Genetics Exam 3

dna is transcribed by RNA which are brought to ribosomes, the RNA is then translated into amino acids which form proteins

sequences of base pairs in dan determines the sequence of base pairs in a colinear manner

implication: the order of gene mutations match the order of amino acid mutations

cellular control of transcription

region of dna that facilitates transcription of dna

it is located at the 5' end and influences upstream dna

strong --> high rate of transcription
weak --> low rate

low in general but can be influenced by promoters

rna ppolymerase

no

chain initiation
elongation
termination

areas where transcription will end

not the same in complimentary strands

holoenzye w/ 6 subunits

generalized base seqeunce related to other sequnces found in other locations in the genome

most similar to the repeated elements in the genome where the consensus sequence is

-35 and -10 sequence (tata box) indicators of transcriptin of downstream elements

rna pol I: exclusive to making rRNA (two types of molecules)

rna pol II: makes all mRNA and some small molecules

rna pol III: makes tRNa and the 5S component of rRNA

proks need little modificatino before becoming proteins (primary transcript is mRNA)

euks need more (primary transcript is not mRNA)

5' end capped with a guanosine
3' end capped with adenine (poly A tail)

occurs in the nucleus

splicing of interviening sequences

addition of introns (non coding)
exons are portions that code for proteins

used for evolutionary mix/match

poly-A recognition site

recognition site
endonuclease cleaves
polyadenylate polymerase uses ATP to ad A t the end

introns are removed by being spliced such that the upstream end will loop to form a 5' to 2' bond

intro leaves as lariat and exons are rejoined

spliceosome

structure of an intron immediately after excision in which the 5' loops back to form a 5' - 2' linkage w/ another nucelotide

if intron fails to be removed, or is partially removed due to a crytpic splice site, it leads to abnormal protiein folding

when the expression of a gene is dependent upon certain conditions

ex: galactose requires the requires 3 enzymes which are only coded for when galactose is present

cis - dna sequences that serve as attachments for dna binding proteins that regulatin the initation of transcription

trans - the dna binding proteins that regulate the intiatino of transcription

the longer the greater the half life

interact w/ RNA to form dsRNA which is targeted for degradation

products of genes gal 3 gal80 gal 4

when galactose is in the cytoplasm gal80p binds with gap3 leading to activation of gal4p protein, if it is not there, there gal80p binds solo and gal4 does not transcribe

a cell type can produce different quantities of protein due to different splicing

ex: for the insulin receptor gene, in the liver, exon 11 is excluded to produce a protein w/ low affinity for insulin, in the muscle, 11 is excluded leading to high affinity for insulin

a gene that is expressed in the same level in all cells for basic processes

gal1, gal7 and gal10 enzymes

initiation: assembly of ribosome complex, aug added

elongation

termination with stop codon, ribosome dissasembles

mRNA to protein

messenger RNA, provides template for amino acid sequence

where protein synthesis takes place

matches w/ mRNA and carries amino acids that match witha codon

introns are marked with proteins to make sure that they are removed; if the intro had not been removed and ther eis a stop codon then the defective mRNA is targeted for breakdown

the carboxy group is bound with the amino group of the next

splicosomes

before mRNA translation

it could decrease affinity for rna polymerase and transcription will not occur

could stop or lead to a change in amino acid seqence

it will have no major effect if it is properly spliced

when an intron is not completely spliced from the genome

leads to changes in reading frame and amino acids that are coded for

there is not going to be a poly A tail which would decrease stability

the control of synthesis of particular gene products

transcriptonal regulation
rna processing -regulation through rna splicing
translational control-polypeptide synthesis
stability of mRNA-long last vs short lived
posttranslational control- affect enzyme activity and activation
dna rearrangements-expression changes depending on the position of DNA sequence sin the genome

default state is that transcription is on and is turned off by a repressor that beinds to the DNA upstream from the transcriptional start site

repressor protein keeps transcription in off state. When transcription needs to take place, an inducer binds tithe the repressor, making it have less affinity for dna

the default state is on. The combination of an aporepressor and a corepressor give it affinity to bind on DNA and stop transcription

default state is off and an transcriptional activator protein turns it on

negative regulation that is repressible

trp codes for enxymes that synthesis typ. When there is enough trp, then the operon is repressed and vice versa.

in the presence of enough trp, trp acts as the corepressor to bind with aporepressor and decrease transcription

5 enzymes that are all located on the genome in the order that they are needed in the biosynthetic pathway

mechanism where transcription is controlled in its depressed state by the quantitative amount of the corepressor in the cell

ex: e coli trp operon can still work with some trp in the cell if the trp concentration is not optimal in order to increase concentrations

28 base region in the leader sequence before a given operon where termination occurs

a small leader sequence before a the translation of a given operon. If this sequence is synthesized then the operon is not translated (because stop codon is hit).

