You have identified two mutant lines of corn that each frow in the presence of a

ID: 57502 • Letter: Y

Question

You have identified two mutant lines of corn that each frow in the presence of a particular herbicide. Line 1 is homozygouse for recessive mutation a, and Line 2 is homozycous for recessive mutation b. You mate the two lines together to create hybrdie AaBb plant. You want to self this plant to creat a new line that will be resistant to both herbicides.

A) if A and B assort independently, what proportion of the F2s would you expect to ba able to grwo inthe presence of both herbicides?

B) Instead, when you grow the F2 seed in the presence of both herbicides, you find that only 1% grow. Based on these data, how closely linked are A and B?

Explanation / Answer

Perform a cross between AaBb x AaBb and calculate frequency of aabb (genotype of corn which can grow in presence of both herbicides) from punnet sqaure

AABB

AABb

AaBB

AaBb

AAbB

AAbb

AabB

Aabb

aABB

aABb

aaBB

aaBb

aAbB

aAbb

aabB

aabb

As you can see expected frequency of aabb = 1 / 16 = 6.25 %

Observed frequency of aabb = 1 %

This seems like a case of linkage because the recombination frequency is less than 6.25 percent expected from independent assortment.

we need to perform a 2 test.

2 = (observed frequency - expected frequency)2 / expected frequency

= (.01 - .0625)2 / .0625

= .044

calculate degree of freedom (df) = n -1 = 2 - 1 = 1

Consult chi square table to calculate probability (P)

The X2 value of .044 and degrees of freedom of 1 are associated with a P value of less than 0.90, but greater than 0.75. This means that a chi-square value this large or larger (or differences between expected and observed numbers this great or greater) would occur simply by chance between 75% and 90% of the time.

By convention biologists often use the 5.0% value (p<0.05) to determine if observed deviations are significant . Any deviations greater than this level would cause us to reject our hypothesis and assume something other than chance was at play.

Because the probability is greater than 5 percent, the hypothesis of independent assortment must be accepted.

Inference: A anb B are not linked