AP BIOLOGY

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HEREDITY

LINKED GENES

Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
If the distance between A/B is 0.12 mu and gene C is on the opposite side of A with respect to B at a distance of 0.23 mu, What is the order of the gene linkage map
A
both A-B-C and C-B-A are possible
B
A-B-C
C
C-A-B
D
The linkage map cannot be detemined as the ratio is not 1:1
Explanation: 

Detailed explanation-1: -The linkage distance is calculated by dividing the total number of recombinant gametes into the total number of gametes.

Detailed explanation-2: -The map distance (30 m.u.) is equal to the recombination frequency, so 30% of gametes will be recombinant, but there are two types of recombinants, so 15% will be F G and 15% will be f g.

Detailed explanation-3: -∙ The more the distance between the gene on chromosomes, then gene shows weak linkage and follow crossing over to form new combination of genetic material and the less the distance between the genes the greater the linkage.

Detailed explanation-4: -Map distance between 2 linked genes = number of recombinants (offspring with certain phenotypes) / total number of offspring (times) 100.

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