Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the $F_{2}$ segregation will show

Higher number of the parental types

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Higher number of the recombinant types

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Segregation in the expected 9 : 3 : 3 : 1 ratio

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Segregation in 3 : 1 ratio

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Solution

The correct option is A Higher number of the parental types
When the linked genes are situated quite close, the chances of crossing over are highly reduced. Due to this, a large number of parental gametes are formed and only a few recombinant gametes are formed. This results in higher number of parental types in an F2 generation as compared to recombinants.

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Similar questions

Q. Two genes R and Y are located very close on the chromosomal linkage map of the maize plant. When RRYY and rryy genotypes are hybridized, then F

_{2}

segregation will show:

Q. Two genes R and y are located very close on the chromosome linkage map on maize plant. When RRYY and rryy genotypes are hybridised, then

F_{2}

segregation will show

Q. Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridised then

F_{2}

- generation will show

Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the F2 segregation will show:

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