A population of 120 pea plants were examined. Three colorations existed within the population; 70 red...

A population of 120 pea plants were examined. Three colorations existed within the population; 70 red (RR), 43 pink (Rr), and 7 white (rr). Calculate the 1) number of each allele and the 2) allele frequencies. Using the Hardy-Weinberg equilibrium equation (p2 +2pq+q2=1.0) calculate the 3) genotype frequencies. Based on a population size of 120 pea plants, 4) what is the predicted numbers of each genotype? 5) How does the predicted compare to the observed population? Show your work.

Solutions

Expert Solution

number of RR individuals =70 ,in this there is 70×2=140 R alleles.

Number of Rr individuals =43 ,in this there is 43×1=43 R alleles.and 43×1=43 r alleles.

Number of rr individuals =7 ,in this there is 7×2=14 r alleles.

1)So number of R alleles =140+43=183

Number of r alleles =43+14=57

2)here there are total of 2×70+2×43+2×7 =240 alleles.

So frequency of allele R=183/240 = 0.7625

Frequency of allele r= 57/240 = 0.2375.

3)here frequency of allele R = 0.7625

Frequency of allele r = 0.2375

So frequency of genotype RR = R×R= 0.7625×0.7625 = 0.5814

Frequency of genotype rr= r×r= 0.2375×0.2375= 0.0564

Frequency of genotype Rr = 2×R×r = 2×0.7625×0.2375= 0.3622

4)predicted number of each genotype are,

RR= 0.5814×120= 69.768

Rr= 0.3622×120= 43.464

rr=0.0564×120= 6.768

5)in predicted genotype in the number of RR individual and rr individuals there is a small decrease and small increase in Rr individuals. (But approximately it is same )

gladiator answered 1 year ago

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