Past Exam Questions for Final
What is the equation to calculate chi-squared
(observed-expected)^2/expected
Characteristics of an X-linked dominate trait
-Does not skip generations -Affected fathers pass their one allele on to ALL daughters but not to sons -Affected mothers pass the allele .5 to daughters and .5 to sons -both males and females affect with females tending to be affected more.
Match each description to the appropriate vocabulary term. -All of the alleles of genes within a population -Random changes in allelic frequency of a population due to chance -Changes in allelic frequency of one population due to the migration of alleles from another population
-Gene Pool -Genetic drift -Gene flow
In a study of flower color in roses, you cross a homozygous white flowered rosebush with a homozygous red flowered rosebush. What outcome would you expect if.... -The trait is controlled by a single gene with incomplete dominance -The trait is controlled by a singe gene with partial penetrance
-You would expect a pink flowered rose bush Incomplete: AaxAa makes 1:2:1 ratio with underlying alleles still discrete.Ex. Color - making pink from red - You would expect some flowers expressing and some not. so some red and some white. Penetrance means you have the genotype but some don't express the phenotype Expressivity is kind of mix of different shades
You are making karyotypes of mice as part of a research project. You will find a mouse that has a sec chromosome genotype of XXY -Would this mouse show a male or female phenotype? -What could cause this karyotype. Name a mechanism that could cause this mutation and describe where and how it happened
-this would show a male phenotype because it follows the mammal rules so if y is present the SRY will be expressed -mutation can cause this or nondisjunction in gamete function so that a game has 2 X's so during meisosi.
You cross an individual of genotype AaBBccDdee with an individual of genotype AabbCcDdEE. What is the probabilty of an offspring with genotype A_B_CcddEe?
0.09375 So look at Aa and Aa and it is 3/4 chance to get A_ BB and bb have probability of 1 to get B_ cc and Cc have probability of 1/2 to get Cc Dd and Dd have probability of 1/4 to get dd and ee and EE have probability of 1 to get Ee so multiply and get 0.09375
In a population of 1000 turtles you observe the following genotypes: TT-580 Tt-320 tt-100 What is the frequency of allele t?
0.26
In a population of 1000 baby kitties, 40 are white because of a recessive allele (b). Non-white cats are either homozygous dominant (BB) or heterozygous dominant (Bb). What is the expected frequency of heterozygous kitties in the population?
0.32
A mutation occurs in one cell of a regenerating liver in an adult human. Cells in regenerating livers are undergoing mitosis. This mutation causes expression of an enzyme that should not be expressed in the liver. 1.) use a term to categorize the phenotypic effects of this mutation. 2.) which cells in this individual's body will express this mutant allele? Should this individual worry that the mutation may be passed to their future offspring? Explain your answer
1.) Gain of function 2.) This mutation can NOT be passed to their offspring due to early separation of gamete cells. The only cells in this persons body that will express this mutant allele are the daughter cells of that original mutation.
In a study of antennae shape in Drosophilia, gene N determines whether antennae are formed, with the presence of at least one N- allele resulting in the absence of antennae. A second gene, L, determines antennae length, with the wildtype long allele (L+) dominant over the mutant short allele (L-). 1.)What term should be used to describe the effect of these genes on the antennae trait. Be as specific as possible. 2.)You cross two heterozygous Drosophila (L+L-N+N-).What is the phenotype of the heterozygotes you crossed? What is the phenotypic ratio of the offspring? For the offspring phenotypic ratio, be sure to include the phenotypes of the offspring with the numerical ratio?
1.)Epistasis - more specifically dominant epistasis 2.)12:3:1 Epistasis: Interaction between two genes where one masks the effect of the other at DIFFERENT locus Ratios of each Typical- 9:3:3:1 aa masks B-9:3:4 A_masks B-12:3:1 Dominant at either-15:1 recessive at either- 9:7
In illumina sequecning of mature mRNA you find all the transcripts of one gene including the first intron (transcripts of other genes are not affected. ) A mutation in what sequences could cause this outcome? How would this mutation affect the protein product of this gene?
5' or 3' consensus site. Since it is only impacting the one mature mRNA and not all of them, the problems isn't with the splicisome. Rather the consensus site since the splicisome won't be able to bind to it and cut out introns. This mutation would affect the protein product of this gene because the introns would also be translated.
A women has blood type A and her child has blood type O. Which of the following could not be the blood type of the child's father?
