An individual who is heterozygous for two linked genes (with alleles A, a and B, b) is crossed with an AB/ab individual, and among the progeny are: 14 AB/ab 36 Ab/ab 34 aB/ab 16 ab/ab What is the frequency of recombination

The correct answer is - 0.30

Explanation:

In case of the same alleles, there will be only one type of gametes as there is no repulsion takes place while in heterozygous 4 types of gametes will form due to repulsion.

Always less number shows recombinant. A higher number shows parental. Since the linkage is possible just when removing between the gens is under 50 So, the number of recombinants are:

16+14 = 30

then:

Recombination recurrence = number of recombinants/Total offspring

Recombination Frequency = 30/100 = 0.30

Thus, the correct answer is - 0.30

answer : haversian

[explanation: -(compact bone is organized as parallel columns, known as haversian systems, which run lengthwise down the axis of long bones. these columns are composed of lamellae, concentric rings of bone, surrounding a central channel, or haversian canal, that contains the nerves, blood vessels, and lymphatic system of the bone)]