(4 points) One day while doing field biology research you discover a new species

ID: 147296 • Letter: #

Question

(4 points) One day while doing field biology research you discover a new species of beetle. You bring it back to the lab and attempt to breed this species to characterize its various phenotypes. You end up isolating some beetle lines that breed true for either purple shells and long antenna or green shells and short antenna. When you then cross these lines, they yield F: progeny with purple shells and long antenna. Upon crossing the Fi progeny with beetles that have green shells and short antenna yields the following progeny: Purple shell, long antenna Green shell, short antenna Purple shell, short antenna Green shell, long antenna Total 82 78 37 43 240 A) Perform a chi-square test to test for independent assortment. What is the resulting chi-square value and how many degree(s) of freedom should be used in its interpretation? B) Assuming that the genes are linked, what is the map distance between them in m.u.?

Explanation / Answer

Let P be the dominant allele for the color of shells, resulting in purple color. Let L be the dominant allele for the length of antenna resulting in the long antenna.

The genotype of F progeny is clearly heterozygous – PpLl

It was crossed with beetles with green shells and short antenna, which have genotype ppll.

Assuming the genes are unlinked, we expect all four possible phenotypes to be present in the ratio 1:1:1:1, i.e. out of total 240, 60 for each phenotype. Performing the chi-square test as below, using the formula, chi-square = sum of (O-E) ^ 2/E of all four phenotypes –

Phenotype

Observed, O

Expected, E

(O-E)^2/E

Purple shell, long antenna

8.0667

Green shell, short antenna

5.4000

Purple shell, short antenna

8.8167

Green shell, long antenna

4.8167

Total

27.1001

Now, the degrees of freedom is one less than a total number of phenotypes, i.e. 3.

When referring to the chi-square table, for 3 degrees of freedom, we see that our value of 27.1001 is greater than that of for probability 0.01, hence we reject our null hypothesis. Hence the genes are linked.

For finding the map units, we subtract the number of total recombinant offspring, i.e. 115 from total offspring, i.e. 240 and multiply it by 100. The value we get is the distance between the genes in map units, which in this case is 47.91 map units.

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