A- A B- B C-AB D-O
Which of the following would not directly contribute to the development of cancer? A- A mutation of one allele of a tumor suppressor gene (not p53) B- Loss of contact inhibition C- A mutation of one allele of a proto-oncogene D- Translocation of a promoter into a cell division gene E- Cells becoming undifferentiated
A- A mutation of one allele of a tumor suppressor gene (not p53)
Which of the following statements about genetic fitness and/or selection is false? A- Individuals with high fitness have greater selection coefficients B- Individuals with low fitness have greater selection coefficients C- Selection against recessive alleles tends to be slower overall than dominant alleles because of the survival in heterozygotes D- Fitness involves both relative viability and reproductive success
A- Individuals with high fitness have greater selection coefficients
Which of the following is true for a species with genic sex determination A- alleles of a gene determine sex B- Presence of differentiated sex chromosomes in the genome C-Sex is determined by the ratio of sex chromosomes to autosomes D- More than one of these may be true
A- alleles of a gene determine sex Genic Sex Determination - where the genotype of an individual at one locus determines sex AND the chromosome carrying that locus is not differentiated. genic balance- is where the sex is determined by ratio of x chromosmes to haploid sets. Ex. XX,2A ratio is 1 or greater so female
Which of the following does not describe the typical eukaryotic genome of a somatic cell during most of the cell cycle? A- condensed B-One chromatid per chromosome C-diploid D-Contained in the nucleus
A- condensed DNA doesn't condense until the Prophase. The majority time of the cell cycle is spent in Interphase Cell cycle: Interphase: G1 Cell growth and protein synthesis G1/S check: checks if cell ready to divide Synthesis: DNA replication and centrosome duplication G2: cell growth and protein synthesis G2/M check: checks if DNA replicated and undamaged Then mitosis
A male mouse is born with no tail which is caused by the expression of a mutant allele at a paternally imprinted gene. From which parent did he inherit the mutant allele? could his offspring be born without a tail? A- it was inherited from his female parent; all his pups will be born with tails B-it was inherited from his male parents; at least some of his pups will be born without tails C-the answer depends on the dominance of the allele D- the correct answer was not available
A- it was inherited from his female parent; all his pups will be born with tails
Which of the following is a noncoding RNA A- rRNA B-intron C-core promoter D-more than one of these
A- rRNA
In eukaryotes, each regulatory transcription factor recognizes its own: A- response element B-enhancer C-core promoter D-regulatory promoter
A- response element Enhancers- bind activator proteins to increase expression core promoter-is that portion of the proximal promoter that contains the transcription start sites. regulatory promoter- They work in cis and are about 40-120 bp upstream. They have response elements bound to it where TF (activators or repressors) can interact with
Which of the following is not a noncoding RNA? A-intron B-telomerase C-rRNA D-snRNP
A-intron A non-coding RNA (ncRNA) is an RNA molecule that is not translated into a protein. Ex: -transfer RNAs (tRNAs) -ribosomal RNAs (rRNAs), -microRNAs, siRNAs, piRNAs, snoRNAs, snRNAs, exRNAs, scaRNAs and the long ncRNAs such as Xist and HOTAIR.
What causes the constricted shape of a eukaryotic chromosome's centromere A-satellite DNA near the centromere is packed as heterochromatin B-minisatellite DNA near the centromere is packed as heterochromatin C-the conserved sequence of the centromere D-the modified nucleosomes at the centromere
A-satellite DNA near the centromere is packed as heterochromatin Satellite DNA: - found near the centromere and subtelomere -short sequences (100bp) repeated in tandem -function would be chromosome structure Short tandem repeats (STR): -found on tips of telomeres,genes, and in 1 million different locations -minisatellites(5-10 bp) and microsatellites (2-4 bp)
What is the difference between an autonomous DNA transposon and a non-autonomous DNA transposon? How does this difference affect the "jumping" of these genes?
Autonomous DNA transposon are able to produce transposonase by itself so TE free to "jump". Non-autonomous DNA transposon can't produce transposonase by itself so has to use neighboring one, but can still jump DNA transposon: -Cut and paste method -Transposase cuts ends of transposons and takes copy out of chromosome then inserts new place -MOVES ORIGINAL RNA Transposon: Make RNA transcription of original copy. then Reverse transcription to make DNA copy by reverse transcriptase. The DNA copy is then inserted.DUPLICATE AND MOVE NEW COPY.
