaments
protofilament
comprised of alpha and beta tubulin in a 9+2, 13 protofilaments make a microtubule
cytoskeleton summary
alpha and beta tublulin form a heterodimer that forms long chains called protofilaments, 13 protofilaments surround a hollow core make up 1 microtubule, 20 microtubules form a nine doublet design (9+2)
cytoskeleton
network of microfilaments, microtubules and intermediate filaments that provides structure and stricture for intracellular transport
spindel aparatus
microtubules that grow out from the polar centrioles during mitosis, it binds the chromosomes during metaphase to the centromere and aids with disjunction of the cell
actin
a protein monomer that polymerizes to form microfilaments, forms filament of the sarcomere (tracks for which filaments (myosin motor proteins) to move on)
intermediate filaments
general class of proteins that polymerize to form filaments that are intermediate in diameter ti microfilaments and microtubules
myosin
motor protein
flagella
whip like, used for movement
cilia
protrusuons, numerous in number used to for many things, like movement and moving debris in the respiratory tract (note important places: cerebrospinal cavity- moves CSF, fallopian tubes - move egg, respiratory tract- clear debris)
9+2 arrangement
found in microtubules, and EUK cilia and flagella, not in proks
flagella
eukaryotic - whipping motion, made of microtubules/tubulin. prokaryotic - spinning/rotating motion, made of simple helices of flagellin
phosphlipids
lipid molecules with non-polar tail and polar phosphate head, think phospholipid bilayer and amphipathic characteristics
integral proteins
proteins that have one or more segments (moieties) embeded in the phopholipid bilayer
transport proteins
cross phopholipid bilayer, transports ions, proteins, etc.. into hydrophobic core
membrane receptor
any protein on a membrane that binds a ligand (signaling molecule)
cholesterol
amphipathic, steroid region, polar region, between phosphlipids in bilayer, add rigidity
fluid mosaic model
bilayer model, with polar heads facing outward, and non-polar tails facing inward
exocytosis
vesicles inside the plasma membrane dumps contents into outside as it fuses with membrane
endocytosis
cell uptake via invagination forming a endosome vesicle
phagocytosis
invagination of large particles, bacteria, etc.
pinocytosis
invagination of extracellular fluid and small particles, non-specific (phagocytosis is specific and is receptor mediated)
passive (simple) diffusion
no atp needed, materials cross with no aid, usually small molecules (water diffusion
facilitated diffusion
no atp needed, a type of passive diffusion, normally just a conformational ion channel that allows for the movement (sodium potassium gradient)
active transport
requires atp, travels against concentration gradient or electrical potential
secondary active transport
no direct coupling of atp required
tight junctions
waterproof barriers, found on skin, and linings of the bladder blood-brain barrier, and tubules of the kidneys
gap junctions
tunnels between cells, ex. between cardiac cells smooth muscle cells and neurons like in the eye
adherens junctions
strong mechanical attatchments, in epithelium and between cardiac cells
desmosomes
strong cellular junction, weld cells together to protect against stress, found in tissues that receive considerable amounts of stress such as the skin
tissues
epithelial, nervous, connective, and muscle
note: blood, fat, and dermis are connective tissue, the epidermis is epithelial
endocrine system
hormone signaling, secreted by cells of endocrine gland, travel through bloodstream,
second messenger systems
g-protein systems,
g protein system
a hormone or molecule binds to a g-protein coupled receptor, causes conformational change, activates alpha subunit (normally off when GDP is bound), alpha subunit separates fro the beta and gamma subunits and activates adenylyl cyclase where it hydrolyzes GTP to GDP and rebinds to the beta and gamma subunits, adenylyl cyclase helps convert ATP to cAMP, cAMP activates protein kinase A which then leads to phosphorylation and a cascade effect
intracellular receptors
lipid-soluble (steroids), do not require membrane surface receptor.They dissolve through the membrane and bind targets in the cytosol. In most cases, their activated target then acts inside the nucleus on the promoter region of a gene to regulate transcription up or down.
paracrine
signal molecules secreted by one cell bind to receptors on local cells, ex. neurotransmitters communicating in the synaptic gap
autocrine
signal molecules secreted by ta cell bind to receptors of the same cell
intracrine
signal molecules (usually steroids) bind to receptors inside the same cell that pruduced them without ever being secreted (like autocrine but without leaving the cell)
juxtacrine
direct cell signaling via contact
cell cycle pi chart
2 rings, outer ring mitosis and interphase only, inner ring: mitosis, G1, S, G2, mitosis
G0
gully differentiated neurons and cardiac muscle cells are frozen in G0 and do not divide, multicucleanted skeletal muscle cells can also be considered to be in G0 phase (AKA quiescent)
For human beings, how many chromosomes are there?: a) before replication, b) after replication, c) during interphase, d) before S-phase, e) after S-phase, f) in a diploid cell, g) in a haploid cell. Describe how the mass of DNA differs for each of the above scenarios.
