Review

1) You are handed a vial of homozygous mutant flies that have "curled wings". a) How would you determine whether the mutation that causes curled wings is dominant or recessive? mate m/m with +/+ flies and analyze the phenotype of the m/+ progeny b) What outcome would tell you that the mutation is dominant? m/+ progeny flies have curled wings c) Give two molecular/genetic reasons why a mutation can be dominant. m may be a loss of function mutation in a haplo-insufficient locus m may be a constitutively active, gain of function mutation, similar to those we studied in cancer and in vulva development 2) You are handed 2 vials of homozygous mutant flies that have "held-out wings". a) How would you determine whether the flies in the two vials have mutations in the same gene or in different genes? mate ho1/ho1 with ho2/ho2 flies and analyze the phenotype of the progeny b) What outcome would tell you that the mutations are in different genes? Write the phenotype and inferred genotype of the relevant flies. seeing the progeny have a wild-type phenotype would suggest that ho1 and ho2 are in different genes - they complement their genotype would be ho1 + / + ho2 if linked and ho1/+; ho2/+ if unlinked

3) A geneticist is interested in the genes responsible for synthesis of histidine (an amino acid) in

the haploid fungus Saccharomyces. He treats haploid cells with UV light to increase the mutation rate and obtains 7 haploid his-requiring mutants (a-g), all of which need histidine to be added to the medium in order for them to grow. (Without histidine, none of them will grow.) Experiment I: He first makes all possible diploid combinations of the mutations and tests them for their ability to grown on medium lacking histidine. He obtains the following results, where “+” indicates that the diploid cells grew, and “-“ indicates that the diploid cells did not grow. a b c d e f g a - + + + + - + b - + + + + + c - + + + + d - + + - e - + + f - + g - Based on the above results, how many genes are defined by the 7 mutations? five: a and f= gene 1 b = gene 2 c = gene 3 d and g = gene 4 e = gene 5 4) 4. Three enzymes (GT, DE, QS) convert colored intermediates to a final red product. Mutants for those enzymes accumulate the colored intermediates. From the data below, figure out the order of gene action in the pathway and then answer the questions below. GT mutants accumulate purple DE mutants accumulate yellow QS mutants accumulate green when GT and QS dihybrids are crossed with each other, they produce 9 red : 4 purple : 3 green when DE and QS dihybrids are crossed with each other, they produce 9 red : 4 green : 3 yellow Draw the biochemical pathway and show the order of gene action. GT QS DE purple green yellow red

5) There is a diploid cell with n=4 and the following genotype: Ab/aB; C/c; D/D; E/e.

a) Draw the cell in metaphase of mitosis. Indicate what genes are on what chromosomes. A A b b a a B B C C c c D D D D E E e e b) Draw the cell in metaphase of meiosis I. Ignore the effects of recombination. A A a a b b B B C C c c D D D D E E e e c) Draw the cell in metaphase of meiosis I if a crossover has occurred in the A-B interval. A A a a b B b B C C c c

D D D D E E e e d) Draw the cell from 5c after it has undergone meiosis I and is in metaphase of meiosis II. A A a a or b B b B C C c c D D D D E E e e

6) You are working with three loci in yeast:

♦ The argT gene is on the right arm of chromosome 3. Mutations make yeast auxotrophic for arginine. ♦ RFLP A/B is also on the right arm of chromosome 3 and maps 20 map units away from argT. Digestion of yeast genomic DNA with the restriction enzyme PstI and Southern blot analysis of digested DNA with a probe that recognizes the A/B region yields two different restriction fragment patterns. The A pattern is an 8 kb fragment. The B pattern is two fragments of 5kb + 3 kb. ♦ RFLP M/N is on chromosome 9. Restriction digestion of yeast genomic DNA with the enzyme PstI and Southern blot analysis of digested DNA with a probe that recognizes the M/N region yields two different restriction fragment patterns. The M pattern is a 19 kb fragment. The N pattern is a 12kb fragment. You mate yeast cells of the following genotypes: argT- A M x argT+ B N (where A, B, M, and N

indicate RFLPs). Draw out chromosomes 3 and 9 in the resulting diploid yeast cell. Draw them during metaphase of meiosis I and label the loci (argT-, argT+, A, B, M, and N). In this cell, show me the arrangement of markers not having been shuffled by recombination.

