BIOLOGY 321 SPRING2013 ANSWERS TO ASSIGNMENT SET #4 Answers to Edition 9 text questions:

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1 BIOLOGY 321 SPRING2013 ANSWERS TO ASSIGNMENT SET #4 Answers to Edition 9 text questions: Chapter 2 NOTE ON 38: substitute dominant for mutant in part a of the question. 38 a. This would be an example of a haploinsufficient gene since one copy of the wild-type allele does NOT produce enough protein for normal function. You do not expect the mutation to be recessive Chapter 6 Comment on Problem 34: avoid using upper and lower case allele symbols in situations of incomplete dominance where neither allele is dominant over the other. It would be better to use a (+) to designate the wild-type allele -- such as f + = wild-type and f= wooly: f + f + = wildtype f + f = frizzled and ff = wooly. 1

2 Chapter 6 Problem 62 (10th edition): HOW to do a systematic analysis of the data First summarize info in problem: Parental: Disc X Long (both truebreeding) à F1 Disk F1 Disk self à Disk 270/480 Sphere 178/480 Long 32/480 How to explain inheritance pattern: First start with the simplest explanations: One gene, two alleles, complete dominance F1 consistent F2 doesn t work because of the sphere phenotype popping up One gene, two alleles, incomplete dominance F1 inconsistent -- we would have seen sphere here Therefore, must be two (or more) genes involved Next, consider two genes, two alleles (A,a & B,b) each with complete dominance and see if you can find a Mendelian 9:3:3:1 ratio (or some modification) with this data How to transform the raw data so that you can work with it easily: Some trial and error works here: 1. Divide 32 into each of the other numbers: Disk 8.4 Sphere 5.6 Long 1 Hmmm not so enlightening but sort of looks like 9:6:1 2. Another way: 9/16 X 480 = 270 = DISK 6/16 X 480 = 180 = close to 178 Sphere 1/16 X480 = 30 = close to 32 long So ratios are modified mendel : 9/16 disk A-B- 6/16 sphere (aab- & A-bb) 1/15 long aabb So at least one dominant alleles for both genes = Disk One dominant allele for either but not both = Sphere Homo recessive for both genes = long 2

3 Answer continues on the next page Can we make some biological sense of the genetic phenomenology? 9/16 disk A-B- 6/16 sphere (aab- & A-bb) 1/15 long aabb So at least one dominant allele for both genes = Disk One dominant allele for either but not both = Sphere homo recessive for both genes = long One possibility is that functional alleles (A & B) of both the A gene and the B genes promote outgrowth of the fruit in an axis perpendicular to the stem-blossom axis. If there are two doses (A-B-), there is more expansion (fruit is wide but not long) along this axis than if there is one dose (either aab- or A-bb). If the plant has only loss-of-function alleles (aabb) for these genes, there is little or no expansion along this axis (fruit is long but not wide). Answers to Additional Study Problems Ì Problem 1 ABO blood group: A and B are codominant alleles, O is recessive to both. MN blood group: M and N are codominant alleles; therefore a blood group of N means the individual is homozygous for that allele. Rh + is dominant to Rh -. Child 1 belongs to the husband; child 2 belongs to the lover; child 3 could be the progeny of either the husband or the lover. For child 1 and child 2, which genotypes are critical in excluding the husband or lover as the parent? Which genotypes are consistent with either being the parent? Ì Problem 2 c ch = chestnut c= cremello a. Chestnut = bbc ch c ch Cremello = bbcc Palomino = bbc ch c Note: that the palominos don t breed true b. bbc ch c X bbc ch c 1/4 bbc ch c ch 1/4 bbcc 1/2 bbc ch c bbc ch c X bbc ch c ch 1/2 bbc ch c 1/2 bbc ch c ch bbcc X bbc ch c 1/2 bbcc 1/2 bbc ch c c. Mate chestnest with cremellos d. 0, 0, 1 3