When there is sufficient Trp then the leader sequence is transcribed and trp is not synthesized. When there is not enough trp then transcription continues.

see page 393 of book

double stranded RNA

RNA inference: the ability of small fragments of double stranded RNA to silence genes who transcripts contain homologous sequences

looking at petunias, 1990s

they added an extra copy of a gene responsible for pigment, but it lead to white flowers. The presence of the extra copy was silent and it causing silencing of wildtype copies

small interfereing RNA: small cleaveage products of dsRNA used to target RNAs containing complementary sequences for destruction or for inhibition of its function

microRNA: small dsRNA that repress translation of mRNAs containing complementary sequences

double stranded RNA that is cut by a dicer enzyme into 25 bp products

stem loop RNA that is diced into 25 bp segments

they are incorporated into the RNA induced silencing complex (RISC). Both strands enter and only on serves as a guide for mathcing up with target RNA through bp matching.

once RISC is established and a matching region is found and that region is cut

guide and target usually have some mismatches the RISC complex attaches tot he target RNA and blocks translation

to ensure that each chromosomal dna is only replicated once

ensure that identical replicas of chromosomes are equally distributed betweeen sister cells

where the spindles connect. known as the microtubule organizing center

microtubule ring structure in the centrosome that help with microtubule organization in animals

the new budding cell must reach a minimal size before synthesis beings. Ceentrosome duplication begins

the duplication of DNA and the duplication of centrosomes

dna replication is fully complete, nuclear envelope breaks down

Saccharomyces cerevisiae - done in early genetic analysis

Schizosaccharomyces pombe - more similar to mammals

in S a new bud begins to emerge and by the onset of mitosis it is almost as big as the mother cell

cell division cycle

there are termperature sensitive. Permissive at 23C but restrictive at 36 degrees C

when the cells are transferred to a restrictive temperature they will freeze at a specific point int he cell cycle

mutant protein is required to complete G1 and S. Ones that are in G1 or S when they are placed in restrictive conditions undergo arrest. Ones that are in a part past S complete their cycle, but undergo arrest in the G1/S of the following cycle and form a quartet

it is required for S phase and some of G2 to M transition. Mutants dhave defects in telomere metabolism and make really long telomere regions

cyclin-CDK (cyclin depednent kinases) complexes

in yeast a single CDK interacts with different cyclin to modify the cycling complex

in euks the CDKs are expressed separately at different stages in the cell cycle

through phosphorlyation of Ser Thr and Try residues of other proteins

cyclin partner tether the protein target and the CDK partner phosphorylates

present in 1/2 G1 to 1/2 S

end of G1 to end of G2

middle of S to end of M

periodic throughout the entire cycle

it allows for a conformational change that allows for phosphorlyation

found in animal cells that holds cells in G1/S restriction by binding and sequestering a transcription factor that initiates the cell cycle

retinoblastoma protein holds onto transcriptoin factor E2F making transcription inactive. Cyclins D-cdk4 and Dckd6 and E-cdk2 phosphorlyate RB to release E2F which allows activation of transcription of DNA and the genes for cylcins and cdks. the Cycl an cdks also go to creating a prereplication complex to replicate complexes needed for transitioning to he S phase

cdc6

cyclin B-cdc 2 complex

after S and G2, cyclin B-cdc2 is in the cytoplasm in its inactive form (bc rate of import into the nucleus is less than export). When cyclin B is phosphorylated, the balance is tipped towards importing cyclin B into the nucleus. cyclinB-cdc2 is then dephosphorylated once it is in the nucleus and it phosphorylates other parts

g1
s
g2

non dividing state of the cell

absence of nutrients or growth factors

dna damage
oxygen depletion
reduced dNTP
loss of normal cellular adhesion
high levels of misfolded proteins

dna damage
centrosome duplication
spindle assembly and position

stops cell when there is dna damage, this allows for repair of the dna.

checkpoints are at g1/s, s, and g2/m boundaries

nicks in dna

blocked movement of the replication fork

dsDNA breaks

a chromosome not properly attaching to the spindle in anaphase

a protein that helps regulate mammal cell response to stress esp dna damage

it is low in concentration and it is taken by mdm2 out of the cell to be broken down by proteasomes

pg 563

p53 is phosphorylated and acetylated t function as transcription factor, making it so mdm2 cannot bind.

