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Final_exam_Su09_KEY

Final_exam_Su09_KEY - GENE 3200 Summer 2009 Bedell...

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GENE 3200 – Summer 2009 – Bedell Cumulative Final – July 31 ID no. 810_________ KEY ___________ 1 Write your ID number (810 number, NOT SSN) on every page . Write your initials on the first page only . All of the multiple choice questions have only one correct answer. The exam should be completed in ink. Regrades will not be given if you write in pencil. All requests to re-grade an exam must be made in writing to Dr. Bedell within 48 hours of the grades being posted on WebCT. 1. (1 pt) True or false . Sixteen genotypes are obtainable from the cross Aa Bb X Aa Bb. 2. In Drosophila, two autosomal genes are known to control wingsize and eye shape. For wingsize, t is the recessive allele for tiny wings and T is the dominant allele for normal wing size. For eye shape, n is the recessive allele for narrow eyes and N is the dominant allele for oval eyes. A male with normal wings and narrow eyes was crossed to a female with normal wings and oval eyes. The offspring of this course were: normal wings, oval eyes 35 normal wings, narrow eyes 29 tiny wings, oval eyes 10 tiny wings, narrow eyes 11 2 (cont). (3 pts) What are the parent's genotypes? Give a brief justification for your answer. Male Tt nn x female Tt Nn 3:1 segregation for normal (35 + 29) vs tiny wings (10 + 11), both parents have to be hets; 1:1 segregation for oval (35 + 10) vs narrow eyes (29 + 11) so male is homozygous recessive x female heterozygous Answer EITHE R question 3 OR 4, not both. (4.5 pts) If you answer both, only the first will be graded. 3. At the right is a pedigree for a recessive human trait (use R for the dominant allele and r for the recessive allele). Individuals I-1 and II-4 are known to be R/R but you should NOT make any assumptions about whether this is a rare or common trait (i.e. unrelated individuals may be carriers of the trait). 3 (cont). (4.5 points) What is the probability that IV-1 will exhibit the trait? Show all of your calculations. On the pedigree or in the space below, write the genotype and probability for each individual needed to calculate the total probability for IV-1. III-3, 1/2; III-4, 2/3; IV-1, 1/4 1/2 x 2/3 x 1/4 = 2/24 = 1/12
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GENE 3200 – Summer 2009 – Bedell Cumulative Final – July 31 ID no. 810_________ KEY ___________ 2 Answer EITHE R question 3 OR 4, not both. (4.5 pts) If you answer both, only the first will be graded. 4. The pedigree to the right is for a rare genetic disorder inherited as an X-linked recessive. You can assume that individuals II-6, III-1, and III-9 are not carriers. Announced in class that individuals II-1, III-1, and III-6 are not carriers. 4 (cont). (4.5 pts) For each of the individuals in generation IV, state the percentages of offspring of each sex that would be affected by the disorder or would be carriers. Individual Affected females Carrier females Affected males IV - 1 None None None IV - 2 None Half 50% IV - 3 None All None 5. (1 pt) True or false . In codominance, heterozygotes have phenotypes that do not resemble either homozygote.
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