a) 46; b) 46; c) 46; d) 46; e) 46; f) 46; g) 23; Before replication we have 46 diads (i.e., NO sister chromatids) or 46 unreplicated chromosomes. Let’s define that as a mass of m. After replication we have 46 tetrads, or duplicated chromosomes, so the mass would be 2m. During interphase the mass would change. Prior to S phase the mass would be m, then during S phase it would increase to 2m and remain so until mitosis. Before S-phase, as already stated, the mass would be m. After S-phase the mass is 2m. In a diploid cell the chromosome number will be 2n (46 for humans) and the mass could be m or 2m depending on whether or not S-phase has occurred. A

haploid cell would have a mass of 1/2m. The take home point is that the chromosome number does NOT change when DNA is replicated–you just end up with twice as much DNA per chromosome. The only time the chromosome number changes is during meiosis, and after fusion of gametes in the production of a zygote!
apoptosis
programmed cell death, autolysis of cell contents by lysosomes
histones
proteins that wrap DNA into compact chromosomes (predominantly his+ AA giving + charge to attract to - DNA backbone)
nucleosomes
a set of 8 histones in a cube shape with DNA coiled around it like thread
chromatin
general term for DNA and protein
diploid
2n chromosomes (46 in humans)
haploid
n chromosomes (23 in humans)
homologues
two related but non-identical chromosomes - one from each parent
sister chromatids
two strands of DNA in a duplicated chromosome attached by a centromere
centromere
region of the chromosome that joins the two sister chromatids
kinetochore
The term kinetochore is often used synonymously with centromere, but they are not identical. The centromere is a region on the chromosome and the kinetochore is a specialized group of proteins to which the spindle fibers attach directly during mitosis/meiosis.
cell division
prophase, metaphase, anaphase, telophase, interphase
prophase
nuclear membrane dissolves and chromosomes condense
metaphase
chromosomes lining up at metaphase plate and formation of spindle apparatus
telophase
nuclear membranes beginning to reform and chromosomes unwinding,
interphase
since cell with well defined nuclear edges and uncoiled chromosomes
know mitosis and meiosis phases (see diagrams)
see diagrams
nondiscuntion
when chromosomes fail to separate properly during anaphase, meiosis 1 &2 or mitosis- results in uneven number of chromosomes (ex. common one - trisomy - 21 chromosomes - downs syndrome)
cross-overs
crossovers occur during prophase of meiosis 1, sister chromatids switch segments or whole parts of their chromosomes with eachother
meiosis centromeres
centromeres do not split in meiosis 1 but split in meiosis 2
A karyotype is somewhat like a photographic list of all of the chromosomes found in a cell. If homologous pairs are present, they appear next to one another on the exposed film. All of the chromosomes for the entire cell are presented on the same slide, making differences in the relative length, size and orientation of the chromosomes readily apparent. If a karyotype were produced for a human cell that had just undergone Telophase I of Meiosis, the entire karyotype slide should contain (assume that the chromosome are still condensed and have not reverted back to chromatin):
Meiosis I takes a cell with 23 pairs of homologous chromosomes, or 46 total chromosomes, and creates two cells, each with 23 non-paired, non-homologous chromosomes. You should also know that chromosomes generally decrease in size, with chromosome One being by far the largest. Thus answer D is the best answer.
nucleotides
triphosphate, sugar, and base
common nucleotides
DNA, RNA, cAMP, NADH, FADH2, FMN, CoA, ATP, GTP, UTP, etc..
DNA bases
Adenine,thymine, cytosine, guanine
adenine
A, pairs with T in DNA and U in RNA, forms 2 hydrogen bonds
thymine
T, analogous with U in RNA, in DNA, forms 2 hydrogen bonds
uracil
U, analogous with T in DNA, in RNA, forms 2 hydrogen bonds
cytosine
C, forms 3 hydrogen bonds with guanine
guanine
G, forms 3 hydrogen bonds with cytosine
Draw a short DNA helix. Include the bases, sugar-phosphate backbone, and exact connectivity (i.e., Which element bonds to which, via which type of bond?).
