Exam2_KEY&GradingRubric

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Genome 361 – Exam 2 Summer 2022 This exam is open-note, open book. However, this exam must be completed solely by you, with no input from other individuals, including via text, email, discord or any other source of online or in-person communication . Affirm that you will follow this rule by adding your name below. Name here: _____________________ Question 1 (15 pts) Researchers studying a population of brown wild-type (WT) mice noticed that mutant mice populations on several nearby islands were white. They crossed white mutant mice from seven different island populations (m1 through m7), all of which were homozygous for autosomal mutations that caused white fur. They also crossed each mutant to homozygous WT brown mice from their original population. The fur color of the progeny are indicated below in the table. A “?” indicates a cross that has not yet been performed. A. The mutation that leads to white fur is dominant in which population(s)? (choose all that apply) (5 points) none m1 m2 m3 m4 m5 m6 m7 can’t determine Grading rubric +2 for each correct mutant +1 for not including any additional answers B. Indicate which mutations fall into the same complementation groups. Place the members of each complementation group on one line for clarity. If there are any mutants you can’t place in a complementation group clearly indicate this in your answer and explain why. (6 points) Complementation Group: m1 Complementation Group: m2 Complementation Group: m3, m5, m7 m1 m2 m3 m4 m5 m6 m7 WT m1 white brown ? ? brown ? ? brown m2 white brown white brown white brown brown m3 white white ? white white brown m4 white white white white white m5 white white white brown m6 white ? white m7 white brown WT brown

You cannot place m4 and m6 into complementation groups because both mutations are dominant and thus offspring will always be white regardless of whether they are in the same or different genes as another mutation. Grading rubric: +1 for group with m1 +1 for group with m2 +1 for group with m3, m5, m7 +1 for not putting any additional mutants in any of the groups (and not making m4 or m6 its own group +1 indicating that m4 and m6 can’t be put into complementation groups +1 for explanation of why you can’t put m4 and m6 into complementation groups C. A graduate student working with this research group wants to do a complementation test with two new mutant populations (m8 and m9) using mice that are heterozygous for the recessive mutations that cause their white fur. Will this work? (choose 1) (4 points) Yes No Explain in 1-2 sentences: You will still be able to tell whether m8 and m9 are in the same or different genes. If the mutations from these populations are in the same gene, then offspring will be in a ratio of 3 brown : 1 white. If they are in different genes, then all their offspring will be brown. Grading rubric: +2 for “Yes” +1 for indicating you’ll still be able to tell whether m8 and m9 are in the same or different genes (essentially understanding the question you are trying to ask with a complementation test) +1 for indicating how you can tell. Question 2 (18 pts) Wild-type cherry-plum trees have red plums and green leaves. The pathway below describes how genes A and I affect whether the red pigment in plums is made from a precursor molecule. If the red pigment is not made the default color is green. A. Eddie isolates a true-breeding mutant plant with a recessive loss-of-function mutation in Gene A . What color are the leaves and plums on this plant? (choose 1) (4 points)

Leaves__ green ______ Plums____ green _____ If A doesn’t function, then the precursor molecule can’t be converted to red and thus the entire plant will be green. Grading rubric: +2 each B. Eddie isolates a true-breeding mutant plant with a recessive loss-of-function mutation in Gene I . What color are the leaves and plums on this plant? (choose 1) (4 points) Leaves____ red ____ Plums_____ red ____ If I doesn’t function, then it cannot block A and thus the whole plant will be red. Grading rubric: +2 each C. Eddie then crosses the two true-breeding mutant plants with each other. List the genotype of the F1 progeny and describe the color of both the leaves and the plums. (4 points) Genotype____ AaIi ______ Leaf color___ green ___ Plum color___ red ___ The true breeding plant from the first part of the question is IIaa. The true breeding plant from the second part of the question is iiAA. Their offspring will be heterozygous at both loci. If I functions normally in the leaves (the only place I is active), then it will block A and thus the leaves will be green. I isn’t active in the plums so it won’t block A and the plums will be red. Grading rubric: +2 for genotype +1 for each color designation D. Eddie then crosses the F1 progeny to a plant with a genotype aaii. What is the phenotypic ratio of the F2 progeny (fill in the blanks with the correct values). (3 points) __ 1 _green leaf, red plum : _ 1 __ red leaf, red plum : _ 2 __ green leaf, green plum Grading rubric: +1 each E. Is gene I epistatic to gene A? (choose 1): (3 points) Yes No Explain in 1-2 sentences.

Gene A is instead epistatic to gene I because if A is nonfunctional (genotype aa) then it will mask whichever genotype is at the I locus. Grading rubric: +1 for “no” +2 for explanation Alternative Answer worth full credit: Saying yes and then explaining how when I is dominant it will mask the effect at the A locus in leaves (this is an example of dominant epistasis), where II?? will always lead to a green phenotype in the leaves (where I is active). Question 3 (17 pts) Dr. Santana is investigating a family with a history of a rare form of muscular dystrophy. She has identified a VNTR locus that is 30cM away from the disease gene. The pedigree below shows the family members affected by the disease (filled symbols). The results of a PCR analysis on the linked VNTR is shown below some family members. Assume nothing aberrant like nondisjunction has occurred. A. What is the inheritance pattern for this disease? (choose any that apply) (3 points) Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive

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