A. C = 0.71 c = 0.29 B. i. cc = 1.0 Cc = 0.6 CC = 0.2 ii. = 0.43 iii. CC = 0.23

Question

A. C = 0.71 c = 0.29 B. i. cc = 1.0 Cc = 0.6 CC = 0.2 ii. = 0.43 iii. CC = 0.23 Cc = 0.58 cc = 0.2 iv. C = 0.52 c = 0.49 v. = 0.58 Those are the answers How does he get .71 and .29 I tried 75/900 it dosent work I'm confused could you explain !! In a population of 900 zeebogs, there are 75 sane critters (homozygous for the recessive allele c). A. Assuming the population is in Hardy Weinberg equilibrium, what are the allele frequencies? B. After a cataclysmic change to their environment (the zeebog stock market crashes, sane zeebogs become more fit than the other genotypes. For every 5 sane zeebogs that reproduce, 3 heterozygotes reproduce and 1 crazy homozygote reproduces. i. What is the fitness of each genotype? ii. What is the mean fitness of the population? iii. What will the genotypic frequencies be after 1 generation of selection' iv. What will the new allelic frequencies be? v. What is the new mean fitness?

Explanation / Answer

A) If the population is in Hardy-Weinberg equilibrium, the allele frequency for the sane critters should be 0.08 (75/900), since it is homozygous recessive and the allele frequency for crazy homozygotes 0.92.

We can assume q= frequency of cc= 0.08

p= frequency of CC= 1-0.08=0.92

For the B part enough information is not given, it has been mentionted that sane zeebogs have become more fit, but what is their abundance? this has ti be known for calculating fitness.

  

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