Question

A geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation spastic (sps) and determines that it is due to an autosomal recessive gene. She wants to determine if the gene encoding spastic is linked to the recessive gene for vestigial wings (vg). She crosses a fly homozygous for spastic and vestigial traits with a fly homozygous for the wild-type traits and then uses the resulting F1 females in a testcross. She obtains the following flies from this testcross.

vg+ sps+ 230     vg sps+ 97

vg   sps   224   vg+ sps 99

Total 650

Are the genes that cause vestigial wings and the spastic mutation linked? Do a chi-square test of independence to determine if the genes have assorted independently. There is a chi-square table on the back page to help with your analysis.

Answer 1

Parental cross vgvg spssps (Female homozygous for spastic and vestigial traits) x vg+vg+ sps+sps+ (male wild type)

F1 vg+vg sps+sps

F1 gametes vg+ sps+, vg+ sps, vg sps+, vg sps

Now, if genes are assorting independently, we will observe each phenotype in 1:1:1:1 ratio in F2.

So, expected phenotypes of each type = Total number / 4 = 650 / 4 = 162.5

Now, = (230-162.5)2/162.5 + (224-162.5)2/162.5 + (97-162.5)2/162.5 + (99-162.5)2/162.5 = 28.04 + 23.28 + 26.40 + 24.81 (Up to 2 decimals) = 102.53

Now, degrees of freedom = n - 1 where, n = number of expected phenotypes

So, degrees of freedom = 4-1 (As 4 different phenotypes are expected) = 3

From the distribution table, we find that probability of = 102.53 with degrees of freedom 3 is much less than 0.05. As probability is less than 0.05, we can say that difference between observed & expected values are not due to chance alone & significant differences exist between them. So, genes are not assorted independently & they must be linked.