Question

1. in caucasian humans, hair straightness or curliness is thought to be governed by a single pair of alleles showing partial dominance. individuals with straight hair are homozygous for the $i_s$ allele, while those with curly hair are homozygous for the $i_c$ allele. individuals with wavy hair are heterozygous ($i_si_c$). in a population of 1,000 individuals, 245 were found to have straight hair, 393 had curly hair, and 362 had wavy hair.

(a) calculate the allelic frequencies of the $i_s$ and $i_c$ alleles.

phenotype (genotype)	# $i_s$ alleles	# $i_c$ alleles
wavy ($i_si_c$)
curly ($i_ci_c$)
total:
frequency

(b) is this population in hardy - weinberg equilibrium? justify your answer. your explanation should include a chi - square goodness of fit test.

individuals	straight	wavy	curly
observed
($o - e$)
chi - square table	degrees of freedom
p	1	2	3	4	5	6	7	8
0.05	3.84	5.99	7.82	9.49	11.07	12.59	14.07	15.51
0.01	6.64	9.32	11.34	13.28	15.09	16.81	18.48	20.09

Explanation:

Step1: Calculate the total number of alleles

The population has 1000 individuals. Since each individual has 2 alleles for this gene - locus, the total number of alleles is $2\times1000 = 2000$.

Step2: Calculate the number of $I_s$ alleles

Individuals with straight hair ($I_sI_s$) have 2 $I_s$ alleles each. There are 245 straight - haired individuals, so they contribute $245\times2=490$ $I_s$ alleles. Individuals with wavy hair ($I_sI_c$) have 1 $I_s$ allele each. There are 362 wavy - haired individuals, so they contribute 362 $I_s$ alleles. The total number of $I_s$ alleles is $490 + 362=852$.
The frequency of the $I_s$ allele, $p=\frac{852}{2000}=0.426$.

Step3: Calculate the number of $I_c$ alleles

Individuals with curly hair ($I_cI_c$) have 2 $I_c$ alleles each. There are 393 curly - haired individuals, so they contribute $393\times2 = 786$ $I_c$ alleles. Individuals with wavy hair ($I_sI_c$) have 1 $I_c$ allele each. There are 362 wavy - haired individuals, so they contribute 362 $I_c$ alleles. The total number of $I_c$ alleles is $786+362 = 1148$.
The frequency of the $I_c$ allele, $q=\frac{1148}{2000}=0.574$.

Step4: Calculate the expected genotype frequencies under Hardy - Weinberg equilibrium

The expected frequency of $I_sI_s$ is $p^{2}=(0.426)^{2}=0.181$. The expected number of $I_sI_s$ individuals is $0.181\times1000 = 181$.
The expected frequency of $I_sI_c$ is $2pq=2\times0.426\times0.574 = 0.489$. The expected number of $I_sI_c$ individuals is $0.489\times1000 = 489$.
The expected frequency of $I_cI_c$ is $q^{2}=(0.574)^{2}=0.33$. The expected number of $I_cI_c$ individuals is $0.33\times1000 = 330$.

Step5: Calculate the chi - square value

The chi - square formula is $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$, where $O$ is the observed number and $E$ is the expected number.
For straight - haired individuals: $\frac{(245 - 181)^{2}}{181}=\frac{64^{2}}{181}=\frac{4096}{181}\approx22.63$.
For wavy - haired individuals: $\frac{(362 - 489)^{2}}{489}=\frac{(- 127)^{2}}{489}=\frac{16129}{489}\approx33.0$.
For curly - haired individuals: $\frac{(393 - 330)^{2}}{330}=\frac{63^{2}}{330}=\frac{3969}{330}\approx12.03$.
$\chi^{2}=22.63+33.0 + 12.03=67.66$.
The degrees of freedom, $df = 3 - 1=2$ (since there are 3 genotypes).

Step6: Determine if the population is in Hardy - Weinberg equilibrium

Looking at the chi - square table with $df = 2$ and a significance level of 0.05, the critical value is 5.99. Since $\chi^{2}=67.66>5.99$, we reject the null hypothesis that the population is in Hardy - Weinberg equilibrium.

Answer:

(a) The frequency of the $I_s$ allele, $p = 0.426$ and the frequency of the $I_c$ allele, $q = 0.574$.
(b) The population is not in Hardy - Weinberg equilibrium. The chi - square value is 67.66 with 2 degrees of freedom. Since $67.66>5.99$ (the critical value at a 0.05 significance level), we reject the null hypothesis that the population is in Hardy - Weinberg equilibrium.