Assignment Problems: Cumulative 0350 Genetics Fall 2020 1

0350 Genetics Fall 2020 Assignment Problems: Cumulative (1) Analysis of vulval development in C. elegans has been important in understanding some key signaling pathways, some that play roles in human cancer. The vulva is the organ through which fertilized eggs leave the mother. Wild-type worms only have one vulva, but mutants have been identified that either have more than one vulva or no vulva. These include mutations in the mpk-1, lin-1, lin-39, let-23 and lin-3 genes. Below is a table that lists single or double mutants and their phenotype.

Mutant (all are homozygous mutants in the indicated gene(s))

phenotype

mpk-1 no vulva lin-1 multiple vulvas lin-3 no vulva lin-39 no vulva let-23 no vulva mpk-1 lin-1 multiple vulvas lin-3 lin-1 multiple vulvas lin-39 lin-1 no vulva let-23 lin-1 multiple vulvas

1 (1a) Who originally chose C. elegans as a model system to study genetics A: Sydney Brenner B: Francis Crick C: Seymour Benzer D: Barbara McCintock 2 (1b) What other major contribution did he make to genetics? A: Uncovered the secrets of the lac operon B: Used T4 phage rII gene frameshift mutants to show that the Genetic Code is composed of triplet codons with no commas C: Used recombination between thousands of T4 phage rII gene mutants to demonstrate that a gene is likely just a linear sequence of nucleotides D: Identified the first gene linked to a chromosome, the white gene of Drosophila

Assignment Problems: Cumulative 0350 Genetics Fall 2020 2

3 (1c) Which of the pathways below is consistent with the data (if you cannot position some genes unambiguously then they must be placed at the same position in the pathway).

4(1d) Which of the following genes is most likely to encode for a transcription factor? A: lin-3 B: let-23 C: lin-39 D: mpk-1 The following table shows where wild-type protein encoded by the genes is expressed during vulval development.

Gene Cells in which the wild-type gene is expressed

mpk-1 Vulva and cells outside lin-1 Cells outside of the vulva only lin-3 Cells outside of the vulva only lin-39 Vulva and cells outside let-23 Vulva and cells outside

5 (1e) Which gene is most likely to encode for a secreted signaling protein? A: lin-3 B: let-23 C: lin-39 D: lin-1 6 (1f) Which gene is most likely to encode for a receptor for the signaling protein you identified in (1e)? A: lin-3 B: let-23 C: lin-39 D: lin-1

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(2) A pure breeding population of reindeers at the North Pole have red noses. Those in Finland all have black noses while those in Russia all have tan noses. To investigate the genetics of this trait, crosses were done between these pure breeding populations; the most revealing of these was a cross between the Finnish and Russian with the F1 progeny all being black, but a cross between these yielding the following: Black: 89 Tan: 41 Red: 30 7 (2a) Explain these results A: Single gene, two alleles, one incompletely dominant over the other B: Single gene, two alleles, each codominant C: Single gene, three alleles, allelic series D: Two genes, each with two alleles, one completely dominant over the other, and one gene recessively epistatic to the other 8(2b) Would all the tan offspring be pure-breeding like the ones from Russia? A: Yes B: No 9 (2c) What color would the offspring be from a cross between reindeer from the North Pole and Russia A: red B: tan C: black D: pink 10 (2d) Reindeers at the North Pole can fly while those from anywhere else cannot; the ability to fly is controlled by a single gene. A cross between reindeers from the North Pole and Finland produced F1 progeny could not fly and had black noses; these F1s were backcrossed reindeers from the North Pole resulted in following F2 progeny: Black flightless 61 Red flier 59 Black flier 39 Red flightless 41 Is the gene controlling black or red noses in Finnish/North Pole, likely to be linked to that controlling the ability to fly? (Table on last page if needed) A: Yes B: No

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(3) A new mutation that results in eight-legged flies was isolated and was found to be recessive, the gene name was designated extra-legs, el, and the wild-type allele, el+ and the mutant, elR. The mutation was mapped approximately to a region on the 2nd chromosome that included the following known mutant markers (the mutant marker alleles are all recessive and their map position on the chromosome is indicated in the table below).

