Question

1. (4 PTS) 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 born; they are classified as follows:

                                     Normal                                                3

                                     Abnormal knees and fingers           2

                                     Abnormal knees only                     17

                                     Abnormal fingers only                    18

            (a) Using the chi square test, determine whether there is significant linkage between the B/b and the N/n gene.                                                                                

                         

                        

            (b)  If you conclude there is linkage, what is the distance between the two genes?

Answer 1

Answer:

a).

Nb/nB x nb/nb --Testcross

If the two genes are not linked, the testcross progeny ratio is 1:1:1:1.

Phenotypes	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
normal	3	10.00	-7.00	49.00	4.900
abnormal knees and fingers	2	10.00	-8.00	64.00	6.400
abnormal knees only	17	10.00	7.00	49.00	4.900
abnormal fingers only	18	10.00	8.00	64.00	6.400
Total	40	40			22.600

Chi-square value = 22.60

Degrees of freedom = no. of phenotypes – 1

Df = 4-1 = 3

Critical value = 7.815

Chi-square value of 22.60 is greater than the critical value of 7.815. So we reject null hypothesis. It means that the genes are linked.

b).

Always recombinant progeny are less numbered.

Recombination frequency (RF) = (no. of recombinants / Total progeny)100

RF = (3+2 / 40) 100 = 12.5%

RF (%) = Distance between genes (m.u).

Distance between genes = 12.5 map units