apoptosis pathyway- cell death
angiogenesis and metastasis pathways- inibition of those pathways
or arrest and repair pathway

increased transcription of specific genes and decrease transcriptoin of transcription b

block G1

no dna synthesis

stops G2 and spindle formation

activation of cyclin b-cdc2 is correlated with centrosome duplication and spindle formation

monitors the assembly of the spindle attachment of the kinetochores to the spindle

diseases with shared attribute of unctrolled cell growth due to genetic mutation

self sufficiency in growth signals
insensitivity
insensitivity to external anti-growth signals
ability to invade tissues and metastasize
limitless replicative potential
sustained angiogenesis
evasion of apoptosis

proliferation of a network of blood vessels that penetrates into cancerous growths, supplying nutrients and oxygen and removing waste products.

one single ancestral cell

alleles of genes that contain gain of function mutations

proto-ongogenes: normal cellular genes

Ras stimulates growth when it is phosphorylated into its active form. In its mutant form it remains in its active Ras-GTP conformation regardless of the presence of growth signals or not

genes that normally control negative cell proliferation or activate apotitc machinery. Loss of function mutations lead to cancer

when loss of function of p53 happens, the cell becomes insensitive to antigrowth signals and evasion of apoptosis. it eliminates dna checkpoints for dna damage

the mutations must all happen in the same cell that they originate from

from the 5' end of the mRNA

3' end

5' end has modified nucleotide cap

Blood: RBrb
eye w/ tumor: rb rb
eye with no tumor: RB rb
gametes: RB or rb

it leads to uncontrolled DNA synthesis

group of organisms of the same species living in some prescribed geographic area

based on sexually reproducing diploid species that undergo random mating

each individual has a genotype composed of two alleles

division of a population into distinct breeding groups that live within the limited geographic region

The CCR5 receptor enable the HIV virus to combine with the plasma membrane and infect the CD4 Tcells.

Many subpopulations have detla 32, which is a 32 bp deletion from the CCR5 sequence. The deletion creates a frameshift mutation. Leads to a truncation of the protein and cannot lead to entry of HIV to CD4. Heterozygous for the mutation is more common than the homozygous mutation.

genotype frequency: # with given genotype/total # in population examined

allele frequency: # of those alleles (remember to count double)/ number of alleles in the population (remember to double)

mating pairs formed without regard to genotype

1

p represents A
q represents a

all alleles must be accounted for and their frequencies must be able to add up to one

genotype frequencies expected within random mating

AA: p2
Aa: 2pq
aa: q2

it assumes that allele frequencies remain constant from generation to generation

mating is random
allele frequenceis are the same in males and females
all genotypes are equal in terms of fitness (ie no natural selection)
the population is sufficiently large to ignore genetic drift
no mutation
no migration

that when there is a rare allele, there is a very high frequency of getting a heterozygous genotype

ex: if q=0.01 then solve from 2pq/q2 = 198:1

1/3 and 2/3

homoz = sqaure of allele frequency

heteroz = 2 x product of allele frequencies

remeber allele frequencies are still conserved to 1 = a + b+ c

you must separate the X from Y with males.

ex: cross X male gametes with female gametes and cross Y gametes male w/ female

In XX: p2 2pq q2

In XY: p and q because there is only one copy of each X linked gene.

males = the frequency of X linked recessive (0.5 = 0.5)

females = 1/2 X linked recessive (05 --> 0.25)

Ne

the number of individuals in a population contributin gametes to the next generation

(4NmNf) / (Nm + Nf)

Nm = number of males
Nf = number of females

progressive changes in the gene pool

some organisms are more highly evolved than others

mutation
population dynamics (recruitment)
recombination
natural selection
genetic drift (undirected changes in allele frequency that occur by chance in all populations)

differing abilities of individuals to survive and reproduce in their envirionment

fuel for evolution and is the biggest source of new genetic variation

the rate at which A undergoes a mutation to a per generation

p(n) = (1 - u)^n

pn = frequency of A (non mutated) at nth generation
u= rate of muation
n= number of generation

when a mutant allele returns to its wildtype form

that forward and reverse mutations will come to a state of equilibrium

random changes in allele frequencies in a population due to sampling error in selecting gametes from the gamete pool

when the population is small

population bottlenecks occur when a population's size is dramatically reduced for at least one population

As a result, there is only a small subset of gametes that survive and the genetic variation of a population is quickly reduced

a theory predicting that the most fit class of individuals in that populatin will be lost at a high rate due to drift alone, leaving the second best class to suffer same fate, etc leading to a gradual decline in mean fitness and eventual extinction

This would occur in organisms reproducing asexually.

