Question

A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap) and grey body (gy). The F1 mice were wild-type for both traits. When the F1 were interbred, the F2 were distributed as follows:

Females

all wild type

200

Males

wild type	91
apricot	11
grey	9
apricot, grey	89

Which of the following statements is correct?

A) ap and gy are unlinked

B) ap and gy are linked on an autosome and 10 map units apart

C) ap and gy are linked on an autosome and 20 map units apart

D) ap and gy are X-linked and 10 map units apart

E) ap and gy are X-linked and 20 map units apart

Answer 1

Correct option is

D) ap and gy are X-llinked and 10 map units apart.

Explanation-

Let, Apt - wold eyes. Cigt – wild body ap - apricot eye. or - grey body ·Giren, torve-breeding wild-tybe storain semale coroes to a male mouse with apricot eyes lap) and grey body gul. amice were wild-type for both traits. When the FI were in bored, F2 distributed as Females all wild tybe - 200 Males wild tybe - 91 apoicot - 11 grey - 9 apricot, grey - 89 This type of inheritance is basically x-linked. To check, Lets cross. Torse breeding wild-tybe female (xx) male mouse with apricot eyes and grey body (x4) ap gy P1 Apt gut Apt ay apgy gametes genoty be F1 Apt gyt / Apt ap gut gy (from & (all wild type) and females (xx) Males (x4) (All wild-type) 01.

F in bed Find & tam a female Ar' " Apt gyt / apgy F! game te geno- type ay be approapgy from f 1 female and to for females i.e. x (**) f 12 Ap gy? Apoy apgy apa Apt Aptatt Aptar gogy Apt apgytgyt Aptapgyon wild Wild Wild Ap 9 - a female a for males (x4) ales i.e. x i.e. x game te gamete from from male F2 e Apt gyt Apt gy apgy OY Apgyt 14 Aptgy/h apgy le wild grey apreest 911- apgy apgule Thus, with the cross, we prove that the ap and on anex-linked. In F2 males population, the recombinants i.e non-parental ones which are quey and apricot are not 50% of the total population (which will occur due to independent assortment). nence, the apand gy are linked. 02.

Map distance between ap and gy is equals to neombination frequency X 100 As, Recombination Frequency Total No. of Recombinant 7° 0 ener Total No. of progenies Now, he wombinant progenies are grey body wild eye (a) and apricot eyes wild body ((1). So, Total No. of Rewmbinant progenies a 9+11 20 Total No. of progenia 917 89 + 9 +11 200 Recombination frequency 2 20 2.00 Recombination frequency 20.10 in Max distance between ap and gy a recombination frequency & 100 z o.10 X 100 10 map unite Hence, the wop connect option is 1) ap and gy ane X-linked and 10 map units apart. 03