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MCDB 101B Final Exam March 21, 2014

DO NOT OPEN THIS EXAM UNTIL YOU ARE INSTRUCTED TO DO SO

This exam should have 13 pages, including a blank page at the end for scratch. Please count them now and put your name at the top of each page. Answer the questions in the spaces provided. •If you make an error, CROSS OUT the mistake. Exams that contain erasures will not be

considered for regrading.

Q1 (15)

Q2 (20)

Q3-5 (28)

Q6 (24)

Q7 (26)

Q8 (16)

Q9 (10)

Q10 (15)

Q11 (23)

Q12 (25)

Total (200)

To receive full credit, all answers MUST include your reasoning

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1). (15 points) Gene B encodes a protein involved in bone formation. It has an allele, Bb that is dominant for brittle bones but is also recessive lethal. Gene M encodes a protein necessary for uptake of calcium from milk. It has a recessive allele, m, that when homozygous has reduced calcium uptake, which also results in brittle bones. The combination of Bb and homozygous mm is lethal, and mice die before birth. The two genes are located on different autosomes.

You do the following cross:

B+ Bb MM X B+ B+mm a). What is the expected frequency of offspring with brittle bones?

b). If you cross brittle-boned males and females, what phenotypes and their frequencies do you expect in the F2?

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2). (20 points) You have a female fly heterozygous for a variety of recessive alleles and also heterozygous for an inversion (the inversion breakpoints are indicated by the arrows:

a). Diagram the pairing configuration seen at pachytene of meiosis in this fly. Label the loci. The diagram must be clear and legible for credit.

b). The b and w genes in wild-type flies are 4.8 map units apart. If the female heterozygote from part “a” was test-crossed (to a strain homozygous for the non-inversion top chromosome above), what would you expect the measured map distance between b and w to be, and why?

c). The p and c genes in wild-type flies are 7 map units apart. If the female heterozygote from part “a” was test-crossed (to a strain homozygous for the non-inversion top chromosome above), what would you expect the measured map distance between p and c to be, and why?

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3). (8 points) Entry into S phase of the cell cycle is controlled by the kinase CDK2. CDK2 is present throughout the cell cycle; what mechanism ensures that it is only active at the transition to S phase?

4). (10 points) Double-stranded DNA breaks, either blunt-ended or with 5’ or 3’ overhangs, cause the activation of DNA damage response pathways. Yet, chromosome ends do not activate these pathways. What mechanism is responsible for this (include a diagram)?

5). (10 points) Researchers believed that microsatellite locus D5S875 is linked to a dominant disease,

and genotyped 5 different pedigrees of families with the disease. The LOD scores for linkage at a map distance of 3 map units for each pedigree are as shown:

Family 1: LOD score = 1.1 Family 2: LOD score = — 0.11 Family 2: LOD score = 0.8 Family 4: LOD score = — 0.2 Family 5: LOD score = 1.3

From these results, what would you conclude about whether D5S875 is linked to the disease. You

must include your reasoning.

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6). (25 points)Brief answers (3 sentences max): a). If a histone acetylase targeted nucleosomes in the promoter region of a gene, what would be it’s

effect on the rate of transcription?

b). Diagram and briefly explain the “torpedo” model of transcription termination

c). DNAse I hypersensitive sites are often seen near the promoter region of actively transcribed genes,

and are not seen near the promoters of inactive genes. What is your interpretation of this (use a diagram to help explain).

d). Phosphorylation of the carboxy-terminal domain of RNA pol II plays at least two roles in transcription. Briefly describe these 2 roles.

e). Diagram the “joint molecule” intermediate in the process of recombination.

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7). (18 points) Microsatellite locus CBGB45 has four alleles in the population, giving PCR bands of 80, 90, 110, and 125 base pairs; The frequency of each allele in the population is 25%.

You wish to test whether a rare recessive liver disease is linked to this locus, but there are too few informative pedigrees available. Instead, you hypothesize that a single mutagenic event was the source of all disease alleles and use haplotype association to test for linkage. You isolate and pool together DNA from a large number of healthy people, and a number of people affected with the liver disease, and PCR each pooled sample for CBGB45.

a). Below, diagram what you would expect the gel to look like if there is no linkage

disequilibrium or if there is high linkage disequilibrium.

b). If your results looked like the left-hand results (no linkage disequilibrium), what would you conclude?

c). If your results looked like the right side results (high linkage disequilibrium), what would you

conclude?

