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MCB164 2005 Page 1 of 6
ID #________________________________ SECOND MIDTERM MAY 16, 2005 Question 1 2 3 4 5 6 7 8 Total v v v v Points 8 12 8 14 12 20 16 10 100 Score
This examination is closed book. There are 6 pages to the exam. Please count them before you start to make sure all are present. Please write your name on each page of the exam. Answer each question in the space provided. If additional space is required use the back of the page and indicate clearly that you continued your answer on the back. Do not attach additional pages. v GOOD LUCK!
Student authorization: I authorize the instructor to return the graded exam to me in the main office or to distribute it in the bin for me to pick up. Signature _____________________________ Date_____________________
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MCB164 Spring 2005 Page 2 of 6
1. (8 points) Mark the following statements as being either T (True) or F (False). ___T/F___ A yeast spore arising from a MI PSSC event (precocious separation of sisterchromatids) would contain one of each of the parentally-derived chromosomes. ___F___Cohesins form a ladder-like structure between sister chromatids within a bivalent while the synaptonemal complex forms a ring-like structure that encircle the bivalent. ___T___Fertilization of a mouse oocyte by a sperm occurs prior to the second meiotic chromosome division. ___T___ Cohesins linking the centromeric regions of chromosomes are released at anaphase II of meiosis.
2. (12 points) Fill in the blanks of the partially composed letter below using vocabulary words covered during this part of the course. Try to be as explicit as possible in your usage (i.e I vs. II) Dear Mom, First of all, thanks for the check I really appreciate it. I have been taking this really interesting course in genetics. You wouldn't believe what chromosomes can do. Did you know that during meiosis more than 200 DOUBLE-STRAND BREAKS are introduced into our chromosomes before they go on to make gametes? These are repaired by a mechanism called homologous recombination. Usually a NON-SISTER chromatid is used as a substrate for repair. Sometimes there is a nonreciprocal exchange of genetic information, which is also known as GENE CONVERSION. These events are often associated with CROSSING OVER (OR CROSSOVERS), but not always. Anyway, all this is important for holding HOMOLOGOUS chromosomes together prior to the MEIOSIS I division. In human females this can last for 40 years. By the way, do you know what kinds of stuff grandma was into when she was pregnant with you? Just wondering...... Love, your F1 p.s. next month can you send $74 so I can get a subscription to Nature?
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3. (8 points) In Saccharomyces cerevisiae, you make a cross involving four linked markers, A B C D X a b c d, in which markers B and C are extremely close together on the DNA. A. Among 200 asci, you find one containing spores of the type A b c D, a b c d, A B C D, a b c d: How can this ascus be explained by a single meiotic double-strand break repair event? What does the type of tetrad with respect to A, a and D, d allele pairs indicate about the outcome of the event (i.e. crossover or noncrossover)? The b and c loci have undergone a gene conversion event. Since the A and D markers have segregated as a PD tetrad there was no apparent crossover associated with this gene conversion event. B. What would be the explanation for an ascus containing A BC d, a b c D, A B C D, and a bcd? There was a cross over between the C locus and the D locus with no gene conversion at BC.