You observe a cell in Prophase II that contains 60 DNA molecules. Which of the following describes the cell? A-n=60 B- Each chromosome is made of two chromatids C- Pairs of chromosomes are aligned along the metaphase plate D- More than one of A-C is correct
B- Each chromosome is made of two chromatids
______________________ describes the imbalance in genetic information due to the removal of controls at cell cycle checkpoints. A- Oncogene B- Genomic instability C- Benign tumor D- Tumor mosaicism
B- Genomic instability
Which of the following statements describing mutations causing cancer is false? A- They can be induced by infectious agents B- They are mostly inherited mutations C- They are common in cells that are highly regulated by hormones
B- They are mostly inherited mutations
Translocation of a proto-oncogene to a chromosome region with reduced methylation increases its likelihood of... A- triggering apoptosis B- becoming an oncogene C- inducing contact inhibition and halting the cell cycle at the G1 and metaphase checkpoints D- causing cell differentiation
B- becoming an oncogene
Which of the following would generally differ between heterochromatin and euchromatin in eukaryotes A- presence of nucleosomes B-histone acetylation C- SNPs D-none of these
B- histone acetylation Heterochromatin: - Dark bands, tightly packed. - Activity of the genes is modified or suppressed. -Fewer genes - Usually centromere and telomere Ex. Inactivated X chromosome and the Y chromosome. Euchromatin: -Not the dark bands. -It represents the major genes and is involved in transcription. -More Genes -Cycles between more and less condensed
A mutation in the gene you are studying changes the amino acid sequence but does not change the function or activity of the protein product. This is most likely a : A- silent mutation B- neutral mutation C-nonsense mutation D-loss of function mutation
B- neutral mutation silent mutation: Alteration of nucleotide that does not alter the amino acid. (the third nucleotide) nonsense mutation: Premature stop codon. Common in transitions loss of function mutation: Protein is not as active or is "dead
which of the following works only in eukaryotes? A- riboswtiches B- siRNA C-asRNA D- none of these
B- siRNA riboswtiches: (only prokaryotes) -Binding of a signal(ligand) within the secondary structure changes the shape and affects the availability of the Shine-Dalgarno sequence to the ribosome . (sometimes makes it unavailable, sometimes makes it available) siRNA:(only eukaryotes) -usually come from transcription of a different gene that codes for inverted repeat so that a stem loops is formed (double stranded RNA on stem part)-Dicer breaks it up -These siRNA's then pair up with RISC that assist in pairing up with target RNA which leads to perfect pairing and leads to degredation. this leads to second source of siRNA since the degradation leads to more siRNAs asRNA: (both pro and euk) Found in prokaryotes and eukaryotes and is the transcription of about 200 bp long asRNA from another site in the genome or from the non-template strandSince asRNA is complementary to the target mRNA it binds to that target and blocks translation (usually binds to the 5' end covering the shine dalgarno)
In E. coli why do we expect a very low level of expression of the lac operon's structural genes when expression of the operon should be repressed A- there is another copy of the lac operon elsewhere in the genome B- the lacI protein may temporarily disassociate from the DNA allowing some transcription C- the lac operon is under the regulation of more than one signal and regulatory protein D- we don't expect very low levels of expression when the operon is repressed
B- the lacI protein may temporarily disassociate from the DNA allowing some transcription
The rentention of a single intron in one gene's mRNA is likely due to a mutation in A- snRNPs B-5' or 3' splice sites of that intron C-ligase D-more than one of these
B-5' or 3' splice sites of that intron
If crossing over does not occur, when can Mendel's Principle of segregation be observed? A-Mitosis B-Meisosis I C-Meiosis II D-none of these
B-Meiosis I Principle of Segregation - the 3:1 ratio is only possible if parents carry two discrete alleles and each offspring has a 50% chance of inheriting each allele from a parent. If crossing over does NOT occurs- segregating occurs in Anaphase I when R and r for example are separated into separate gametes If crossing over occurs- segregating occurs in Anaphase II when R and r are separated into separate gametes
Your aunt's pet goat has blue eyes, which is an autosomal dominant phenotype in goats. How could you determine whether the goat has a homozygous or heterozygous genotype? A- a reciprocal cross B-a test cross C- a monohybrid cross D-a parental cross
B-a test cross test cross- is a cross to a homozygote recessive individual. The outcome of this cross would differ based on the dominant's genotype. A reciprocal cross - switching the identity of the parent with the recessive phenotype will not affect the inheritance pattern of autosomal traits but will affect the pattern of sex-linked traits. i) Ie, if you start with a white eyed male and red eyed female in the P generation of one cross, the reciprocal cross would start with a white eyed female and red eyed male. A monohybrid cross- is an experiment started by crossing two parental organisms (in the P generation) that differ in one trait and are "true-breeding"
What does the enzyme ligase do? A- makes hydrogen bonds B-makes phosphodiester bonds C-makes both hydrogen and phosphodiester bonds D-breaks bonds
B-makes phosphodiester bonds
How is the template strand identified in prokaryotic transcription of an mRNA? A-The TATA box and other consensus sequences are on the nontemplate strand of the promoter B-the -10 and -35 regions on the sense strand of the promoter C-the -10 and -35 regions on the antisense strand of the promoter D-Either strand might be used as template in transcription
B-the -10 and -35 regions on the sense strand of the promoter In eukaryotes- A would be correct
DNA transposons "jump" using the enzyme: A- reverse transcriptase B-transposase C-DNAse D-more than one of these
B-transposase
What is the name of the repair pathway that: Uses the glycosylase and AP endonuclease enzymes
Base Excision Repair Excises the purine or pyrimidine first, then replaces the entire nucleotide.So takes out ONLY the altered base. A specific glycosylase removes the bad base, the AP endonuclease cuts out sugar-phosphate, polymerase(pol 1 in pro and pol beta in euk) replaces nucleotide, ligase re-seals phosphodiester backbone.