see next page You sporulate lots of these diploid yeast heterozygotes, spread the spores on plates lacking arginine, and

allow the arg prototrophs to grow into colonies. You prepare genomic DNA from 100 different colonies, digest the DNA samples with PstI, prepare Southern blots, and hybridize them simultaneously with probes for the A/B and M/N regions. Draw a Southern blot and show in different gel lanes all of the different banding patterns that you would observe (label the sizes of the bands). Below each gel lane indicate how many of the 100 colonies you would expect to show each banding pattern.

see next page

7) Two compounds, GDP-mannose and GDP-galactose, are thought to be intermediates in the

biosynthesis of vitamin C in higher plants. The order of the compounds in the biosynthetic pathway is unknown. You have isolated 2 "nv" (for "no vitamin C") mutants: - the nv1 mutant fails to synthesize vitamin C and instead accumulates GDP-mann - the nv2 mutant fails to synthesize vitamin C and instead accumulates GDP-gal Two kinases, ERK and MPK, act in a signaling pathway for formation of the vulva. Their order of action is unknown. You have isolated mutations in each gene: - erk(lf) mutants are Vul, erk(gf) mutants are Muv - mpk(lf) mutants are Vul, mpk(gf) mutants are Muv a) Briefly describe how you would use genetics to determine the likely order of gene action in each pathway.

Perform an epistasis test - generate double mutants using mutations that confer different phenotypes, and compare the double mutant phenotype to each single mutant.

So compare nv1/nv1; nv2/nv2 doubles to nv1/nv1 and to nv2/nv2 singles and compare erk(lf); mpk(gf) to each single or compare erk(gf); mpk(lf) to each single b) Give one possible result (genotype and phenotype) from such a genetic test of the nv mutants, and the pathway order you would infer from it (show the intermediates and final product of the pathway, and where each enzyme acts). The nv1/nv1; nv2/nv2 double mutant accumulates GDP-mann. This would suggest that NV1 is upstream of NV2 in the bioynthetic pathway. NV1 enz NV2 enz GDP-mann GDP-gal vitamin C Alternate answer: the nv1/nv1; nv2/nv2 double mutant accumulates GDP-gal. NV2 enz NV1 enz GDP-gal GDP-mann vitamin C c) Give one possible result (genotype and phenotype) from such a genetic test of the mpk and erk mutants, and the pathway order you would infer from it (show the order of kinases in the pathway, and the normal outcome of the pathway). The erk(lf); mpk(gf) double mutant is Muv, or erk(gf); mpk(lf) double mutant is Vul This would suggest that mpk is downstream of erk in the signaling pathway. ERK MPK vulva formation Alternate answer: erk(lf); mpk(gf) double mutant is Vul, or erk(gf); mpk(lf) double mutant is Muv This would suggest that erk is downstream of mpk in the signaling pathway. MPK ERK vulva formation

Recombination and Mutation

1. The R/r and S/s genes are linked and 10 map units apart. In the cross Rs/rS x rs/rs what fraction of the progeny will be RS/rs?

a. 5% b. 10% c. 25% d. 40% e. 45%

2. In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:

sn ct 13 sn ct+ 36 sn+ ct 39 sn+ ct+ 12

What is the map distance between sn and ct? a. 12 m.u. b. 13 m.u. c. 25 m.u. d. 50 m.u. e. 75 m.u.

