Question

In a test cross of Pp Qq plant, the following progeny is obtained:
Pp Qq = 500
Pp q9 = 50
pp Qq = 100
pp qq = 350
What is the distance between genes p and q?
(1) 15 cm (2) 10 cm (3) 25 cm (4) 50 cm​

Answer 1

Answer:

here is your answer :))

Explanation:

Answer 2

Answer:

The distance between the genes p and q is 15cM, son the correct answer is (1).

Explanation:

This problem is belonging to gene recombination mapping. According to that type of problem there are some steps have to follow to solve the problem as here under --

Step 1

here it is stated that a test cross has been occurred, the cross would be

$Pp$  $Qq$        ×         $pp$ $qq$

and these are the parental type

Step 2

as mentioned in the question there are four different genotype come at the first progeny and all are $F_{1}$ generation.

Now, we need to separate the parental from the recombinant by simply matching the genotype to their parents.

Step 3

only two genotypes are recombinant -- $Pp$ $qq$ and $pp$ $Qq$

simply adding these two we can obtain -- $100+50=150$

the total number of individuals are -- $500+50+100+350=1000$

Step 4

finally, to obtain the distance between p and q we divide the recombinant by total number of individuals.

$\frac{150}{1000} \\=0.15$

and then convert it into percentage to get in cM unit

$0.15$ × $100$

$15 cM$

So, the correct option is (1) [15 cM]

for more tap the links below

https://brainly.in/question/19947225

https://brainly.in/question/43152946