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(C) The distance between two genes is calculated by the frequency of recombination.Total individuals = $1000$.Parental types = $700$.Recombinant types

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$30$ map units

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(C) The distance between two genes is calculated by the frequency of recombination.Total individuals = $1000$.Parental types = $700$.Recombinant types = $	ext{Total} - 	ext{Parental} = 1000 - 700 = 300$.Recombination frequency = $(	ext{Recombinant types} / 	ext{Total individuals}) 	imes 100$.Recombination frequency = $(300 / 1000) 	imes 100 = 30\%$.Since $1\%$ recombination frequency is equal to $1$ map unit (or centimorgan), the distance between genes $a$ and $b$ is $30$ map units.

$A$ cross is made between two homozygous individuals, one with the wild type $(+, +)$ and the other with the mutant type $(a, b)$. In the $F_2$ generation (or test cross progeny), out of $1000$ individuals, $700$ are parental types. Calculate the distance between genes $a$ and $b$.

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