A polyploid individual in a species with a haploid number of 5 would not have: A- 10 chromosomes B-15 chromosomes C- 11 chromosomes D-there is no one correct answer
C- 11 chromosomes Polyploidy types Where there is 3n=9 Could be 1.) Autopolypoloidy2.Allopolypoloidy Autopolyploidy -When an individual has more than two sets of one species chromosomes (>2n), triploids(3n),tetraploids(4n),pentaploids(5n),hexaploids(6n),etc. Allopolyploidy -When individual has haploid sets from more than one species also termed a hybrid
A genetic map shows that genes H and R are 15 cM apart. In a heterozygote individual with allele in replusion, what percent of offspring should inherit the alleles Hr from this heterozygous parent? A- 7.5% B-15% C-42.5 % D-85%
C- 42.5 % Repulsion- when H and R phenotypes stay apart in parentals so (Hr/hR) Coupling- when H and R phenotypes stay together in parentals so (HR/hr) Question says repulsion meaning (Hr/hR) and since replusion this means 15% of them would be hr and HR while the other 85% would be Hr and hR. So since asking for Hr, divide 85 in half and get 42.5%
Which of the following is true about a linkage group? A- Every pair of genes in a linkage group are linked genes B-A linkage group is measured in units of base pairs C-A linkage group can be longer than 50 cM D-More than one of answers A- C is correct
C- A linkage group can be longer than 50 cM Map units are roughly additive, which means that map distances calculated from different experiments that share at least one gene can be joined to create a linkage group A pair of genes on the same chromosome may be within the same linkage group but unlinked to each other. This happens because each of those two genes is linked to one or more genes between them. Genetic distances (measured in cM) are not equivalent to physical distances (in base pairs).
Which of the following scenarios best exemplifies the bottleneck effect? A- A small group of individuals leaves their tribe to form a new civilization. B- Minions with larger eyes have greater fitness in the population. C- A natural disaster drastically reduces the population. D-An entire population from North America migrates down to South America.
C- A natural disaster drastically reduces the population.