4. (14 points) Match the two metaphase I bivalents being pulled by the spindle to the possible crossover types listed below and shade the areas distinguishing replicated sister chromatid strands from nonsister chromatid strands. The centromere is marked with a circle. ___B__ centromere proximal crossover _____ centromere proximal double-crossover ______ telomere distal double-crossover __A___telomere proximal crossover
A
B
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5. (12 points) The cascade of regulatory interactions involved in sex determination in the nematode C. elegans is illustrated below for wild type genes and their products (bars indicate negative regulation by genes immediately upstream in the pathway). The levels of predicted gene activity in wild-type animals of either XX or XO genotypes are given based on epistasis analysis of various double mutant phenotypes. X/A XX XO her-1 low high -----| tra-2 high low -----| fem-1 low high -----| tra-1 low
high
hermaphrodite male
Using this information, predict the phenotypic sex for both XX and XO genotypes for the following double mutant combinations. Assume recessive (rec.) alleles are loss-of function and dominant (dom.) alleles are gain of function. A. her-1 (dom.) tra-2 (rec.) XX XO MALE MALE
B. fem-1(rec.) tra-1 (rec.) XX XO MALE MALE
C. tra-2 (dom.) fem-1(rec.) XX XO HERMAPHRODITE HERMAPHRODITE
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6. (20 points) You are describing how gene conversion occurs in the context of the double-strand break repair model to your little sister. You explain how you can cross two strains of Neurospora crassa, a filamentous fungus, to form a heterozygote AB/ab and infer molecular details from the ratios of the segregating A:a and B:b phenotypes. A. Diagram for her a new molecule where a double strand break has been introduced on the recipient chromatid and to what extent it would be resected based on the intermediate shown below. (assume all the markers line up vertically) B. Given the donor chromosome (thin line with a and b alleles) show her where A/a and B/b information is on the double-Holliday junction intermediate before any mismatch correction has occured. 3' 5'
A A
B B
B A B
a a a a 5' 3' a A
+
b b B b B b
3' C. Correct any mismatches on this double-Holliday junction intermediate and and report the ratio for A:a and B:b for the resultant octad. 5' 5' 3'
a a a a
b b b b
As shown, assuming no gene conversion on the sister chromatids, both the A:a and the B:b ratios would be 2:6. (If you converted the A/a heteroduplex to A/A then you're A:a ratio would be 4:4. If you converted both B/b heteroduplex to B/B the ratio would be 6:2 and if you converted them oppositely, i.e. B/B and b/b, then the ratio would be 4:4 (but not aberrant 4:4). D. Now go do something fun together
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7. (16 points) In Drosophila, two X chromosomes can be attached to the same centromere to form an attached-X chromosome. Shown below one X chromosome bearing the markers a+ and b+ is fused to an X chromosome bearing markers a and b. The two chromosomes undergo crossing over and then sister chromatids undergo subsequent disjunction. Suppose that you have an attached X chromosome of the following genotype: a+ b+ b
a
Diagram the products of a single crossover between a and b that would definitively show that crossing over occurs at the four strand stage of meiosis I. What feature of the product(s) (i.e. genotype inferred by phenotype) allows you to make this conclusion? a a+ b+ b b a
b+ a+
The products from a single meiotic crossover between loci a and b are shown above. The new attached X chromosomes in one of the gametes are homozygous at the a locus. Resultant progeny containing this chromosome will be homozygous for the a allele and will show a new mutant phenotype. If the crossover had occurred at the two-strand stage (i.e. before replication of sister chromatids), however, all products from this crossover would retain heterozygosity at both loci and no mutant phenotypes would be observed. Many people disjoined the attached X chromosomes into 4 meiotic products, however the attached X chromosomes stay intact throughout the second meiotic division. Also many people stated that the products formed a tetratype. Not all meiotic products are visible in flies so you do not know if a tetratype exists. Homozygosity at the a locus is the only visible evidence that the crossover occurred at the 4 strand stage. 8. (10 points) List something that appears to be in common to all centromeres and how the centromeres of mammals, plants, Drosophila and even S. pombe differ from budding yeast. Many answers were possible for this question. Features common to all centromeres include the following: confer genomic stability, bind microtubules/kinetochore proteins, always segregate at meiosis I, contain a variant histone H3, do not undergo recombination, are AT rich and gene poor. Budding yeast centromeres differ from the centromeres of mammals, plants, Drosophila, and S. pombe mainly because they are very small, 125 bp, and contain a specific DNA sequence
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(including the CDE I-III core elements). Other species have much larger centromeres without conserved sequences. C. elegans , not budding yeast, contain holocentric centromeres. Also, the statement that "a centromere is a centromere because it was a centromere" refers to centromeres that are defined by epigenetic marks and not specific sequences.