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8). (23 points) A dominant autosomal trait has been localized to a region of chromosome 5. Eight microsatellite loci are in this region as shown at right. The disease locus is somewhere between markers D507 and D2725.

To localize the disease gene, you have genotyped an affected family as shown:

a). In the boxes to the right of each individual’s genotype, put in their two haplotypes b). What is the disease haplotype in generation I?

c). Identify any individuals that received at least one recombinant gamete. For each gamete, what is your best estimate of where the recombination event took place?

d). What is your estimate of where the disease locus could lie?

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9). (16 points) The pedigree below shows a family affected by recessive cystic fibrosis (cf). PCR was used to test a microsatellite locus known to be 4 m.u. from the cf gene. Assume outsiders marrying into the family are not carriers of the disease unless the genetics indicate otherwise. In generation IV, none of the first 4 children have the disease; IV-5 is in utero and its disease status is unknown, but could be genotyped.

a) What are the possible genotypes and their probabilities of III-2 including phase relationships?

b) What is the probability that IV-5 (an unborn child) will develop cystic fibrosis?

c) What is the probability that IV-1, who is healthy, is a carrier for cystic fibrosis?

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10). (17 points) Shown below is a recombinant clone of genomic DNA for the mouse SCUD gene, whose function is unknown. The intron-exon structure of the gene is shown.

Researchers generated a targeted knock-out for this gene, but mice homozygous for the knock-out

die early in development, so no information was gleaned. It is known that the gene is expressed both very early in development, and later is expressed in the embryonic retina.

a). You have access to a mouse strain that express Cre protein only in the retina. Diagram the

targeting vector you would create as the first step to generating an eye-specific conditional knockout of the SCUD gene. Clearly label all components of the targeting vector in your diagram.

b). After eventually getting mice homozygous for your desired engineered SCUD region, diagram what the region will look like, including any remaining portions of the SCUD gene.

c). Diagram what the above region would look like AFTER exposure to Cre.

5’ 3’AAAAAAAAA

AUG codon exon intron stop codon

NotI EcoRI BamHIEcoRI HindIII

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11). (26 points) Below is the structure of a mammalian gene that has 2 alternative promoters and alternative splicing. The sizes of the mRNAs and their encoded proteins are shown on the right. In wild-type cells there are equal levels of each mRNA and each protein. The open boxes represent the open reading frame, and the horizontal lines the 5’ and 3’ untranslated regions. Also marked are the sites of 4 nonsense mutations resulting in a change from an amino acid codon to a stop codon (marked “a”, “b”, “c”, and “d”).

In a cell homozygous for each mutation, how much of each mRNA and protein would you expect to see, relative to a wild type cell. If different from wild type, briefly justify your answer. If any expected size would change, note this and explain.

b). Cell homozygous for mutation “b”

c). Cell homozygous for mutation “c”

d). Cell homozygous for mutation “d”

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12). (12 points) You cross a pure-breeding strain showing two recessive traits, wilco and gaga, to a pure-breeding wild-type strain. All progeny are wild-type. You then test-cross the F1 back to the parental strain with both traits.

wilco, gaga 108

wild-type 96

wilco 80

gaga 72

total 356

Devise a genetic model to explain these results, and test your model using Chi Square. Clearly state your hypothesis, your conclusion, and show all work.

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Critical Values of the Chi-Square distribution

Probabilities d.f 0.9 0.5 0.1 0.05 0.01 0.001 1 0.02 0.46 2.71 3.84 6.64 10.83 2 0.21 1.39 4.61 5.99 9.21 13.82 3 0.58 2.37 6.25 7.82 11.35 16.27 4 1.06 3.36 7.78 9.49 13.28 18.47 5 1.61 4.35 9.24 11.07 15.09 20.52 6 2.20 5.35 10.65 12.59 16.81 22.46 7 2.83 6.35 12.02 14.07 18.48 24.32 8 3.49 7.34 13.36 15.51 20.09 26.13

χ2 = Σ (observed –expected)2/expected

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