Which of the following is a trasnversion mutation of the DNA codon CGA A- CAA B-TGA C- CCA D- there is no one correct answer
C- CCA
You are studying a mutant phenotype and do a complementation test. The results show wildtype offspring. Which of the following is correct? A- The mutations are in the sae gene B-The mutations may be either dominant or recessive C- The mutations are complementary D- More than one of A-C is correct
C- The mutations are complementary Complementary test: If cross mutants and get:offspring are different than parents (show wildtype)- then complementary and in DIFFERENT genes offspring are the same as parent (show mutation)- then not complementary and in SAME gene
How does DNA methylation affect transcription ` A- it causes the nucleosomes to be more tightly packed B- it causes the nucleosomes to be more loosley packed C- it blocks the major groove of the DNA D- DNA methylation affects translation, not transcription
C- it blocks the major groove of the DNA
A codon is 5'-ACG-3'. What is the anticodon? Select the answer that shows the wobble position with * A- 5'-CGU*-3' B-5'-CGT*-3' C-5'-C*GU-3' D-3'-C*GU-5'
C-5'-C*GU-3'
In a study of wing shape in drosophila , gene R determines wing shape with elongated wings (R+) dominant over round wings (R-). Gene W affects wing development, two copies of the W- allele result in the absence of wings. You cross two heterozygotes (W+W-R+R-). What phenotypic ratio do you expect in the offspring? A- 12 wingless:3 elongated wings:1 round wings B- 9 wingless: 3 elongated wings:4 round wings C-9 elongated wings:3 round wings:4 wingless D-the correct answer is not available
C-9 elongated wings:3 round wings:4 wingless
What molecule creates primers during eukaryotic DNA replication? A-DNA primase B-DNA polymerase C-DNA polymerase alpha D-RNA polymerase
C-DNA polymerase alpha In prokaryotes: -DNA primase- during elongation, RNA polymerase (DNA primase) creates short stretch of RNA called primers. Then DNA polymerase can do its things. -DNA polymerase III-extends along DNA adding from 3' to 5' until reach double stranded part such as primer. From there it detaches and -DNA polymerase I- comes and removes RNA primers with DNA. Then ligase comes and seals the knick. -Exonuclease done by DNA polymerase I and III -DNA polymerase II,IV, and V repair DNA damage Eukaryotes: -multiple origin (with license) -Primers made by DNA polymerase alpha -delta does elongation -other enzymes required to replace RNA primers -exonuclease done by delta-beta repairs DNA damage
In Anaphase I, what molecule is responsible for keeping sister chromatids together? A- Separase B-Cohesin C-Shugoshin D-There is no such molecule
C-Shugoshin In Anaphase I - during this stage, separase selectively digests the cohesion rings holding homologous chromosomes together. Cohesin holding together sister chromatids is protected from separase by the binding of the protein shugoshin. As a result, homologous chromosomes separate and move towards opposite ends of the spindle. Separase- enzyme that breaks down cohesin Cohesin- proteins that join chromatids during S
The single strand of DNA shown below forms a stem-loop structure with a stem that is 5 nucleotides long. What are the rest of the nucleotides? 5'-TATCGCTA _ _ _ _ _ _ _ _ -3' A- -ATAGCATC- B- -AATCGTAG- C- -TAGCGATA- D- - CTACGATA-
D- - CTACGATA-
How does the ribosome find the start codon during initiation of translation in eukaryotes A-Based on the location of the core promoter B-Based on the location of the Shine-Dalgarno sequence C-By binding the 5'cap and then using the first three bases after the cap D- By binding the 5' cap and then scanning for the kozak sequence
D- By binding the 5' cap and then scanning for the kozak sequence For prokaryotes B would be correct
Which of the following statements is correct about assortative mating? A- An example of random mating. B- Increase heterozygosity. C- Preferential mating between unlike individuals. D- Increase homozygosity.
D- Increase homozygosity.
You are studying the recessive wingless mutation in Drosophila (fruit flies). You set up a series of crosses between wingless flies and normal flies. You notice that the offspring phenotype ratio from crosses where the male was wingless is different than the offspring phenotypic ratio from crosses where the female was wingless. What does this tell you about the wingless phenotype? A- Recessive B-Dominant C-autosomal D-Sex-linked
D- Sex-linked
Which of the following is not true about translocations A- a reciprocal translocation may explain differences in synteny between species B- a robertsonian translocation may explain differences between species in haploid number c- a robertsonian translocation occurs between acrocentrick chromosomes D- during meiosis, crossing over in a heterozygotes translocated chromosomes may result in gametes with deletions and duplications
D- during meiosis, crossing over in a heterozygotes translocated chromosomes may result in gametes with deletions and duplications Translocations Movement of part of one chromosome to a non-homologous chromosome
In E.Coli. DNA polymerase I corrects misincorporations using: A- 5' to 3' exonuclease of the newly synthesized strand B- 5' to 3' exonuclease of the nucleotides it polymerized C- 3' to 5' exonuclease of the newly synthesized strand D-3' to 5' exonuclease of the nucleotide it polymerized
D-3' to 5' exonuclease of the nucleotide it polymerized DNA pol I -Slower (adds 16-20 bp per sec) -Works shorter amount of time -Can exonuclease in 5' to 3' direction (needed because removes RNA when taking off promoter) DNA polII -Faster (adds 250-1000 bp per sec) -Can work longer Same: -polymerize 5' to 3' direction -Can exonuclease (remove nucleotides) in 3' to 5' direction
Which of the following does not describe transcriptional termination in eukaryotes? A-Rat1 degrades the RNA at the DNA-RNA hybrid B-RAT1 is an exonuclease C- The RNA affected by Rat1 is downstream of the genes coding sequence D-Rat1 works in a 3' to 5' direction
D-Rat1 works in a 3' to 5' direction Complex protein machinery binds and cuts at cleavage site and Rat1 attaches to 5' end. Rat1 is exonuclease that goes 5'->3' and degrades tail so nothing left for RNA polymerase II
Under what conditions would neither LacI nor CAP bind to the DNA of the lac operon A- Low lactose, low glucose B- low lactose, high glucose C-high lactose, low glucose D-high lactose, high glucose
D-high lactose, high glucose
Which direction is DNA synthesized
DNA is synthesized 5 ' to 3'. So that's the direction the DNA is going but in the replication fork the template DNA would then be 3' with the 5' and 5' with the 3'
In a karyotype, why do the stained chromosomes appear to have dark and light bands.
