Discussion 2 key Spring 2010

Discussion 2 key Spring 2010 - D's mother must be...

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Genetics (BIO 325) Spring 2010 Discussion 2 key _______________________________________________________________________ _ 1. (a) not possible (b) not possible (c) meiosis II (d) meiosis I (e) mitosis 2. (a) 16 (b) 16 (c) 32 (d) 64 3. (a) (1/2)(3/4)(1)(1/2)(1) = 3/16 (b) (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32 4. (a) True (b) False (c) False (d) True (e) True (f) True (g) True (h) False 5. X-linked recessive pattern of inheritance P: Female yellow X y X y x Male wild-type X Y Y F 1 : Males: all yellow X y Y Females: all wild-type X Y X y F2: Males: ½ Yellow X y Y ½ Wild-type X Y Y Females: ½ Yellow X y X y ½ Wild-type X Y X y 6. (a) Autosomal recessive (b) I A/-; A/- II A/a; A/a; A/-; A/-; A/a; A/a III a/a; A/-; a/a; A/-; A/-; a/a; A/- 7. X-linked recessive I X R Y; X r X R II X R Y; X r X R ; X R Y; X r Y; X R /-; X r Y; X R X R III X r Y; X R X r ; X r Y; X R X r ; X R X r ; X R X r ; X R X r ; X R Y; X R Y 8. (a) Choosing M for unaffected and m for the disorder, male B must be M/m , and female A has a 2/3 chance of being M/m . The overall chance of an affected child is 1 × 2/3 × 1/4 = 1/6. 1
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(b) If C's mother A is heterozygous, C stands a 1/2 chance of being heterozygous.
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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 late-onset 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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