The degree of freedom table is available online. Please answer A&B for question 2 In humans, the dominant allele N causes an abnormal shape of the patella in the knee (n is the normal allele). A separate gene affects finger length, and the dominant allele B causes abnormally short fingers, whereas b gives normal length. A study focused on people who have both abnormal patellae and short fingers (they most likely have the genotype N/n B/b). They inherited the N allele from one parent and the B allele from the other parent. These N/n B/b individuals mated with normal spouses. (The spouses had no history of abnormal patellae or short fingers in their families; they can be assumed to be homozygous normal.) 40 progeny were boom; they are classified as follows. Using the chi square test, determine whether there is significant linkage between the B/b and the N/n gene. X^2 = P = linked? If you conclude there is linkage, what is the distance between the two genes? Solution Answer: a). N/n; B/b x n/n; b/b N/n B/b : N/n b/b ; n/n B/b ; n/n b/b 1:1:1:1 Chi-square vale = 22.60 The degrees of freedom = 4-1 = 3 p value = 0.00004893 The genes are not linkedPhenotypeObserved(O)Expected (E)O-E(O-E)2(O-E)2/ENormal310- 749.004.90Abnormal knees and fingers210-864.006.40Abnormal knees only1710749.004.90Abnormal fingers only1810864.006.404022.60.

1. The degree of freedom table is available online. Please answer A&B for question 2 In humans,
the dominant allele N causes an abnormal shape of the patella in the knee (n is the normal allele).
A separate gene affects finger length, and the dominant allele B causes abnormally short fingers,
whereas b gives normal length. A study focused on people who have both abnormal patellae and
short fingers (they most likely have the genotype N/n B/b). They inherited the N allele from one
parent and the B allele from the other parent. These N/n B/b individuals mated with normal
spouses. (The spouses had no history of abnormal patellae or short fingers in their families; they
can be assumed to be homozygous normal.) 40 progeny were boom; they are classified as
follows. Using the chi square test, determine whether there is significant linkage between the
B/b and the N/n gene. X^2 = P = linked? If you conclude there is linkage, what is the distance
between the two genes?
Solution
Answer:
a). N/n; B/b x n/n; b/b
N/n B/b : N/n b/b ; n/n B/b ; n/n b/b
1:1:1:1
Chi-square vale = 22.60
The degrees of freedom = 4-1 = 3
p value = 0.00004893
The genes are not linkedPhenotypeObserved(O)Expected (E)O-E(O-E)2(O-E)2/ENormal310-
749.004.90Abnormal knees and fingers210-864.006.40Abnormal knees
only1710749.004.90Abnormal fingers only1810864.006.404022.60