Question

Individuals homozygous for $cd$ genes were crossed with wild type $(++)$ . The $F_1$ dihybrid thus produced was test crossed. It produced progeny in the following ratio-

$++$	$900$
$cd$	$880$
$+d$	$115$
$+c$	$105$

What is distance between $c$ and $d$

• A. $5.75$ unit
• B. $11$ unit
• C. $27$ unit
• D. $88$ unit

Answer 1

The correct option is A $5.75$ unit

People homozygous for compact disc qualities were crossed with wild kind ++ the f1 dihybridtus and album qualities were crossed with wild kind ++. the f1 dihybridtus created was test crossed. it delivered descendants in the accompanying proportion ++900,+d115,cd880, c+105. The separation among c and d will be 11 unit.

So, the correct option is '11 unit'.