Darker bands mean more DNA in that area since more dense look darker. So lighter bands have less DNA in that area
A spontaneous deamination mutation occurs. What type of nucleotide mutation does this cause? What additional effect would you expect if the deamination occurred in a methylated CpG site?
Deamination is: Loss of an amine group (NH2) to double bonded O. When this happens on cytosine, the modified base resembles a uracil so when new complementary strand is made the DNA polymerase would pair this base to an adenine. If happens to adenine pairs with cytosine. - deamination is when amino group is now double bond to O so if odne on a C it makes it look like a U and pairs with A. This is substitution mutation .If this happened on methylated CpG site then there wouldn't be a Cpg island so no longer CpG site and no methylation would take place.
You are growing eukaryotic cells in a petri dish. You apply a chemical that inactivates Rat1. What does Rat1 usually do? How does the deactivation of Rat1 affect transcription in the cell
During transcription termination in eukaryotes: Complex protein machinery binds and cuts at cleavage site and Rat1 attaches to 5' end. Rat1 is exonuclease that goes 5'->3' and degrades tail so nothing left for RNA polymerase II. This would then terminate transcription For Prokaryotes: Rho dependent termination: -rho binds to rut signal-secondary structure (hairpin) slows down RNA polymerase-When RNA pol reaches hairpin the rho interacts and causes RNA polymerase to release and stop Rho independent termination: -Secondary structure causes RNA polymerase to slow down and then the adjacent A-U rich region not strong enough so breaks off from tension
EXAM 3
EXAM3
EXTRA CREDIT QUIZ
EXTRA CREDIT QUIZ
For Sanger Sequencing what pre-existing knowledge does a scientist need to have about the DNA sequence they are interest in sequencing before using the technology? Why is this necessary? How does this compare to illumina sequencing?
For sanger sequencing you would need locus specific primers. Illumina sequencing uses general primers. Sanger Sequencing: It works by making a regular deoxyribonucleoside into dideoxyribonucleoside so 3' carbon goes from OH to H so nothing can be attached to its end. Run this through a PCR with different bases of this so can get all the different stopping points. Then put through gel with different base in different wells. Reading from bottom up can figure out the complimentary strand sequence which is starting from 5'. So if want original make sure to give the complimentary of the complimentary and write other way so 5' to 3'. Illumina: You fragment the DNA sample into 100 bp in length and tether them onto the slides with universal primers. Add all the dNTPs and take photograph after each addition. Will glow certain color when nucleotide there. Remove the fluorescense so can take the next one.
What is the name of the repair pathway that: Repair double-strand breaks using a sister chromatid as a template
Homologous Recombination Repair
You are studying fur color in squirrels, coded by gene F. In one species, some squirrels have black fur and others have grey fur. How can you determine if grey is the dominant phenotype?Describe the experiment you would do, inlcuding the phenotype and genotype of squirrels in that cross, and the outcome you would expect if grey fur is dominant
I would do a test cross where I cross a true breeding black squirrel with a true breeding grey squirrel and if all the offsprings are grey, grey is dominant.
In environmentally conscious bacter, certain strucural genes encode the enzymes (PHAs) necessary to break down plastics. THis operon is not normally expressed but the presence of plastic results in expression of the operon The operon is under the regulation of a regulatory protein RCYCL. A nonfunctional RCYCL allele results in no expression of the PHA enzymes even when plastics are present. What type of regulation is shown in this operon in response to the signal of plastics? Explain how you know?
