Question

In the absence of recombination, how many genetically different types of gamete can an organism with six homologous chromosome pairs produce?

A) 6
B) 12
C) 36
D) 64
E) none of the above

Answer 1

In the absence of recombination, an organism with six homologous chromosome pairs can produce 64 genetically different types of gametes. So the correcct option is D.

Step-by-step explanation:

In the absence of recombination, an organism with six homologous chromosome pairs can produce 26 genetically different types of gametes. This is because during meiosis, each homologous chromosome pair separates, and the different combinations of chromosomes from each pair can result in different gametes. Therefore, the number of genetically different gametes is equal to 2 raised to the power of the number of chromosome pairs.

So, in this case, 2⁶ = 64 different types of gametes can be produced.

Therefore, option D) 64 is the correct answer.

Answer 2

(D) 64

Step-by-step explanation:

We have given number of chromosome pair that is haploid number = 6

We have to calculate the number of possible gamete in absence of recombination

The number of garnets is given by
`2^n` where n is haploid number

In question we have given haploid number, that is n=6

So the number of gametes without recombination
`=2^6=64`

So option (d) is the correct answer