Question

11. A true breeding line of pygmy ewoks with pointy ears (E), long-hair (H) and black heads (B) was crossed with a true breeding line of round eared (e), short-haired (h) and grey-headed (b) pygmy ewoks. The F1 was crossed back to the round eared, short-haired and grey-headed parentals. Traits were observed in the offspring in the following proportions. EHB 580 е НЬ 115 Eh b 3 eh B 184 EhB 130 ЕНЬ 213 ehb 661 i. Is there any evidence of physical linkage between the genes E, H and B? ii. If possible, can you deduce which gene is in the middle iii. What is the distance between the two genes in centimorgans?

Answer 1

Answer 11: (i) To determine whether the two genes are linked or not, we first should know what is the distance between the genes. If the genes are close tofther, i.e., if they are on the same chromosome, they're said to be linked, but if the genes are far away and are on different chromosomes, they're said to be unlnked. So to calculate the distance between the genes, we must find out recombination frequency.

• Recombination frequency between gene E and H:

Now when we're looking for crossing over between these two genes, if there's 'E' from 1st true breeding parent, then in the same set there must be 'h' OR 'e' and 'H'. Because if it is 'e' and 'h' OR 'E' and 'H', then they're true breedings and there is no crossing over.

Considering this fact, data set 2, 3 and 5 with values 115, 3 and 130 respectively are recombinant offsprings.

Now the formula for recombination frequency is #of recombinant offsprings/ #total offsprings × 100

So, 115 + 3 + 130/1886 (total offsprings) × 100 = 13% = 13cM (1% RF = 1cM = 1 map unit)

• Simillarly, with the same rule, RF between H and B = 115 + 184 + 130 + 213/1886 × 100 = 34% = 34cM
• The RF between E and B = 3 + 184 + 213/ 1886 × 100 = 21% = 21cM

The distance between E, H and B is small and so yes, the genes are linked.

(ii) Which gene is in middle?

Here the distance between E and H is 13 cM, the distance between H and B is 34cM and the distance between E and B is 21 cM . The distance between H and B is 34 which is the total of distance between E and H (13) and E and B (21) (21+13 = 34). So, H and B are the farthest and so E is in between H and B.

H----13cM------E----------21cM-----------B (map)

(iii) Distance between genes:

E and H : 13 cM

H and B : 34 cM

E and B : ​​​​​​​21 cM