It is going to be positive inducible because the presence of the signal starts it, and the protein needs to be functional to activate expression. signal-plastic-present-start-inducible protein-RCYCL-nonfunctional-noexpression
What stage of mitosis or meiosis does the image below show if.. Image is cell with 6 chromosomes in X form lined down middle -the haploid number is 6 -the haploid number is 3
Meiosis metaphase II mitosis metaphase
What is the name of the repair pathway that: repairs thymine-dimers in humans
Nucleotide Excision Repair Scans DNA for bulky damage and removes segment of DNA, then polymerase will replace that section and ligase seal nicks. (works for a lot of different types of damage- thymine dimers)
How do you calculate degrees of freedom
Number of groups minus 1
What is the name of the repair pathway that: Harnesses engery from light to break bonds
Photo Reactivation Repair "Direct" repair of thymine dimers.Blue light is harnessed by PHOTOLYASE enzyme to break crosslinking bonds in thymine dimers. No nucleotides are removed.(not in mammals)
Complete the table below the represent whether each feature would be visible in typical eukaryotic gene's DNA coding strand, primary transcript, and mature mRNA DNA coding strand, primary transcript, mature mrNA start codon Intron Shine-dalgarno core promoter
Present, present, present Present , present, absent absent, absent, absent present, absent ,absent
You make a mutation in an E.coli cell that causes the cell's DNA polymerase I to lose it's 3' to 5' exonuclease ability. What effect does this have on the enzyme's activity? How will this affect the DNA synthesized in this cell
This means it can't go back to fix an error if one occurs. This would lead to more rror and or mutation in DNA synthesized in this cell.
Explain the process of how you would make a genetic map of genes R,Q and D.
So based off the data you can identify which are the parents since they are the highest number of results, and the 2 lowest are the 2 doubles. the rest are single. From there you take the parental code and do different double crosses switching up the one in the middle everytime so you can figure out which gene is in the middle. whichever one matches the double is the real one. from there you compare the parental to the singles to find which allele was the one switched for example was it r-d or d-q. Then based off what that is to find the length on the map you add up the 2 single with that same single crossing over with the 2 doubles and divide this by the total. Do the same of the other group so those with d-q for example, then those are the lengths. you multiple this by 10 to convert to cM and put them on the map
TEST 2
TEST2
Which Okazaki fragment would be made first
The one at the very end farthest from the replication fork because they have to be made in hunks since new pieces have to be put on 3' end. So since fork is moving to the left, when first opened, farthest right would be made first.
A patient at a fertility clinic is heterozygous for two inversion. The inversion on chromosome 9 is a 70 cM pericentric inversion. The second inversion on a chromosome 12 is a 5 cM paracentric inversion. Which of these inversions will have a greater negative effect on the patients fertility? Why?
The pericentric inversion would since it would cause duplication AND deletion when crossing over and has larger inversion region. Paracentric- causes dicentric bridge and deletion Pericentric- no dicentric bridge but duplication and deletion.
Imagine that another study measured only the recombination frequency between the two genes on the outside edge of your map from the previous question (skipping the middle). How would the observed data (in the table) be different.
There wouldn't be double recombination since there are only 2 genes.
The imagine on the rights shows a homologus pair of chromosomes observed in a single karyotype. The chromosome shown on the left is wildtype. Name the type of chromosomal rearrangment oberved in the chroosome on the right. Describe one way that this rearrangement could have occurred de novo. Image shows two chromosomes with this gene order: A B C centromere D E F G H A B F E D centromere C G H
This shows pericentric inversion since it includs the chromosme in the inversion. This could have occured de novo due to crossing over or Double stranded repair
A wild type mRNA sequence is shown below. A mutation occurs at the 10th position of the wildtype mRNA. What base(s) might you find at the 10th position as a result of a transition mutation? What base(s) might you find at the 10th position as a result of a transversion mutation Wildtype: 5'-AUG CUA CCA GAG UAU-3'
Transversion- T Transition- A
In a eukaryote, the binding of a transcription factor named tweedle to the regulatory promoters of two genes (Dee and Dum) results in increased transcription of those genes. How would a mutation that deleted the activation domain of the tweedle protein impact tweedle's activity? How would this mutation change expression of Dee and Dum?
Tweedle would bind DNA at each regulatory promoter but can't activate. Dee + Dum would have less expression than wildtype The deletion of the activation domain would decrease tweedle's activity significantly since it would no longer be able to interact with the basal transcription factor and attached on DNA. Due to tweedle being unable to bind Dee and dum's expression woul dbe decreased significantly and be very low.