It basically says that deleterious mutations occur in higher number with more generations.

process where small populations accumulate harmful mutations. Small more susceptible

increased inbreeding

mortality is greater than birth rate

species that are harmed so that they are smaller in number can be subject to mutational meltdown

where an animal evolves traits better suited for its environment

the driving fore of adaptive evolution

differences in genes give advatages to survival

1: More offspring are produced than survive and reproduce
2: organisms differ in their ability to survive and reproduce, some of these differences are due to genotype

in every generation, genotypes that promote survival in an environment must be present in excess among individuals of reproductive age, and the favored genotype will contribute disproportionately to the offspring of the next generation

a measurement of the comparative contribution of each parental genotype to the pool of offspring genotypes produced each generation.

represented as w

(pn/qn) = (po/qo) * (1/w)^n

pn/qn = frequencies of genotypes A and B after n generation
po/qo = initial frequencies

the most favored genotype

heterozygous genotype is greater in fitness than their homozygous genotypes

sickle cell anemia.

AA wildtype: 0.85
AS: 1
SS: 0

AS is more resistant to malarial infections than either of the homozygous genotypes

where natural selection acts against a genotype/phenotype change

natural selection promotes genotype/phenotype changes in a trait

used flour beetles that were heterozygous for a lethal allele "l"

+/l

let them evolve from 12 generations and measured frequencies of alleles

frequency of the lethal allele decreased over generations while the frequency of the dominant allele increase

replacement (nonsynonymous) because it changes the amino acid coded, while the other does not

the number of replacement substitutions per replacement site

the number of silent substitutions per silent site

the ratio of replacement site substitution to silent-site substitution

neutral selection

negative/purifying selection

positive selection

the number of individuals in a population contributing gametes tto the next generation

(1-w) = s = 1/(2Ne)

s proportional to amount of the phenotype that is considered less fit to fertile progeny

s = 0 selectively neutral compared to the dominant phenotype

s=1 complete lethality of the lesser phenotype

measure of the extent that natural selection is acting to reduce to the contribution of a given genotype

>0.0005 = natural selection
<0.0005 = genetic drift (by chance that the studied genotyped is being selected against)

mutation that changes the codon, but not to the extent that an amino acid is changed

mutation that changes the codon such that the amino acid is changed

see silent

see replacement

a fragment of DNA is inserted into the genome of a plasmid. the plasmid is transformed into a bacterium. As the bacteria replicates, the plasmid will replicate as well.

the vehicle for DNA cloning

when restriction enzymes cleave at a specific sequence in a dna

sticky ends have an overhang
blunt end go right down the middle

stick ends are produced in palendrome situations

T4 ligases

1. vector dna can be inserted into host cell easily
2. vector molecule can be replicated in host cell
3. cells harboring the vector can be IDed in a straightforward fashion

if you use the same restriction enzyme to cut your target region and plasmid the sticky ends will stick together properly and can be ligated

the larger the dna target, the larger the vector needed

when a genome is broken into multiple pieces by a restriction enzyme it can lead to random matching of sections when things are ligated, so long as the ends fit properly

cleavage reaction is stopped when only a fraction of the restriction sites have been cleaved

this is done when you're target region may have an area that could be cut by the enzyme you are using, but it is undesirable for that region to be cut in the middle of it.

1. Use PCR to obtain amplified segments of target region

2. run gel electrophoresis and cut of the desired fragment with a scapel and purify it

to detect if bacteria successfully incorporated the plasmidan

an anibiotic resistance marker is often added to selcet for transformed bacteria on a medium with antibiotic.

the use of the MCS site allows for cleavage at many different areas, a gene inactivation sequence is inserted. When the gene with the MCS region is not activated you can select for plasmids that successfully incorporated the DNA you are trying to amplify

BD- downstream bait domain
AD - activating domain

BD is responsible for binding to the upstream activating sequence and the activating domain responsible for starting transcription

bait and prey refer to two target proteins you are trying to see that interact. The bait is fused with gal4. The prey is fused with the transcriptional activator.

In order for transcription to occur, the prey and bait must interact in order for the RNA pol complex to be recruited.

compares a network of interactions with given nuclear proteins

a yeast protein

red- lethal loss of function
green - mutation that is not lethal

the more interactions and connection

can be large cale
requires no protein purification
detects interations that occur in living cells
requirse no information about the proteins being tested

weak interactions cannot be determined from strong interactions
concentration enhancement
requirse that interaction takes place in the nucleus
hybrid proteins may fold differently than in their native conformation