3. The measured distance between genes D and E in a two point test cross is 50 map units. What does this mean in physical terms?

a. D and E are on different pairs of chromosomes. b. D and E are linked and exactly 50 map units apart. c. D and E are linked and at least 50 map units apart. d. either a or b e. either a or c

4. Genes Q and R are 20 map units apart. If a plant of genotype QR/qr is selfed, what percentage of the progeny will be qr in phenotype?

a. 4% b. 10% c. 16% d. 20% e. 40%

5. Crossing over takes place in bivalents consisting of ______ chromatids, and involves _______ of the chromatids.

a. 2, 2 b. 2, 4 c. 4, 2 d. 4, 4 e. 8, 4

6. The cross Lpq/lPQ x lpq/lpq is carried out, and the L gene is found to be in the middle.

What would be the genotypes of the double crossover gametes in this cross? a. LPQ and lpq b. LpQ and lPq c. lpQ and LPq d. Lpq and lPQ e. cannot be determined

7-8. Females heterozygous for the recessive second chromosome mutations pn, px and sp are mated to a male homozygous for all three mutations. The offspring are as follows:

px sp cn 1,410 px sp + 3,498 px + cn 1 px + + 11 + sp cn 8 + sp + 0 + + cn 3,483 + + + 1,489

9,900 7. What is the genotype of the females that gave rise to these progeny? sp- px- cn+ / sp+ px+ cn- 8. What are the distances among these three genes? px-sp 0.2%, px-cn 29.3%, sp-cn 29.5%, 9. 50 map units is the maximum measurable map distance between two genes in a two-point cross, yet some human chromosomes are over 200 map units in length. Explain this discrepancy. 50 map units is the most that can be measured by a 2-point cross. Determining distances by addition of 2-point cross values can give >50 map units 10. The term mutation refers to:

a. only changes in the DNA that result in new phenotypes. b. only changes in the DNA that result in novel proteins. c. any change in the DNA of a cell. d. a heritable change in the DNA of a cell. e. any change in the cell that changes its survival chances.

11. Indicate the statement that is most accurate regarding mutations.

a. Most mutations alter protein structure and phenotype. b. Only those mutations that alter protein structure will alter phenotype. c. Mutations altering a region that does not code for a protein may alter phenotype. d. All mutations that alter protein structure will alter phenotype. e. All altered phenotypes result from altered proteins.

12. A complementation group is:

a. a group of mutations that produce the same phenotype. b. a group of mutations that are in the same gene and complement each other. c. a group of mutations that are in the same gene and do not complement each other. d. a group of mutations in two different genes that complement each other. e. a group of mutations in two different genes that do not complement each other.

13. Mutations that abolish the function encoded by the wild-type allele are known as:

a. null mutations. b. hypomorphic mutations. c. hypermorphic mutations. d. conditional mutations. e. neomorphic mutations.

14. A neomorphic mutation results in an allele that:

a. produces no gene product. b. produces a nonfunctional gene product. c. produces novel proteins or cause inappropriate expression resulting in a new phenotype. d. produces proteins that aggregate with wild-type subunits, inactivating them. e. produces an altered protein that results in a wild-type phenotype.

15. A mutation that is characterized by a change in the DNA sequence, but no change in the resulting protein sequence, is called a:

a. frameshift mutation. b. missense mutation. c. silent mutation. d. nonsense mutation.

16. A mutation that changes a codon sequence, and subsequently changes the amino acid that should have been placed at that point in the polypeptide chain, is called a:

a. frameshift mutation. b. missense mutation. c. silent mutation. d. nonsense mutation.

17. A mutation that changes a codon that originally coded for an amino acid into a stop codon is called a:

a. frameshift mutation. b. missense mutation. c. silent mutation. d. nonsense mutation.

18. A mutation that occurs when one or more bases (but not a multiple of 3) is inserted into or deleted from a DNA sequence, completely altering an entire amino acid sequence, is called a:

a. frameshift mutation. b. missense mutation. c. silent mutation. d. nonsense mutation

19. A man whose grandfather on this mother's side had favism (a rare X-linked recessive sensitivity to broad beans) marries a woman whose uncle on her mother's side also had favism. a. Draw the pedigree as described. b. What is the probability that their 1st child will have favism? 1/16 c. If the 1st child does have favism, what is the probability that a 2nd child will have favism? 1/4