Question

A trait in garden peas involves the curling of leaves. A dihybrid cross was made involving a plant with yellow pods and curling leaves to a wild-type plant with green pods and normal leaves. All F1 offspring had green pods and normal leaves. The F1 plants were then crossed to plants with yellow pods and curling leaves. The following results were obtained: 117 green pods, normal leaves 115 yellow pods, curling leaves 78 green pods, curling leaves 80 yellow pods, normal leaves.

(a) Conduct a Chi-Square analysis to determine if these two genes are linked.
(b) If they are linked, calculate the map distance between the two genes.

Answer 1

a) Chi-Sq = 14.08. Degrees of freedom = 3. Genes are linked.

b) Distance= 40.5 map units.

Step-by-step explanation:

All the offspring from the dihybrid cross of plants of different phenotype for both traits had green pods and normal leaves, so these two are the dominant traits.

The alleles for the two genes are:

• Y_= green pods, yy= yellow pods
• C_=normal leaves, cc=curling leaves

The parental plant with yellow pods and curling leaves (yc/yc) was crossed with the wild-type plant (YC/YC). The F1 plants were heterozygous YC/yc.

The F1 plants were then test crossed: YC/yc X yc/yc.

The resulting F2 was:

• 117 green pods, normal leaves (YC/yc)
• 115 yellow pods, curling leaves (yc/yc)
• 78 green pods, curling leaves (Yc/yc)
• 80 yellow pods, normal leaves (yC/yc)

Total: 390

a) If the genes were independent, the four gametes produced by the F1 individual would have the same frequency: 1/4. The test cross individual can only produce 1 type of gamete: yc.

The expected offspring would be:

• 1/4 x 390 = 97.5 green pods, normal leaves
• 1/4 x 390 = 97.5 yellow pods, curling leaves
• 1/4 x 390 = 97.5 green pods, curling leaves
• 1/4 x 390 = 97.5 yellow pods, normal leaves

Null hypothesis: genes are not linked.

The Chi square can be calculated as follows:

`ChiSq=\sum((O-E)^2)/(E)`

Where O is the observed number of offspring of a particuar phenotype and E is the expected number of offspring showing the same phenotype.

`ChiSq=((117-97.5)^2)/(97.5) + ((115-97.5)^2)/(97.5) + ((78-97.5)^2)/(97.5) + ((80-97.5)^2)/(97.5) \\\\ChiSq=14.08`

For dihybrid crosses, the degrees of freedom (DF) can be calulated as number of phenotypes - 1. In this case there are 4 possible phenotypes, so 3 DF. If you take a look at a Chi-Square Table, for 3 DF the Chi-Sq must be greater than 7.815 (p < 0.05) in order be statistically significant.

In our case, Chi-Sq was 14.08, so we can reject the null hypothesis. The genes are linked.

b) Distance = Frequency of Recombination x 100

The recombinant gametes are Yc and yC, so there are a total of 78+80=158 recombinant individuals in the offspring. Frequency of Recombination= 158/390=0.405

Distance=0.405 x 100 = 40.5 map units.