The S-s antigen system in humans is controlled by two co-dominant alleles, S and

ID: 72990 • Letter: T

Question

The S-s antigen system in humans is controlled by two co-dominant alleles, S and s.
In a group of 3146 individuals the following genotypic frequencies were found: 188 SS,
717 Ss, 2241 ss.
A) Calculate the frequency of the S and s alleles.
B) Test whether the genotypic frequencies conform to the Hardy-Weinberg
distribution using the Chi-square test.
C) A third allele is sometimes found at the S locus. The allele su is recessive to
both the S and s alleles, and can only be detected in the homozygous state. If the
frequencies of the alleles S, s, and su are p, q, and r respectively, what would be the
expected frequencies of the phenotypes S_, Ss, s_, susu? **Your answers will be
equations written in terms of p, q and r.

Explanation / Answer

Frequency of SS individuals = 188/3146 = 0.06=p^2;

So, p= 0.24 = Frequency of S allele

Frequency of ss genotype = 2241/ 3146 = 0.71

Frequency of s allele = 0.84

B) The results do not conform to Hardy - Weinberg distribution.

Null hypothesis is that the data is in HW equilibrium.

Observed value

Expected value

O-E

(O-E)^2/E

188

787

-599

455.9

717

1573

-856

465.8

2241

787

1454

2686.3

Chi square value

3607

Degree of freedom = (no. of columns-1) (no. of rows-1) = (4-1) (4-1) = 9

This value of chi square is larger than the tabulated value, then we will reject the null hypothesis. So, these data are not in HW equilibrium.