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Unformatted text preview: D's mother must be heterozygous, and D stands a 1/2 chance of inheriting that heterozygosity. The overall chance of an affected child is 2/3 × 1/2 × 1 × 1/2 × 1/4 = 1/24. (c) The probability is still 1/24. (d) Now that we know individuals C and D must both be M/m , the chance of the second child's being m/m is 1/4. 9. Apparently the father and grandfather did not show symptoms of Huntington's disease, but since it is a lateonset disease, they probably would not have expressed it in their early twenties. There is a 1/2 chance that the grandfather inherited the allele, a 1/2 chance that the father received it from him, a further 1/2 chance that the woman received it, and then a further 1/2 chance that her future child would receive it. The overall chance that the woman's child will develop the disease is 1/2 × 1/2 × 1/2 × 1/2 = 1/16. 2...
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This note was uploaded on 12/30/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas.
 Spring '08
 SAXENA
 Genetics

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