1. If 16% of the individuals in the population have recessive phenotypes, what i

Question

1. If 16% of the individuals in the population have recessive phenotypes, what is the percentage of heterozygous genotypes?

2. If 16% of the individuals in the population have recessive phenotypes, what percentage of the TOTAL number of recessive genes exist in the heterozygous condition?

3. A sample population in West Virginia show that 152 residents had type M blood (I^M I^M), while 259 had type MN blood (I^M I^N), and 89 had type N blood (I^N I^N). What is the frequency of the type M allele (I^M) in this population?

Explanation / Answer

1. According to the hardy weinberg law

p+q=1

p2+q2+2pq=1

So q2= 16% or 0.16

q=0.4

p=1-q

p=0.6

Percentage of heterozygote population is 2pq

2 x 0.4 x 0.6= 0.48 or 48%

2. Total number of recessive implies, two recessive genes in recessive condition and one recessive gene in heterozygote

So 0.4 x 0.4 x 0.4=0.064 or 6.4%

3. The total population size is 500 residents, out of which IM IM phenotype are 152 residents which means they are 30.4%

p2= 0.304

p= 0.551 or 55.1%

So the frequency of IM allele is 0.551

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