Gene Wild-type

allele (dominant)

Wild-type phenotype

Mutant marker allele

(recessive) Mutant

phenotype Map position in map units

wingless wg+ Full sized wing wgT No wing 25

orange or+ Red eyes or5 Orange eyes 37

short bristles sb+ Long bristles sbS Short bristles 52 Three crosses were set up: (i) wgT or5 elR/ wgT or5 elR x wg+ or+ el+/ wg+ or+ el+ (ii) or5 sbS elR/ or5 sbS elR x wg+ or+ el+ / wg+ or+ el+ (iii) wgT sbS elR/ wgT sbS elR x wg+ or+ el+/ wg+ or+ el+ The F1 progeny from these crosses were then crossed separately to: wgT or5 sbS elR/ wgT or5 sbS elR (i) F1 heterozygous parent: wgT or5 elR/ wg+ or+ el+ Full wing, red eyes, 6 legs 884 Full wing, orange eyes, 6 legs 46 No wing, red eyes, 6 legs 66 No wing, orange eyes, 6 legs 4 Full wing, red eyes, 8 legs 2 Full wing, orange eyes, 8 legs 23 No wing, red eyes, 8 legs 29 No wing, orange eyes, 8 legs 455 95/1509=0.629 (ii) F1 heterozygous parent: or5 sbS elR/ wg+ or+ el+ Red eyes, long bristles, 6 legs 872 Red eyes, short bristles, 6 legs 70 Orange eyes, long bristles, 6 legs 8 Orange eyes, short bristles, 6 legs 50 Red eyes, long bristles, 8 legs 26 Red eyes, short bristles, 8 legs 5 Orange eyes, long bristles, 8 legs 39 Orange eyes, short bristles, 8 legs 433 (iii) F1 heterozygous parent: wgT sbS elR/ wg+ or+ el+ Full wing, long bristles, 6 legs 736 Full wing, short bristles, 6 legs 200 No wing, long bristles, 6 legs 61 No wing, short bristles, 6 legs 14 Full wing, long bristles, 8 legs 8

Assignment Problems: Cumulative 0350 Genetics Fall 2020 5

Full wing, short bristles, 8 legs 23 No wing, long bristles, 8 legs 103 No wing, short bristles, 8 legs 322 11 (3a) Between which two marker genes is the el gene located A. wg and or B. wg and sb C. or and sb 12 (3b) What is the distance between wg and el A. 7 m.u B. 9 m.u C. 13 m.u. D. 16 m.u. 13 (3c) Is there anything unusual about the numbers of any of the classes of phenotype in the three crosses? A There are a lot more parentals that would be expected B There are a lot fewer double recombinants that would be expected C Both classes of parentals should be approximately the same but are not D The total number of recombinants is too high compared to the parentals

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(4) If you were to generate mutations for the his gene in Salmonella gene with X-rays and with the chemicals EMS and profavine (i.e. separately) and then sequenced the gene, what might you expect to find in each case? 14 (4a) X-rays A: No sequence because the whole gene is deleted B: Point mutations C: Small insertions and deletions D: Several pyrimidine dimers 15 (4b) EMS A: No sequence because the whole gene is deleted B: Point mutations C: Small insertions and deletions D: Several pyrimidine dimers 16 (4c) Proflavine A: No sequence because the whole gene is deleted B: Point mutations C: Small insertions and deletions D: Several pyrimidine dimers 17 (4d) Which of these mutations might be useful to use in the Ames test? A: All three B: X-ray only C: EMS only D: EMS and proflavine 18 (4e) As well as a mutation in the his gene, the strain of Salmonella used in the Ames test also usually carries a mutation in what other gene? A: Antibiotic resistance B: uvrA or uvrB C: DNA polI D: RND efflux pump

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(5) Different temperature sensitive mutants of E. coli and yeast in genes encoding proteins involved in transcription were grown at the permissive temperature for several generations and while still actively growing the temperature was raised to the restrictive temperature. After raising temperature, the synthesis of new RNA was assessed using radioactively labeled nucleotides. Assume the mutations completely inactivate the proteins at the restrictive temperature and that function of the mutant protein was abolished instantly temperature was raised. What would happen to the level of new RNA synthesized (i.e. radioactive incorporation into RNA) in the following mutants after raising the temperature?) A Transcription would stop instantly B Some RNA would still be synthesized but will be reduced and eventually stop C Levels would probably increase D No effect on the level of transcription 19 (5a) Sigma factor (E. coli) 20 (5b) Subunit ß of RNA polymerase (E. coli) 21 (5c) RNA Pol II (yeast) 22 (5d) Rho protein (E. coli)