Regulatory RNAs can regulate transcription or translation. Give an example of a type of reglatory RNA that affects transcription and briefly explain how it works to specifically regulate transcription. Then, give an example of a different type of regulatory RNA that affects translation and briefly explain how it works to regulate translation
Type of regulatory RNA that affects transcription would be siRNA because when siRNa is attached to RITS it goes to DNA and recruits methylation. THis methylation would inhibit transcription. Type of regulatory RNA that affects translation would be miRNA because when attached to RISC it does imperfect pairing on RNA which inhibits translation
A patient of yours is heterozygous for a chromosomal rearrangement that is unbalanced in his somatic cells. What does it mean that this chromosomal rearrangement is unbalanced? What chromosomal rearrangements are unbalanced in a heterozygote's somatic cells?
Unbalanced means that genomic information was lost or added so basically more or less genomic information than started with normal. Chromosomal rearragments that are unbalanced are duplication and deletion in somatic cells. Balanced ones would be: inversions, translocation
You are creating a transgenic drosophila where the gene EY is regulated by a conditional promoter under control of the LacI protein. You want the gene EY to be expressed during embryonic development but then not expressed in adults. What sequences would you need to add tot he drosophila genome to regulate EY expression with the lac I protein? How would you turn EY's expression on? How would you turn it off?
You would need a LacI operator. You would turn on EY expression by adding allolactose to diet and turn off expression by removing allolactose to the diet
An mRNA codon reads 5'-CUU-3'. What anticodon would pair to this codon perfectly? Be sure to label 5' to 3' ends on the anticodon. Circle the position of the anticodon where a wobble pairing might occur
We know in the ribosome it goes EPA with 5' being towards E and 3' being towards A. So the mRNA is 5'-CUU-#3' then the tRNA anticodon would be 3'-GAA-5' and the wobble position would be the A in the 3rd base pair position.
Say if if functional, nonfunctional, or no transcription when lacZ and lacY when lactose is absent and present for the following cases IsP+OcZ+Y-/I-P+O+Z-Y+
When lactose is absent: Lac Z- functional LacY-nonfunctional When lactose is present LacZ- functional LacY-non functional
Say if if functional, nonfunctional, or no transcription when lacZ and lacY when lactose is absent and present for the following cases I+P-O+Z-Y+/I-P+O+Z+Y-
When lactose is absent: Lac Z- no transcription LacY-no transcription When lactose is present LacZ-functional LacY-nonfunctional
How does tryptophan affect the binding activity of the TrpR protein? Through what mechanism does this change the binding of TrpR protein to DNA
When trp is present TrpR binds DNA and changes shape of DNA binding domain. So Tryptopan acts a sthe signal to activate the TrpR protein. Due to the activation of the inactive protein it is able to bind to the operator DNA. So without tryptophan the TrpR protein is inactive and ubale to bind to the operator. So once the troptophan binds to the TrpR protein it is then active and its binding domain is able to bind to the operator DNA.
A mutation occurs in a gene on the y chromosome of one individual of a species. This mutant gene acts as a segregation distorter (breaking mendel's law of segregation). How does the is affect meisosis in that at individual? How does it affect the phenotypes of their offspring?
Would kill the sperm without selfish gene
What does it mean to accept the null hypothesis
You can accept the null hypothesis as variation due to sampling
If looking at a karyotype of a newly discovered bird species. What would you circle if asked to circle a single haploid set of chromosomes. What would the haploid number be
circle on sister cromatid for each set and n would be the total number of chromatids/2
Nontemplate strand other names
coding or sense
Template strand other names
non-coding strand or antisense
What is the name of the repair pathway that: Works immediately after sunthesis when nick are still present in the newly synthesized strand.
mismatch repair Repairs soon after DNA replication to catch substitution errors. So although DNA polymerase usually fixes, this can come in after and fix soon after.Can recognize old and new strand to repair NEW strand using "MUT" protien.Bacteria- there is a chemical modification on the old strandEukaryotes- the presence of nick (before ligase fixes)So removes REGION of that strand then lets DNA polymerase 3 (in prokaryotes) go back and fill in the gap hopefully correctlyMutation in Mut genes associated with cancer risk
What does it mean to reject the null hypothesis
too different from expected more than sampling would explain
When do you accept or reject the null hypothesis
when your chi-squared value is greater than the chi-square critical value you reject the null. If your chi-squared value is less than the chi-squared critical value then you "fail to reject"/ accept the null.
