The map distance between genes $A$ and $B$ is $3$ units, between $B$ and $C$ $10$ units and between $C$ and $A$ $7$ units. The order of the genes in a linkage map constructed on the above data would perhaps be

  • A

    $A, B, C$

  • B

    $A, C, B$

  • C

    $B, C, A$

  • D

    $B, A, C$

Similar Questions

How many types of gametes will be produced by a $O$ Drosophila having following arrangement of
two genes ($y*$ and $w*$) on $X-$chromosome?

The distance between two genes in a chromosome is measured in cross-over units which represent

Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their $F_{1}$ progeny. He observed that

$(a)$ $F_{2}$ ratio was deviated very significantly from the $9 : 3 : 3 : 1$ ratio

$(b)$ Both genes did not segregate independently of each other

$(c)$ Recombinant types are not obtained in $F_{2}$ generation

$(d)$ Both genes segregate independently of each other

Select the correct set of statements :

Given below are two statements : one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.

Assertion $(A)$:

Mendel's law of Independent assortment does not hold good for the genes that are located closely on the same chromosome.

Reason $(R):$

Closely located genes assort independently.

In the light of the above statements, choose the correct answer from the options given below:

  • [NEET 2022]

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 $b$) should be

  • [AIPMT 2003]