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(6) Hedgehog (Hh) is a secreted signaling protein that activates the Hh signaling pathway in responding cells resulting in the activation of Hh-target genes that can control cell activities such as proliferation. The pathway is inappropriately activated in some cancers. Key proteins in the Hh pathway are as follows: Hedgehog (Hh): secreted signaling protein found outside cells Patched (Ptc): transmembrane receptor, externally portions of the protein bind Hh, other regions bind to Smo. Note, Ptc is the Hh receptor, but unlike most receptors its activity is inactivated by its ligand, Hh. Smoothened (Smo): transmembrane protein that positively regulates the Hh pathway by binding and inactivating Cos2. Smo activity is inhibited by Ptc. Costal-2 (Cos2): cytoplasmic protein that has 3 separate binding domains. It binds to Smo and that blocks its activity. It binds to GSK/PKA and that forms an active complex. It binds to Gli when the latter active complex forms and that results the processing of Gli to the repressor form. GSK3 and PKA: kinases that phosphorylate Gli to result in its partial degradation Glioblastoma (Gli) Transcription factor that can act as a repressor or activator of transcription. In the absence of Hh, it is partially degraded to a repressor form. In the presence of Hh this does not happen and the full length form is an activator. Absence of Hh: Ptc inactivates Smo by binding to it and preventing it interacting with Cos2, which then binds to Gli, and GSK3 and PKA. The latter 2 phosphorylate Gli and this targets it for partial degradation. The smaller from of Gli enters the nucleus, binds to Hh-target genes and represses their expression. Presence of Hh: Hh binds to its receptor which is Ptc and this prevents Ptc from inactivating Smo, which can now bind Cos2 and preventing GSK3 and PKA from phosphorylating Gli. Gli then remains full length and can enter the nucleus and bind to and activate Hh-target genes. (N.B. some liberties have been taken with pathway to simply the question so don’t take it all completely literally)

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23 (6a) Which of the simple genetic pathways below is correct based on the information above.

The table below lists mutations in different regions of proteins in Ptc, Smo and Cos2 in the Hh pathway. (Determine whether the pathway will be on or off in each mutant, i.e. is the gene G is expressed in a homozygous mutant. (c) Determine if the mutation will be a simple loss of function (SLF), possibly a dominant negative (DN) or probably a gain of function (GOF). (0.33 points each) To work out answers to (c) you need to first determine the loss of function phenotype for the gene in question (one mutation of each is obviously a SLF), then determine if each of the other mutations in the same gene would result in the same phenotype or not – if the same then the mutation is either SLF or DN, if the opposite then it would be a GOF. To work out if it is SLF or DN you have to consider what would happen in a heterozygote: can the mutant protein interfere with wild-type protein in some way? Could it block the wild-type protein or would it compete with the wild-type for a protein the wild-type needs to interact with? Note, a mutation in one part of a protein could be DN, while mutation in a different part could be GOF. Hint…only one of the mutants is predicted to act as a DN

Protein Mutation (region inactivated) (b) Is pathway always on or off in homozygous mutant (i.e. is gene G expressed)

(c) Is the mutation SLF, possibly DN or probably GOF?

Ptc Deletion of gene 24 6b 32 6j

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Smo Deletion of gene 25 6c 33 6k

Cos2 Deletion of gene 26 6d 34 6l

Ptc Hh binding domain 27 6e 35 6m

Ptc Smo binding domain 28 6f 36 6n

Smo Ptc binding domain 29 6g 37 6o

Cos2 Smo binding domain 30 6h 38 6p

Cos2 Gli binding domain 31 6i 39 6q

24-32 (6b- 6i) Determine whether the pathway will be on or off in each mutant, i.e. is the gene G is expressed in a homozygous mutant. A = On B = Off 33-41 (6j- 6q) Determine if the mutation will be a simple loss of function (SLF), possibly a dominant negative (DN) or probably a gain of function (GOF). A = SLF B = DN C = GOF To work out answers you need to first determine the loss of function phenotype for the gene in question (one mutation of each is obviously a SLF), then determine if each of the other mutations in the same gene would result in the same phenotype or not – if the same then the mutation is either SLF or DN, if the opposite then it would be a GOF. Remember, a GOF mutation is only in relation to the function of the protein itself, not as to whether the pathway is activated by the mutation. So for a gene that functions to inhibit the pathway a SLF mutation will result in the pathway being on all the time, but a GOF would result in the pathway being off all the time. To work out if it is SLF or DN you have to consider what would happen in a heterozygote: can the mutant protein interfere with wild-type protein in some way? Could it block the wild-type protein or would it compete with the wild-type for a protein the wild-type needs to interact with? Note, a mutation in one part of a protein could result in a DN mutation, while mutation in a different part could be GOF. Hint…only one of the mutants is predicted to act as a DN 40 (6r) Based on what you know about other signaling pathways, why might a GSK3 mutant have more complicated phenotypes than would be predicted based on its role in Hh signaling? (2 points) A. It is redundant

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B. It is also involved another pathway – in the Wnt pathway – so loss of GSK3 would affect both pathways C. It is haploinsufficient D. Mutations are always dominant negative