Below is a diagram of a section of DNA helix. Be sure students can visualize and easily draw all connections—especially the phosphate connection between the 3’ hydroxyl group and the 5’ carbon. A common mistake students make when drawing these structures is to have too many bonds to the phosphate, too few, the incorrect charge, etc. Also make sure they have drawn the ribose WITHOUT a 2’ hydroxyl group. It is NOT required knowledge to be able to draw each base (A,T,C,G,U) from memory, but drawing a helix a few times will solidify the concept of how hydrogen bonds form between strands.
Which of the four DNA bases are purines and which ones are pyrimidines?
Adenine and Guanine are purines, Cytosine and Thymine are pyrimidines. Later in this lesson we will discuss Uracil, which is also a pyrmidine.
purines
A and G
prymadine
C and T, and U
origin of replication
location on the chromosome where replication begins, human chromosomes have multiple origins
bidirectional
refers to the fact that replication in both directions simultaneously from the origin
semi-conservative
refers tot he fact that each newly formed daughter helices are made up of one old strand and one new strand
semi-conservatie
refers to the fact that one strand is senthesized continuously and the other strand is synthesized in okazaki fragments
DNA replication
DNA replication begins at an origin of replication. Helicase unzips the double-helix. Immediately, single strand binding proteins coat the individual strands and prevent them from re-annealing. Simultaneously both strands are fed through a replication complex that contains all of the proteins necessary for replication. Because DNA polymerase can only add to an existing 3’ OH group, primase (an RNA polymerase) first constructs short RNA primers on both strands. Two DNA polymerase molecules then begin building new complementary DNA strands. In doing so, they must “read” (i.e., move along the strand) in the 3’ to 5’ direction and are therefore building the new strands in the 5’ to 3’ direction. The sliding clamp is a protein that helps keep the DNA polymerase tightly associated with the strand. Because both enzymes must move along the strand in the 3’ to 5’ direction, they will be moving in opposite directions. If this continued indefinitely the two enzymes would move farther and farther apart. Instead, all enzymes and proteins remain closely associated with the replication fork in what is often called the “replication complex.” As a result, the enzyme working on the lagging strand must copy short segments downstream, release from the strand, move upstream, and copy another short segment downstream—and then repeat. This also means that while the leading strand requires only a single primer, the lagging strand requires multiple primers—one for each these short segments called Okasaki fragments. After this initial replication step, the enzyme RNase H removes all RNA primers. DNA polymerase then fills in the gaps. However, remember that DNA polymerase can only add nucleotides to existing 3’ OH functional groups. Therefore, although it can add a nucleotide to fill the last missing base pair in a gap, it cannot connect that last nucleotide to its downstream neighbor. This functionality is performed by DNA ligase. DNA ligase creates the last necessary phosphodiester bond, completing the strand.
What causes the daughter strand to be shorter after replication?
The DNA polymerases require an existing 3’ hydroxyl group to which they can add their first nucleotide—they cannot set down a nucleotide with a free 5’ end. For this reason, an RNA primer must be placed at the 5’ end of any DNA strand. Later in the process all primers are removed and the gaps are filled in by DNA polymerase and DNA ligase. At the 5’ end, however, there will still be no existing 3’ hydroxyl group and so DNA polymerase cannot replace that section of primer. As a result, every time a chromosome is replicated the new daughter strands will be slightly shorter than the parent strands—by an amount exactly equal to the RNA primers that were in place on both ends of the chromosome.
telomeres
long sections of repetitive DNA nucleotides found at both ends of each chromosome that they provide a buffer region of non-coding DNA so that these repetitive losses in length do not impact a gene sequence, approximately 50 replication cycles will consume the entire theorem region and any subsequent replications will result in the loss of gene sequences
telomerase
an enzyme that adds length to the telomeres,
Telomerase is active in somatic cells early in development, but is turned off in somatic cells thereafter. Why would a mature somatic cell with telomerase activity be potentially harmful?
Because telomeres are shortened by each round of cell division, they provide somewhat of a “time clock” for cells. After the telomeres are gone, subsequent division will quickly damage important coding sections of the DNA and presumably the cell could not survive very many additional divisions without being directed into apoptosis. This would act to prevent the uncontrolled cell division found in tumors. However, if the enzyme telomerase were present, the cell could replace the telomere as it was being used up—essentially removing the “clock function” of the telomeres and theoretically allowing for unlimited cell division.
DNA Damage and Repair
spontaneous hydrolysis of DNA, damage by external chemicals or radiation, mismatch base repairs
