Question

You cross a heterozygous female with a homozygous recessive male for three traits - A B C. You see the offspring ABC 233, abc 239, ABc 231, abC 241, aBc 12, AbC 14, aBC 14, Abc 16. Which of the genes - if any - are linked? How far apart are any/all linked genes? What is the order of genes on the chromosome if applicable?

Answer 1

So , in order to calculate distance between genes , it is important to know the parental genotype.

In this question the parental genotype is ABC and abc. Further calculations are shown here.

Since , the recombination frequency between A and B is quite low and high no. of ABc 231 and abC 241 also suggests that A and B genes tend to remain together hence they are linked.

Order of the genes :

---A---B----------------------------------------------------------------C--------

0.056 0.498

ABC A b a B c C = 14 = 12 7 Double cross over products. (D.C.o.) a b c (D.c.o) are usually the genotype lowest in number. ② We calculate the recombination frequency (r.f.) • for this we need to find single cross over (s.c.o.) S.C.O vw A and B = A B C Abc : 16 a b c a BC (3 R.S. = No. of scot No. of Dco x100 -- - (a) to tal no. of progeny . (16+14) + ( 14+12) x100 1000 = 5.6% sco wo B and c: O B C A B C = 1317 a b c = 241 Total - 472 = formula Applying = 472 + (14+12) x 100 = 49.8% 1000