Question

A dihybrid plant with genotype PpNn
produces four types of gametes in following
number PN = 200, pn = 200, Pn = 800, pN
800 then what is the distance between linked
genes ?
(1) 30 CM
(2) 10 CM
(3) 25 CM
(4) 20 CM​

Answer 1

Answer:

option (2)

Explanation:

Total possible type of gametes produced by an organism is 2

n

where "n" is the number of genes for which the person is heterozygous. A dihybrid is heterozygous for two chromosomes and thus forms four different types of gametes 2

2

= 4 two parental types and two recombinant types. Here, the percent of parental types will be higher due to the linkage. On test cross, these four gametes may give rise to four different phenotypes. To form 8 different gametes, individual should be trihybrid; 2

3

= 8. A monohybrid produces two types of gametes 2

1

= 2 and homozygotes (AA) forms single type of gametes.

So, the correct answer is option B.

Answer 2

Answer:

distance between linked Gene's is calculated by finding percentage recombinants produced so of the 2000 gametes produced 400 are recombinants thus 20% is the recombinant frequency

according to alfred Stutewart

1% recombination = 1 centimorgan

thus 20 CM is answer