4 Ì Problem 3 correct answers: a and d Ì Problem 4 Correct Answers: a and d Scenario a: If a loss-of-function is recessive, then addition of one wild-type gene copy should restore a normal phenotype Scenario b: untreated hets for a recessive allele will show no phenotype so the gene therapy test would not be instructive Scenario c: By definition, two copies of a haploinsufficient gene are required for a normal phenotype. So the addition of a single copy would not restore a wild-type phenotype. Scenario d: The het will show a mutant phenotype. Restoration of two copies of the wild-type gene (one introduced) should result in a wild-type phenotype Ì Problem 5 Circle True or False or I (not enough info to decide) F This loss-of-function mutation represents a polymorphism for the pax-6 gene. T The pax-6 gene is haploinsufficient. F The pax-6 mutation is pleiotropic (True) and completely dominant to the wild-type allele (False it is incompletely dominant). Ì Problem 6 T F I The gene that codes for the apob-100 protein is haploinsufficient (note hets show a higher plasma cholesterol level compared to normal) T F I The gene that codes for the ARH protein is haploinsufficient. (mutant allele is recessive) T F I In the general population, the frequency of the mutant LDL receptor allele is 0.1%. This mutant allele would be considered a polymorphism in this population. T F I This data in this table indicate that hypercholesterolemia (high plasma cholesterol levels) is genetically heterogeneous. Ì Problem 7 ANSWER IS B Ì Problem 8 This is an example of additive gene action (red + green = brown) N= chlorophyll is lost from mature pepper (gene might code for an enzyme that is directly or indirectly involved in its degradation?) n= chlorophyll not lost (loss-of-function in enzyme required for degradation?) R= red pigment produced (enzyme converts yellow pigment to a red pigment?) r = yellow pigment produced (loss-of-function mutation in enzyme required to convert yellow pigment to red pigment) Cross #1 Red NN RR X brown nn RR (green chlorophyll + red pigment) F1 = NnRR F2 3/4 red (N-RR) and 1/4 brown (nnrr) F2 ratios in quarters since parental genotypes differ by only one gene Cross#2 Parental strains must differ in two genes since F2 ratios in 16ths Parental: Yellow NNrr X nnrr brown F1 Red NnRr F2 N-R- red N-rr yellow nnr- brown nnrr yellow-green 4

5 Ì Problem 9 UNINFORMATIVE: O X O In either scenario, the trait will breed true. POTENTIALLY INFORMATIVE: AB X O. In the one gene scenario, the kids would have to have either A or B blood types. In the two gene scenario, if the AB parent is AABB, all kids would be AaBb and have the AB blood type. If parent is AaBb, then basic mendel tells you that the kids could be AaBb, or aabb or aabb or aabb. I ll let you sort through the rest of this question on your own. Ì Problem 10 a. normal, albino, albino, albino, normal b. complementation; the two albino individuals must have mutations in two different genes: AAbb X aabb most likely genotypes Ì Problem 11 a. The experiment is essentially an in vitro complementation test. Lines H and C complement each other. In these two strains the mutations that affect anthocyanin biosynthesis are in different genes. b. 9/16 of the F2 will be pigmented and 7/16 will have no pigment. c. Lines H and W fail to complement each other. The mutations in these strains are allelic -- that is, in the same gene. Ì Problem 12 (i) c (ii) True, one X-linked (strain #3) and four autosomal. No conclusion can be drawn about mutant line #5. (iii) b Mutant 5 does not show X-linked inheritance. Mutant 3 is X-linked Ì Problem 13 Answer is 1/2. The aa genotype is epistatic to genotypes at the B locus. NOTE: this is a test cross, not a self cross. Ì Problem 14 NOTE: a cat must have a dominant allele of the agouti locus in order to show tabby stripes. If a cat is non-agouti, it will be completely black. a. t B t B PP AA male X t M t M pp aa female F1 t B t M Pp Aa F2 black kitten t-t- P- aa aa is epistatic to the tabby locus b. 9/64 c. 1/4 5

6 Ì Problem 15 A. Gene B: presence or absence of scales B + = scales form (dominant) b = no scales Gene A: controls arrangement of scales a + = wildtype (scattered) A = linear arrangement (dominant) B: The bb genotype is epistatic to genotypes at the A gene. Aa + bb a + a + B + B + Parental naked X wild-type scales ê (truebreeding) F1 1/2 linear scales A a + B + b 1/2 wild-type scales a + a + B + b F1 Cross A F1 linear X F1 linear A a + B + b X A a + B + b ê F2 9/16 linear A - B + - 3/16 wild-type a + a + B + - 4/16 naked -- bb (any genotype at gene A) F1 Cross B F1 linear X F1 wild-type A a + B + b a + a + B + b ê F2 3/8 linear A a + B + - (1/2)(3/4) 3/8 wild-type a + a + B + - (1/2)(3/4) 1/4 naked Aa + bb or a + a + bb 2[(1/2)(1/4)] F1 Cross C F1 wild-type X F1 wild-type a + a + B + b X a + a + B + b (only one gene segregating) ê F2 3/4 wild-type a + a + B + - 1/4 naked a + a + bb 6

