Question

1. in humans, the rh factor genetic information is inherited from our parents, but it is inherited independently of the abo blood - type alleles. in humans, rh+ individuals have the rh antigen on their red blood cells, while rh - individuals do not. there are two different alleles for the rh factor known as rh+ and rh. assume that a dominant gene rh produces the rh+ phenotype, and that the recessive rh allele produces the rh - phenotype. in a population that is in hardy - weinberg equilibrium, if 160 out of 200 individuals are rh+, calculate the frequencies of both alleles.
2. a population has 300 homozygous dominant, 150 heterozygous, and 250 homozygous recessive genotypes.

a. calculate the p and q frequencies for this original population.
b. after 5 generations, the genotype frequency changes. the new population has 290 homozygous dominant, 150 heterozygous, and 260 homozygous recessive genotypes. is this population in hardy - weinberg equilibrium?

1. you collect 100 samples from a large butterfly population. 50 specimens are dark brown, 20 are speckled, and 30 are white. coloration in this species of butterfly is controlled by one gene locus: bb individuals are dark brown, bb are speckled, and bb are white.

a. what are the allele frequencies for the coloration gene in this population?
b. after 10 generations, the population now is as follows: 125 dark brown butterflies, 50 speckled butterflies, and 75 white butterflies. is this population in hardy - weinberg equilibrium?

Explanation:

1.

Step1: Calculate the frequency of Rh - individuals

The total number of individuals is $n = 200$, and the number of Rh+ individuals is 160. So the number of Rh - individuals is $200 - 160=40$. The frequency of the recessive phenotype ($q^{2}$) is $\frac{40}{200}=0.2$.

Step2: Calculate the frequency of the recessive allele ($q$)

Since $q^{2}=0.2$, then $q=\sqrt{0.2}\approx0.447$.

Step3: Calculate the frequency of the dominant allele ($p$)

Using the Hardy - Weinberg equation $p + q=1$, so $p = 1 - q=1 - 0.447 = 0.553$.

Step1: Calculate the total number of alleles

The total number of individuals is $300 + 150+250 = 700$. The total number of alleles is $2\times700 = 1400$.

Step2: Calculate the number of dominant alleles

The number of dominant alleles from homozygous dominant individuals is $2\times300 = 600$, and from heterozygous individuals is $150$. So the total number of dominant alleles is $600 + 150=750$.

Step3: Calculate the frequency of the dominant allele ($p$)

$p=\frac{750}{1400}\approx0.536$.

Step4: Calculate the frequency of the recessive allele ($q$)

$q = 1 - p=1 - 0.536 = 0.464$.

Step1: Calculate the new $p$ and $q$

The new total number of individuals is $290+150 + 260=700$. The total number of alleles is $2\times700 = 1400$. The number of dominant alleles is $2\times290+150 = 730$, so $p=\frac{730}{1400}\approx0.521$. Then $q = 1 - p=1 - 0.521=0.479$.

Step2: Calculate the expected genotype frequencies

The expected frequency of homozygous dominant is $p^{2}=(0.521)^{2}\approx0.271$, the expected frequency of heterozygous is $2pq=2\times0.521\times0.479\approx0.499$, and the expected frequency of homozygous recessive is $q^{2}=(0.479)^{2}\approx0.230$.

Step3: Compare with the observed genotype frequencies

The observed frequencies of homozygous dominant, heterozygous and homozygous recessive are $\frac{290}{700}\approx0.414$, $\frac{150}{700}\approx0.214$ and $\frac{260}{700}\approx0.371$ respectively. Since the observed and expected genotype frequencies are different, the population is not in Hardy - Weinberg equilibrium.

Answer:

The frequency of the Rh+ allele ($p$) is approximately $0.553$ and the frequency of the rh allele ($q$) is approximately $0.447$.

2.a