ProblemEx3

# ProblemEx3 - 0.50 0.25 0.5 0.5 yes yes OK b 0.36 0.55 0.09...

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Problem session (Lab 7) 14.4 14.6 Heterozygotes = 2 pq , therefore we need to know p and q . We know that q 2 = 0.10, so q = 0.10 0.5 = 0.316. That means that p = 1 – 0.316 = 0.684. Frequency of heterozygotes is 2 x 0.316 x 0.684 = 0.43. Normal cattle = 90% of herd so proportion of heterozygotes among normals is 0.43 / 0.90 = 0.48. 14.7 I A : p = 0.16 I B : q = 0.10 I O : r = 0.74 Blood type Genotype Expected Frequency A I A I A I A I O p 2 = 0.16 2 = 0.026 2 pr = 2 x 0.16 x 0.74 = 0.237 0.263 B I B I B I B I O q 2 = 0.10 2 = 0.010 2 qr = 2 x 0.10 x 0.74 = 0.148 0.158 O I O I O q 2 = 0.74 2 = 0.548 0.548 AB I A I B 2 pq = 2 x 0.16 x 0.10 = 0.032 0.032 Total genotype = 1.001 (close enough) 14.9 AA Aa aa p q p + q = 1? Aa frequency = 2 pq ? Hardy- Weinberg a 0.25
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Unformatted text preview: 0.50 0.25 0.5 0.5 yes yes OK b 0.36 0.55 0.09 0.6 0.3 no no NO c 0.49 0.42 0.09 0.7 0.3 yes yes OK d 0.64 0.27 0.09 0.8 0.3 no no NO e 0.29 0.42 0.29 0.54 0.54 no no NO 14.11 Frequency of heterozygotes is lower than expected in inbred populations. 14.12 q 2 = 8.5 x 10-6 = 0.0000085 q = 0.0000085 1/2 = 0.00292 First cousins = q 2 (1 – F ) + qF = 0.0000085(1 – 1/16) + 0.00292(1/16) = 0.00019 Second cousins = q 2 (1 – F ) + qF = 0.0000085(1 – 1/64) + 0.00292(1/64) = 0.000054 A alleles a alleles 10 AA 20 15 Aa 15 15 4 aa 8 Allele frequency 35/58 = 0.603 23/58 = 0.397...
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