Give scientific reasons : Linkage and crossing over of genes are alternatives of each other.
 

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The gene for haemophilia is present on $X$-chromosome. $A$ male has only one $X$-chromosome which he receives from his mother and $Y$-chromosome from father. The human male passes the $\mathrm{X}$ chromosome to his daughters but not to the male progeny (son).

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In Drosophila, long wing and broad abdomen characters are dominant over to characters vestigeal wing and narrow abdomen. A pure breeding Drosophila having long wings and broad abdomen is crossed with a pure breeding vestigeal-winged and narrow-abdomened Drosophila. Two of the $F_1$ individuals were crossed and the following results were produced in the $F_2$ generation Long wing, broad abdomen in $482$ Drosophila. Vestigeal wing, narrow abdomen in $154$ Drosophila. Which one of the following is illustrated by these results

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

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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

Linkage in Drosophila was first discovered by