a) failure of activity of maintenance methylases following DNA replication

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1 GENETICS 603 Exam 2 October 31, 2014 In Class Exam NAME kinds of answers sought: many possibilities received credit 1. Data has suggested that the DNA in the maternal nucleus in a mouse egg is hypomethylated (low amounts of 5MC) by a passive mechanism, while the DNA of the male nucleus enters hypermethylated but is actively and rapidly demethylated before fusing to form a zygote. a) Why might it be important to demethylate DNA at this time? Development is likely to require activation of many previously silenced genes b) Propose mechanisms, including roles of enzymes (by function, if not actual name) that are likely to be involved in a) passive and b) active demethylation. a) failure of activity of maintenance methylases following DNA replication b) changing 5mC to T by dekamination or adding OH or =O etc. to the CH 3 (TET enzymes) which will cause mismatch repair to replace the modified 5MC with C 2. What does the yeast 2- hybrid protocol detect, and how does it work? (a diagram should help answer the question) It detects protein/protein interactions. The figure would show a bait protein fused to the DNA recognition component of the Gal system in yeast with a library of fish proteins fused to the detached activator domain, so that only when interactions occur would expression of the reporter gene be activated. 3. The human lactase gene promoter has 5 motifs known to interact with at least 3 transcription factors in the 150 bases just prior to the transcriptional start site. There are no CpG dinucleotide pairs in those 150 bp. The enzyme which is made only in intestinal wall cells requires glycosylation to function in its membrane bound form. A) Based just on this information list hypothetical means by which levels of lactase activity might be regulated. - Transcription via interacting TFs that are present/activeated in the intestine wall cells (but not affected by methylated DNA)? - Chromatin remodeling based on histone methylation, acetylation, etc? - Hormone signaling affecting presence of transcription factors - mrna stability & translation as affected by, microrna IXs, RNAi, rare codons, alternate splicing etc? - protein stability, ubitquination? - protein modification: transport to ER, glycosylation steps? Ie, all sorts of possible points of regulation of the amount/activity of the lactase enzyme

2 B) Via sequencing, it has been shown that individuals with a C at upstream position - 13,910 and a G at position - 22,018 do not have lactase activity as adults, but those with T and A, respectively do. Both enzyme assays and Northern blots are compatible with higher and adult- persistent levels of expression in the latter individuals. a) What is a Northern blot and what did it measure in this case? Amount of lactase mrna present b) What kind of a regulatory model fits best given this additional information? Transcription level: likely chromatin remodeling due to the distance from the promoter. 4. Use the circles below to show labeled line drawings of chromosomes #1 (long, metacentric) and #2 (short, acrocentric) for an organism at metaphase that is heterozygous for genes Q and R, which are on chromosomes 1 and 2, respectively. MITOSIS Daughter genotypes Qq, Rr R q Q r Qq, Rr MEIOSIS I Daughter genotypes: assume only 1 crossover between Q and its centromere Q r Q/q: rr q R q/q, RR

3 MEIOSIS II (either one of above daughters only) Daughter genotypes Q, r Q q r r q,r 5. The following are legends for inherited traits in Zebrafish: Bd /Bd normal development V V long tail (veil) Z _ Zebra stripes H_ normal colors Bd /Bd slow development V V medium tail zz, plain hh half black Bd/Bd lethal, not hatched VV short tail A female heterozygous for all 4 genes is crossed to a male that had developed slowly, had a short tail, zebra stripes and normal colors, although one of his parents was half black. Some of the progeny were plain. a. Give the genotype of the male Bd /Bd, VV, Zz, Hh b. How many genotypes would be found in gametes from the female 16 male 8? c. What fraction of the progeny from the cross would be Bd /Bd, V V, zz, Hh? 2/3 X ½ X ¼ X 1/2 d. What fraction of the progeny will have normal development, short tail, zebra stripes and be half black? 1/3 X ½ X 3/4 X 1/4 e. Two heterozygous zebra-striped parents are crossed. If 16 progeny are born give numerical formulas that predict: e1) the odds that 12 will be zebra striped and 4 plain? 16!/12! 4! X (3/4) 12 (1/4) 4 e2) the odds that at least one will be plain? 1- (3/4) 16

4 6. Testcrosses of A b/a B individuals to a b/a b are made. Genes A and B are only 0.2 cm apart. A very efficient screening procedure will identify all the a b/a b recombinants. If 2000 progeny are examined, how many a b/a b genotypes are expected. Give a simple formula that shows the probability that no double recessive recombinants will be found. The frequency of ab /ab progeny should be 1/1000 (half of the 2/1000 recombinants) Thus we would expect 2 ab/ab births in this case, so m=2. Usng the Poisson for m=2 and X=0 gives m 0 /0! e -2 or simply e In 1928, Sturtevant reported the following cross results from Drosophila. Males with the 3 sex linked genes, y (yellow body), ca (carmine eyes) and tb (tiny body were crossed to pure breeding wild type females. The female F1 progeny were then crossed to the y, ca, tb males with the following outcome ca y tb 29 y ca tb ca + tb 54 + ca tb + y y tb tb ca y + 45 y ca + ca ca + + y tb 8 y + tb Note that among all testcross progeny there were 309 ca + : 139 ca, 256 y + : 192 y, and 333 tb + : 107 : tb a) Do any of the traits show the expected test cross ratio? Expect 1 to 1 ratios, so testing the closest (y + vs y) gives a Chi Square value of with 1 df of 9.13 suggesting that all the mutant alleles are less viable than the + allele. b) Give the genotype of the F1 female, in correct gene order / y ca tb (since the 3 mutant alleles came in on the males X we know the exact c) Map the genes y 41.6 ca 18.2 tb d) Calculate the coefficient of coincidence and value for interference. C of C = 19/33.3 = 0.57 I = 0.43 e) Are you confident in the map? Explain why or why not. Probably not; if there is differential survival of some genotype combinations, the number of recombinants will be skewed

