Biology 234 – Summer 2014 Tutorial Day 11 June 6th Day 11 Tutorial Answers for BIOL 234 section 921 summer 2014 students ONLY © J. Klenz L. McDonnell Not for sale or duplication. Learning Goals: ● ● ● ● ● Given the functions of two or more gene products that control the same phenotype (genetic interactions) predict the phenotype if there are changes to one or more of the genes involved. Recognize a modified dihybrid ratio and state the specific genotypes that give rise to each specific phenotype. Given sufficient information (such as the function of gene products), create a model to explain the gene interactions resulting in the modified ratio. Integrate concepts from the course to analyze complex scenarios. Show how you check your hypotheses/explanations to verify it supports data you are given. Be able to translate information into a model. Part 1 – Please complete before coming to tutorial 1. Ch 6. #17 In sweet peas, the synthesis of purple anthocyanin pigment in the petals is controlled by two genes, B and D. The pathway is a. What colour petals would you expect in a pure-breeding plant unable to catalyze the first reaction? White b. What colour petals would you expect in a pure-breeding plant unable to catalyze the second reaction? Blue c. If plants in parts a and b are crossed, what colour petals will the F1 plants have? Purple (complementation) d. What ratio of purple: blue: white plants would you expect in the F2? 9 purple : 3 blue : 4 white Biology 234 – Summer 2014 Tutorial Day 11 June 6th 2. Revisit your answers from your Mutation tutorial. Modified from Chapter 6#8, Refer to Fig 6-15. Refer to Figure 6-15. Assume that the active protein is responsible for rapid metabolism of caffeine. Those that metabolise caffeine fast do not feel the jittery effects after drinking coffee, those that metabolize caffeine slowly report feeling jittery after drinking coffee. r+ = functional r allele in panel (a) a+ = functional a allele in panel (a) rb = mutant r allele in panel (b) ac = mutant a allele in panel (c) rd = mutant r allele in panel (d) Predict what the phenotypes and in what proportions will be observed in the offspring produced by the following crosses: i) Cross: r+ /rd; a+/ ac x r+ /rd; a+/ ac 9:3:3:1 genotypes: 9 r+/_; a+/_ wildtype 3 r+/_: ac/ac jittery 3 rd/rd: a+/_ jittery 1 rd/rd; ac/ac jittery 9:7 phenotypes (9/16 nonjittery : 7/16 jittery) ii) Cross: r+ /rb; a+/ ac x rb/rb x ac/ac 1 r+/rb; a+/ac (wildtype) 1 rb/rb; a+/ac (jittery) 1 r+/rb; ac/ac (jittery) 1 rb/rb; ac/ac (jittery) ¼ non-jittery: ¾ jittery Biology 234 – Summer 2014 Tutorial Day 11 June 6th 3. modified Chapter 6 #54 In minks, wild types have an almost black coat. Breeders have developed many pure lines of color variants for the mink-coat industry. Two such pure lines are platinum (blue gray) and aleutian (steel gray). These lines were used in crosses, with the following results: a. Devise a genetic explanation of these three crosses. Show complete genotypes for the parents, the F1, and the F2 in the three crosses, and make sure that you show the alleles of each gene that you hypothesize for every mink. Platinum is recessive to wildtype It is due to a single recessive nuclear gene. PP x pp  Pp  ¾ P_ ¼ pp (platinum) Aleutian is recessive to wildtype. It is due to a single recessive nuclear gene AA x aa  Aa  ¾ A_ , ¼ aa (Aleutian) Platinum and Aleutian are two separate genes. In the F2 we see that the homozygous recessive has a new phenotype (sapphire), and a 9:3:3:1 ratio. ppAA x PPaa  PpAa  9 P_A_ wildttype, 3 P_aa Aleutian, 3 ppA_ platinum, 1 aapp (sapphire) b. Predict the F1 phenotypic ratios from crossing sapphire with platinum and with aleutian pure lines sapphire is homozygous for both Aleutian and platinum (separate genes). a/a; p/p x a+/a+; p/p  a/a+; p/p = all platinum F1 a/a; p/p/ x a/a; p+p+  a/a p+/p = all aleutian F1 c. If these F1s self, what will the phenotypic ratios in the two F2 populations be? ¾ platinum, ¼ sapphire OR ¾ Aleutian. ¼ sapphire 4. In Drosophila, the autosomal recessive brown eye color mutation displays interactions with both the X-linked recessive vermilion mutation and the autosomal recessive scarlet mutation. Flies homozygous for brown and simultaneously hemizygous or homozygous for vermilion have white eyes. Flies simultaneously homozygous for both the brown and scarlet mutations also have white eyes. Predict the F1 and F2 progeny of crossing the following true-breeding parents: 1. vermilion females × brown males 2. brown females × vermilion