Question

You are a researcher who is interested in three genes in drosophila: The a gene, the b gene and the c gene. You think the genes may be linked, so you decide to do a cross using a true-breeding a+/b+/c+ fly and an a/b/c fly. The F1 generation is all a+/b+/c+. However, when the F1 generation is bred to a true-breeding a/b/c fly, it yields the following offspring phenotypes: 589 a+/b+/c+ ; 603 a/b/c ; 183 a+/b/c ; 176 a/b+/c+ ; 92 a/b+/c ; 95 a+/b/c+ ; 4 a+/b+/c ; 6 a/b/c+. Is this the expected ratio for unlinked genes (hint: the critical chi-squared value is 7.815)?

Answer 1

The parental plants are true-breeding and hence are homozygous all three loci.

Hence one parent is a+b+c+/a+b+c+. The other parent is a b c/a b c.

The cross of these will give all individuals in F1 of the genotype a+b+c+/a b c. Phenotypically these will be a+/b+/c+.

Now these F1 individuals are crossed with true-breeding a b c flies.

The gametes produced by a b c flies will only be a b and c.

Now, if the genes are unlinked, then the F1 individual can produce all combinations of genes in the gametes with equal proportions. These gametes will be: a+b+c+, a+b+c, a+b c+, a+b c, a b+c+, a b+c, a b c+, a b c.

The offsprings then can be seen using the Punnett square:

	a b c
a+b+c+	a+b+c+/a b c
a+b+c	a+b+c /a b c
a+b c+	a+b c+/a b c
a+b c	a+b c /a b c
a b+c+	a b+c+/a b c
a b+c	a b+c /a b c
a b c+	a b c+/a b c
a b c	a b c /a b c

Thus, the expected frequency of each of the genotypes is equal and since there are 8 genotypes, the expected frequency becomes 1/8.

The observed numbers of each genotype are:

Genotype	Observed Numbers
a+b+c+/a b c	589
a+b+c /a b c	4
a+b c+/a b c	95
a+b c /a b c	183
a b+c+/a b c	176
a b+c /a b c	92
a b c+/a b c	6
a b c /a b c	603
TOTAL	1748

Now, we can calculate the expected numbers by multiplying the total by 1/8 (expected frequency). Next, we calculate (#Observed-#Expected)^2/#Expected

Genotype	Observed Numbers	Expected Numbers	(#Observed-#Expected)^2/#Expected
a+b+c+/a b c	589	218.5	628.2391
a+b+c /a b c	4	218.5	210.5732
a+b c+/a b c	95	218.5	69.8043
a+b c /a b c	183	218.5	5.7677
a b+c+/a b c	176	218.5	8.2666
a b+c /a b c	92	218.5	73.2368
a b c+/a b c	6	218.5	206.6648
a b c /a b c	603	218.5	676.6144

Now, the chi-square test can be performed to check if the observed are statistically different from the expected numbers which assume that there is no linkage.

Substituting the values from the table, we get the sum as:

The critical value mentioned in the question is 7.815.

Hence we need to reject the null hypothesis and assume the alternate hypothesis to be true.

Here the alternate hypothesis is that there is linkage in these genes.

Thus, we conclude that genes a, b, and c are linked.