Solutions For An Introduction to Genetic Analysis Chapter 19 Problem 29P

(b) The frequency of the alleles in the SNP (single nucleotide polymorphism) locus can be calculated as follows.

The total number of alleles for 1000 individuals for this trait = 2×1000= 2000

Every homozygous gene will have two copies of an allele and the heterozygous will have one copy of that allele.

The alleles in SNP – OCA2 are A and G.

The frequencies of allele A in OCA2 (in A/A and A/G) =

The frequency of allele A = 0.895

The frequency of allele G in OCA2 (in G/G and A/G) =

The frequency of allele G = 0.105

The alleles in SNP- MCIR are C and T.

The frequency of allele C = (in C/C and C/T)

= (2×192) + (2×448) +89+231/2000

The frequency of allele C = 0.80

The frequency of allele T = In T/T and C/T

= (2×21)-(2×19) +89+231/2000

The frequency of allele T = 0.20

(c) Using chi-square test we can test the null hypothesis.

In the following table the total values of rows and columns are shown. The total value of rows for both the SNPs is the same. So it is added only one time.

These are observed values for the combination of trait and genotype.

To know expected values of genotype and traits combination we have to multiply three values. They are the total frequency of the genotype and the total frequency of the trait and the number of individuals in the population.

In the case of Burning for the homozygous A/A the expected value

=0.302×0.800×1000

= 241.6

In the case of Tanning for the homozygous A/A the expected value

= 0.698×0.800×1000

= 558.4

In the case of burning for the homozygous C/C the expected value

= 0.302×0.640×1000

= 193.3

The expected values are calculated for all the genes and are shown in the table below.

For the SNP- OCA2 the expected values are as follows:

For the SNP- MCIR, the expected values are:

Now we have to calculate the deviation.

The deviation = (O-E)²/E

Where,

O = observed value

E = expected value

For OCA2 the deviations for A/A

In the following table of deviations, the values are rounded up to the nearest 100^th number.

The deviation values are added to get the chi-square.

For the OCA2 the chi-square = 2.02

For MCIR, the chi-square = 10.25

The df is 2 or (columns-1) (rows-1) for this particular test.

With df = 2, the critical value for a chi-square is 5.99.

In the case of OCA2, the chi-square is 2.02. So, null hypothesis can be accepted.

For MCIR, the chi-square is 10.25 and is greater than 5.99. This can be due to random association only. In this case, the null hypothesis which says that there is no evidence of association between tendency to burn and MCIR is not accepted.

(d) In the case of SNP- MCIR, there is evidence for an association between the gene and the trait there by rejecting the null hypothesis.

From the data, we could infer that the individuals in the population which has the tendency for burning are T/T. Heterozygous with single T do not show any marked effect as the trait T is recessive. Here, the mode of gene action is recessive.

(e) It the P value is greater than 0.05 this does not prove that the gene does not contribute to the variation of trait and in this case it is sun sensitivity.

But, it shows us that the difference between observed and expected values is large as this much and this must be only due to random chances, but not certainly. This greater P value is not a tool to prove that the gene does not contribute in the trait variation.