males Biology 234 – Summer 2014 Tutorial Day 11 June 6th 3. scarlet females × brown males 4. brown females × scarlet males b/b = brown, b+/_ = WT; XvXv or XvY = vermillion, Xv+_=WT; sc/sc = scarlet, sc+/_ = WT b/b; XvXv =white eyed female, b/b;XvY = white eyed male b/b;sc/sc = white-eyed male or female **I have assumed that you have to carry at least one b+ and one v+ to be WT (e.g. XvXvb+b = vermillion female, not WT) 1. vermilion females × brown males Cross is: XvXv b+/b+ x Xv+Ybb F1: XvY b+b – vermillion males; and XvXv+b+b – WT females F2: both males and females have the same ratios: 1/8 white, 3/8 WT, 1/8 brown and 3/8 vermillion Xvb+ Xvb Xv+b+ Xv+b Xvb+ vermillion vermillion WT WT Xvb vermillion white WT Brown Yb+ vermillion vermillion WT WT Yb WT white WT brown Cross 2 – reciprocal of #1: brown females (v+v+bb) x vermilion males (vYb+b+)  F1 = wild type females (XvXv+ b+b) and WT males (Xv+Yb+b)  F2: females ¾ v+_b+_ (WT): ¼ v+_bb (brown) Males 3/8 v+Yb+_ (WT): 3/8 vYb+_ (vermilion): 1/8 v+Ybb (brown): 1/8 vYbb(white) Cross 3: scarlet females x brown males sc/sc;b+/b+ x sc+/sc+; b/b *I’m assuming that sc/sc;b+ is scarlet, that you need at least one sc+ and one b+ to be WT F1: sc/sc+; b+/b – all are WT F2: no x-linked alleles in this cross. 9 WT: 3 scarlet: 3 brown: 1 white scb+ scb sc+b+ sc+b scb+ scarlet scarlet WT WT scb scarlet white WT brown sc+b+ WT WT WT WT sc+b WT brown WT brown Cross #4: brown females (b/b; sc+/sc+) x scarlet males (b+/b+; sc/sc)  F1 b+/b ; sc+/sc (all WT)  F2 is the same as cross #3. Part 2 questions 1. Chapter 6 #68 Assume that two pigments, red and blue, mix to give the normal purple color of petunia petals. Separate biochemical pathways synthesize the two pigments, as shown in the top two rows of the accompanying diagram. “White” refers to compounds that are not Biology 234 – Summer 2014 Tutorial Day 11 June 6th pigments. (Total lack of pigment results in a white petal.) Red pigment forms from a yellow intermediate that is normally at a concentration too low to color petals. A third pathway, whose compounds do not contribute pigment to petals, normally does not affect the blue and red pathways, but, if one of its intermediates (white3) should build up in concentration, it can be converted into the yellow intermediate of the red pathway. In the diagram, the letters A through E represent enzymes; their corresponding genes, all of which are unlinked, may be symbolized by the same letters. Assume that wild-type alleles are dominant and encode enzyme function and that recessive alleles result in a lack of enzyme function. (Note: Blue mixed with yellow makes green; assume that no mutations are lethal.) Deduce which combinations of true-breeding parental genotypes could be crossed to produce F2 progeny in the following ratios: a. 9 purple : 3 green : 4 blue Suggest half the class work backwards from the data and the other half make all the combinations of dihybrid crosses from each different pair A xB, AxC Strategy: determine what alleles must be present to get the phenotypes listed. From that work backwards to parents. For these phenotypes blue must always be present (EE). Upon examination it will also become clear that C and D are not involved. So, that leaves A and B. Purple = A_B_ Green = A_bb Blue = aaB_ and aabb Parents must have been AAbb x aaBB or AABB x aabb giving F1 of AaBb, which when crossed would result in the F2 above. b. 9 purple : 3 red: 3 blue : 1 white Biology 234 – Summer 2014 Tutorial Day 11 June 6th c. 13 purple : 3 blue d. 9 purple : 3 red : 3 green : 1 yellow 2. Modified Chapter 6 #66 The allele B gives mice a black coat, and b gives a brown one. The genotype e/e of another, independently assorting gene prevents the expression of B and b, making the coat color beige, whereas E/– permits the expression of B and b. Both genes are autosomal. In the following pedigree, black symbols indicate a black coat, gray symbols indicate brown, and white symbols indicate beige. Biology 234 – Summer 2014 Tutorial Day 11 June 6th a. Draw the pathway to illustrate how the B and E genes interact to control coat colour. E could either be the first gene or the last gene in the pathway and either way it could mask the B alleles. E gene B gene 1. (E at the beginning) Beige pigment  conversion to other colours dark-coloured fur OR B gene E gene 2. (E at the end) Brown  Black  transported into beige fur to turn dark b. What are the genotypes of the individual mice in the pedigree? (If there are alternative possibilities, state them.