Gen380SpHW_3 (1) - occurring together is the product of...

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Genetics 380: Problem set #3: Chapter 3: problems 7, 18, 20, 29, 32, 33, 34, 37, 38. Ch3 #7 : Give the genotypic ratios that may appear among the progeny of simple crosses and the genotypes of the parents that may give rise to each ratio. Genotypic ratio Parental genotype 1:2:1 Aa × Aa 1:1 Aa × aa AA x Aa Uniform progeny AA × AA aa × aa AA × aa Ch3 #18 : Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains one-half white kittens and one-half black kittens. When the black kittens are interbred, they produce all black kittens. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait? Explain your reasoning. The black coat color is likely recessive. When Sam was crossed with a black cat, one- half the offspring were white and one-half were black. This ratio potentially indicates that one of the parental cats is heterozygous dominant while the other parental cat is homozygous recessive—a testcross. The interbreeding of the black kittens produced only black kittens, indicating that the black kittens are likely to be homozygous, and thus the black coat color is the recessive trait. If the black allele was dominant, we would have expected the black kittens to be heterozygous, containing a black coat color allele and a white coat color allele. Under this condition, we would expect one-fourth of the progeny from the interbred black kittens to have white coats. Because this did not happen, we can conclude that the black coat color is recessive. Ch3 #20 : In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. a. Give the genotypes of Sally, her mother, her father, and her brother. Sally’s father, who has alkaptonuria, must be aa. Her brother, who also has alkaptonuria, must be aa as well. Because both parents must have contributed one a allele to her brother, Sally’s mother, who is phenotypically normal, must be heterozygous (Aa). Sally, who is normal, received the A allele from her mother but must have received an a allele from her father. The genotypes of the individuals are: Sally (Aa), Sally’s mother (Aa), Sally’s father (aa), and Sally’s brother (aa).
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b. If Sally’s parents have another child, what is the probability that this child will have alkaptonuria? Sally’s father (aa) × Sally’s mother (Aa)Sally’s mother has a 1⁄2 chance of contributing the a allele to her offspring. Sally’s father can contribute only the a allele. The ratio for a cross between a heterozygous and a homozygous is 1:1. The probability of an offspring with genotype aa and alkaptonuria is therefore 1⁄2. c.
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This note was uploaded on 02/09/2012 for the course GENETICES 380 taught by Professor Ohylmeyer during the Spring '11 term at Rutgers.

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Gen380SpHW_3 (1) - occurring together is the product of...

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