7 Ì Problem 16 a. Since the strains were known to carry mutations in different strains, they should show complementation. b. The friend overlooked the fact that, by definition, a loss-of-function mutation in a haploinsufficient gene will not be recessive to the wild-type allele. Therefore a complementation test will be invalid Ì Problem 17 (i) answer is d (ii) Mutant #1 females aabb X Mutant #3 males Ab â F1 females AaBb â haploid male progeny: 1/4 AB purple 1/4 Ab black 1/4 ab black 1/4 ab black (ii) Genes A and B are both required for biosynthesis of scarlet pigment. a and b are recessive, loss-of-function mutations which affect scarlet biosynthesis. The synthesis of the black pigment is unaffected by these mutations. Ì Problem 18 GENOTYPES Heavy No-spots (true-breeding) Medium (true-breeding) Hypothesis #1 One gene 2 alleles called a 1 and a 2 Incomplete dominance a 1 a 1 a 1 a 2 a 2 a 2 Hypothesis #2 One gene, 3 alleles a 1 = heavy a 2 = medium a 3 = no-spots > = Complete dominance a 1 > a 2 > a 3 Hypothesis #3 Gene A involved in anthocyanin biosynthesis; a is a recessive allele that produces a nonfunctional enzyme. Gene S determines the extent of pigmentation in leaf: S- heavy ss medium a 1 a 1 a 2 a 2 a 3 a 3 a 2 a 3 SSAA ssaa ssaa ssaa Ssaa* SSaa *operationally true breeding since genotype hidden 7

8 Ì Problem 18 continued (ii) Cross heavy and no-spot true-breeding strains and self F1 If hypothesis #1 is correct: F1 will be all medium and F2 1/4 heavy, 1/2 medium and 1/4 no-spot If hypothesis #2 is correct: F1 will be all heavy and F2 3/4 heavy and 1/4 no-spot (iii) a. not possible to generate a true-breeding medium strain if hypothesis #1 is correct b. If hypothesis #2 is correct, you can t generate medium plants from your original true-breeding heavy and no-spot strains. Go back to the field and collect medium spotted plants: (iv) SSAA X ssaa à F1 SsAa heavy à F2 9/16 heavy S-A- 3/16 medium ssa- 4/16 no-spot ssaa, S-aa (v) Yes, this is consistent with your hypothesis. Your second no-spot strain was SSaa in genotype: SSAA x SSaa à F1 SSAa heavy à F2 3/4 SSA- heavy 1/4 SSaa no-spot (vi) Accidentally omitted from original problem: Assuming hypothesis #1 is correct, explain the variation in spotting: Variable expressivity due to one or more of the following variation in genetic background: a modifier gene segregating in the population environmental effects on expression of the trait inherent randomness in the formation and positioning of the spots during leaf development 8

9 Ì Problem 19 a. complementation test b. This test is used to determine whether mutants with the same (recessive) phenotype have defects in the same gene or different genes c. The mutant Strains C and B fail to complement and have mutations in the same gene. Strain C complements both D and O, which also complement each other. In summary, these 4 strains carry mutations in three different genes. Mutations in any one of these genes can cause a Rex phenotype. d. The simplest explanation for the W data is that this Rex strain is due to a dominant mutant allele. In this case, the complementation test gives us no information about allelism with the recessive mutations in the Rex genes defined in part c. (NOTE: A formal, but very, very, very unlikely possibility is that Rex W is homozygous for recessive mutations in all three of the Rex genes.) Ì Problem 20 From text of problem: colorless compound à red à purple gene A gene B Since F1 is purple, parents must be AAbb (red) X aabb (white) F2 = 9/16 A-B- 3/16 A-bb 3/16 aab- 1/16 aabb The aa genotype is epistatic to allele of the B gene Answer is d Ì Problem 21 ANSWER is b c. shows an additive gene scenario which is inconsistent with the data a. predicts a different pattern of epistasis yellow genotype epi to blue answer to 22 is on the next page. 9

10 Ì Problem 22 Define allele symbols in this box. Indicate dominance and which trait(s) the gene(s) controls. BELOW: Indicate all genotypes in each box. W = prevents formation of pigment (dominantly epistatic to pigment color gene) w= allows pigment formation M= dark magenta m = light magenta Parental: dark magenta (truebreeding) X white ww MM X Ww mm à F1 1/2 dark magenta 1/2 white ww Mm Ww Mm Self F1 dark magenta plants: ww Mm 3/4 dark magenta ww M- 1/4 light magenta ww mm Self F1 white plants: Ww Mm self 12/16 white W- - - = W- M- & W- mm 3/16 dark magenta ww M- 1/16 light magenta ww mm Predict outcome (phenotypes & ratios) of the following cross: F1 dark magenta plants X F1 white plants ½ white 3/8 dark magenta 1/8 light magenta ww Mm X Ww Mm WwM- = white = ½ X ¾ =3/8 white Ww mm = white = ½ x ¼ = 1/8 white wwm- = ½ X ¾ = 3/8 dark magenta ww mm = ½ X ¼ = 1/8 light magenta 10