Question

Genes a, b and c are linked on a particular chromosome in this order. The distance between a and b is 9.0 map units, while b and c are 30 map units apart. What is the probability of a double crossover if interference in this chromosomal region is 70%

Answer 1

Answer:

The  value is $G = 0.81 \%$

Step-by-step explanation:

From the question we are told that

    The distance between a and b is  e =   9.0 map unit

    The distance between b and  c is  f = 30 map unit

      The interference in this chromosomal region is L  = 70% =  0.7

Gnerally the  first single cross-over in this chromosomal  region  is mathematically represented as

        $k = \frac{e}{100} \%$

=>    $k = \frac{9}{100} \%$

=>    $k = 0.09 \%$

Gnerally the  first single cross-over in this chromosomal  region is mathematically represented as

        $J = \frac{f}{100} \%$

=>    $J = \frac{30}{100} \%$

=>    $J = 0.30 \%$

Gnerally the expected double cross-over in this chromosomal  region  is mathematically represented as

       $D = k * J$

=>   $D = 0.09 * 0.30$

=>   $D = 0.027$

Generally the coefficient of  coincidence is mathematically represented as

      $C = 1 -L$

=>   $C = 1 -0.7$

=>   $C =0.3$

Generally this coefficient of  coincidence can also be mathematically represented as

     $C = \frac{G}{D}$

Here G is  the probability of a double crossover (the observed double cross over)

So

     $0.3 = \frac{G}{0.027}$

=>  $G = 0.0081$

Converting to percentage

     $G = 0.0081 * 100$

     $G = 0.81 \%$