CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Prophase II

If the four g...

Question

If the four genes A, B, C and D are 5, 10 and 30 Map units away from each other respectively and arranged in the same manner in a chromosome, which two genes would likely show more recombination?

A

A and B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

B and C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

C and D

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

A and C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C C and D
Crossing over is the exchange of genetic materials between two non-sister chromatids of two homologous chromosomes by forming chiasmata.
Linkage is the tendency of genes in a chromosome to stay together and crossing over is inversely proportional to the linkage. More the distance between the linked genes, more will be the chances of crossing over, and hence more recombinants will be produced.
In the question given, genes C and D have the highest distance in between them, so they will show the maximum recombinations among the 4 linked genes.

Suggest Corrections

thumbs-up

0

Similar questions

Q. A scientist performed the gene mapping experiments in maize. He mapped the genes on chromosomes on the basis of % crossing over between different genes. One map unit corresponds to one % crossing over or recombination. The genes showing more than 50% recombination were not supposed to be linked on same chromosome. In crossing over studies on maize, scientist observed the following % crossing over between genes A, B, C, D: between A and D 10%, between A and C 3%, between genes C and D 7%, between genes A and B 5%, and between genes C and B 8%. On the basis of above observation, find out the correct sequence of genes A, B, C and D on chromosomes.

Q. The linkage map of X-chromosome of fruit fly has 66 units with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) would be

Q. Who used in frequency of recombination between gene pairs on the same chromosome as a measure of distance between genes and mapped their position on chromosome.

Q. How many of the following statements are correct?
Statement I: Recombination of genes is directly proportional to the strength of linkage.
Statement II: Recombination of genes is inversely proportional to the strength of linkage.
Statement III: More the distance between the two genes on a chromosome, less the strength of linkage.
Statement IV: More the distance between the two genes on a chromosome, more the strength of linkage.
Statement V: Recombination frequency is inversely proportional to the distance between the genes present on a chromosome.

Q. Linkage map of X-chromosome of fruitfly has

66

map units with yellow body gene(y) at one end and bobbed hair(b) at the other. The recombination frequency between y and b gene would be

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard XII Biology

Join BYJU'S Learning Program

CrossIcon