UT EID: 24 Multple Cholce Questions (2.5 pt each)-Pick the best answer and fill
ID: 152340 • Letter: U
Question
UT EID: 24 Multple Cholce Questions (2.5 pt each)-Pick the best answer and fill in the scantron. 1. What is the probability of producing a child that will phenotypically resemble either one of the two parents in the cross Aa Bb Cc dd x aa bb Cc dd if alleles of each gene show complete dominance and four genes are unlinked? A. 1/8 B. 1/4 D. 3/8 E. 1 2. How many phenotypically different kinds of progeny could potentially resultfrom the cross Aa Bb Cc dd x aa bb Cc dd if alleles of each gene show complete dominance and four genes are unlinked? A. 1 B. 2 C. 4 D. 8 E. 16 3-4. Piebald spotting is a condition found ih humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment-producing cells to migrate properly during development 3. Two adults with piebald spotting have one child who has this trait and a second child with normal skin pigmentation. Which of these is true if the condition is completely penetrant? A. The piebald spotting trait is dominant and the parents are both PP B. The piebald spotting trait is dominant and the parents are both Pp. C. The piebald spotting trait is dominant and the parents are PP and Pp. D. The piebald spotting trait is recessive and the parents are both pp E. It cannot be determined. 4. Assume the mode of inheritance you answered for Q3. When two heterozygotes marry, what is the probability that their child will have piebald spotting if the condition is 50% penetrant? A. 1/4 B. 3/4 C. 1/8 D. 3/8 For 5-6: In a species of tropical fish, a colorful orange and black variety called montezuma occurs When two montezumas are crossed, 2/3 of the progeny are montezuma and 1/3 are wild Montezuma is a single-gene trait, and montezuma fish are never true-breedingExplanation / Answer
Solution: The probability that the child will resemble quadruply heterozygous parent is 1/2*A * 1/2*B * 1/2*C * 1/2*D = 1/16 and the probablity of the child resembling quadruply homozygous recessive parent is 1/2*A * 1/2*B * 1/2*C * 1/2*D = 1/16 so the probability that child will resemble any parent is 1/16 + 1/16 = 1/8.
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