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(D) The distance between linked genes is calculated by the percentage of recombinant offspring.Total offspring = $9 + 9 + 41 + 41 = 100$.Recombinant o

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$18\, cM \, (trans)$

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(D) The distance between linked genes is calculated by the percentage of recombinant offspring.Total offspring = $9 + 9 + 41 + 41 = 100$.Recombinant offspring are those that show a new combination of alleles,which are $++/ab$ and $ab/ab$ (total $9 + 9 = 18$).Recombination frequency = $(	ext{Number of recombinants} / 	ext{Total offspring}) 	imes 100$.Recombination frequency = $(18 / 100) 	imes 100 = 18\%$.Since $1\% 	ext{ recombination} = 1\, cM$,the distance is $18\, cM$.In the $F_1$ fly $+a/+b$,the wild-type alleles $(+)$ are on different chromosomes (one with $a$ and one with $b$),which indicates a $trans$ (repulsion) arrangement.

$A$ test cross of an $F_1$ fly $+a/+b$ produces the following offspring:
$++/ab = 9$
$ab/ab = 9$
$+b/ab = 41$
$a+/ab = 41$
What is the distance between the linked genes?

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$A$ test cross of an $F_1$ fly $+a/+b$ produces the following offspring:$++/ab = 9$$ab/ab = 9$$+b/ab = 41$$a+/ab = 41$What is the distance between the linked genes?