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Question
In a cross between genotype AB and ++,650 out of 1000 individuals were of parental type. The distance between A and B is?
In a cross between genotype AB and ++,650 out of 1000 individuals were of parental type. The distance between A and B is?
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Solution
The correct option is A 35 map units
Genes that are sufficiently close together on a chromosome will tend to "stick together," and the versions (alleles) of those genes that are together on a chromosome will tend to be inherited as a pair more often than not. This phenomenon is called genetic linkage. When genes are linked, genetic crosses involving those genes will lead to ratios of gametes (egg and sperm) and offspring types that are not what we'd predict from Mendel's law of independent assortment. That is the reason why the frequency of offsprings with the parental phenotype is different.
The distance between a and b = no. of recombinants/total no. of offsprings * 100
= 350/1000 *100
= 35 map units.
So, the correct answer is '35 map units'.
Genes that are sufficiently close together on a chromosome will tend to "stick together," and the versions (alleles) of those genes that are together on a chromosome will tend to be inherited as a pair more often than not. This phenomenon is called genetic linkage. When genes are linked, genetic crosses involving those genes will lead to ratios of gametes (egg and sperm) and offspring types that are not what we'd predict from Mendel's law of independent assortment. That is the reason why the frequency of offsprings with the parental phenotype is different.
The distance between a and b = no. of recombinants/total no. of offsprings * 100
= 350/1000 *100
= 35 map units.
So, the correct answer is '35 map units'.
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