5 8. At least 3 genes (AGA1, 2 & 3) have been associated with male pattern baldness, including a gene on the X (recessive in females), one on chromosome 3 (dominant in males) and another on chromosome 20, also androgen sensitive so dominant in males. A woman is heterozygous for all three genes marries a man with no family history of pattern baldness. a) Give legends that explain the inheritance pattern for the three AGA genes AGA2 and AGA3 X B _ Hair X B Y Hair AGA /AGA Hair hair X b X b bald X b Y bald AGA /AGA hair bald AGA/AGA BALD bald b) Predict the ratio of male pattern baldness versus hair in her sons. X B X b, AGA2 AGA2, AGA3 AGA by X B X B, AGA2 /AGA2, AGA3 /AGA3 ½ X ½ X ½ = P(hair) so P bald = 7/8: ratio is 7 bald: 1hair 9. Very briefly, differentiate between the following: dioecious and monoecious plants dioeicious = 2 houses each sex on a different plant monoecious = both sexes on same plant (usually used when at different locations as in maize) X-inactivation and parental imprinting X inactivation is the random condensation of all X s >1 in animals that occurs in early embryology and where the same X remains inactive through mitosis. Imprinting is specific loci that are inactivated in each sex during gametogenesis so that only the allele from the other parent is express in the embryo (and beyond, at least in some tissues) Both involve DNA and histone methylation XXY phenotype & fertility in flies and humans Flies = fertile female Humans = sterile male (Klnefelters syndrome) 10. Predict the progeny phenotypic ratios, including sex, of the cross in chickens: Z B Z b X Z b W where allele B produces barred feathers rather than plain feathers. 1 barred female : 1 plain female : 1 barred male : 1 plain male TAKE HOME 1. You are asked to clone a transcription factor for a eukaryotic organism and have several bits of information that may be useful. Assume you have all the vectors, enzymes and an antibody or can devise another method for detection.

6 Information available includes a) data for close and a not so closely related organism that has the sequences shown below (dots show identical bases). TGGCGCCGGTGCCGTACCCCTTTGAGGGGGTGATAAGGGGGAGGAGGAGCGGTGGTAACG...A...A...A...CAGGT. b) purified TF seems to have about 300 amino acids with the following sequence determined from a fragment near the middle: thr asp phe trp met gln asn gly arg ser c) knowledge that the TF is relatively highly expressed in a specific tissue when stressed Design cloning strategies to take advantage of the 3 types of available information, separately or together, as a starting point, then outline how the process would be completed with the production of a verified clone. Mention all enzymes and other tools needed for each of the steps. How would the process differ if you just needed a fragment of the gene to use as a probe versus having a functional product.. Suggest other ways that a transcription factor might be screened than with an antibody. For expression it would be best to use cdna cloning, so I expected to see all the tools for doing so, such as Reverse TCase, RNAase H, oligo dt, a vector, etc. I expected one of two approaches to take advantage of the available information: 1) make a cdna library and then screen it using the related sequences as a probe, Western blots, etc. Pichia might be nice if there is no activity in E. coli and sequencing should show if it encodes the given aa sequence. 2) use the given sequences to specifically target the TF gene using 5 and 3 RACE (neither sequence was long enough for 2 primers) which could then be combined (useful to have overlapping primers in this case). Lots of room for other approaces too. Other ways to screen might include complementation if a known mutant is available, a modified yeast 2 hybrid screen, a gel shift assay for ability to bind DNA etc. 2. for simplicity, assume the following pigment pathway exists in a flower that can readily be self or cross- pollinated with the genes P and R encoding the enzymatic steps: P R white pink red A true breeding white is crossed to a true breeding pink and all the F1 flowers are red. When the red F1 was self pollinated the following results were obtained: F2 progeny: 822 red : 380 pink : 398 A) write genotypes for the truebreeding parents and F1. The cross would be PP, rr X pp, RR creating a Pp, Rr F1. Selfing would give a 9 Red : 3 Pink : 4 white ratio, or expected progeny of 900 : 300 : 400 in the 1600 under the hypothesis the genes are independent.

7 B) Test the hypothesis that genes P and R are independent. The Chi square value with 2 DF is over 28 so does not support the hypothesis. C) Test the hypothesis that the genes are linked at 20 cm. if the genes are linked at 20 map units the repulsion phase would give gametes in the frequencies of 0.1 PR 0.4 Pr : 0.4 pr : 0.1 pr resulting in expected progeny in the ratio of 8216 red : 384 pink : 400 white. In this case the X 2 value is 0.92 so support the hypothesis. 3. Suppose 1 in 100 persons are born with a homozygous defect (pp) such as PKU and the screening test is 99% accurate in identifying those individuals. However, the test also comes out positive 2% of the time in individuals who are PP or Pp, (ie, false positiives occur). What are the odds that a baby with a positive test is actually pp? P (true with a positive test) = 0.99 X 0.01 All probabilities of a positive = 0.99 X X 0.02 So there is a 1 in 3 chance that a positive test sample is actually affected