Suppose two AaBbCc individuals are mated. Assuming that the genes are not linked, what fraction of the offspring are expected to be homozygous recessive for the three traits?A) 1/4B) 1/8C) 1/16D) 1/64

When we say homozygous recessive, you are looking for a pair of alleles that are both in lower case. This means that the recessive trait will be shown phenotypically. So the genotype that you are looking for in the Punnet would be:

aabbcc

Because it says reecssive for all three traits. To get the ratio, all you have to do is count how many times the combination occurs in the Punnet out of the total number of outcomes.

Attached is a Punnet for this scenario.

As you can see it only occurs once out of the 64 possible combinations.

So the answer would be 1/64.

marianaegarciaperedo marianaegarciaperedo · Answer 2 · 2022-04-05T13:20:26+02:00

Independent genes assort independently during meiosis. The fraction of the offspring expected to be h0m0zyg0us recessive for the three traits is Option D) 1/64

What are independent genes?

Independent genes are those that are not linked to each other. They assort independently during meiosis. These genes do not depend not influence the segregation of other genes.

Independent genes are located in different chromosomes or very separated from each other in the same chromosome.

Cross:

Parentals) AaBbCc x AaBbCc

Gametes) ABC, ABc, AbC, Abc, aBC, abC, aBc, abc

ABC, ABc, AbC, Abc, aBC, abC, aBc, abc

Since these genes assort independently, we do not need to make a Punnett square, which would be too complex. We can just perform multiplication between the probabilities of getting h0m0zyg0us recessive individuals for each og the traits.

Gene A ⇒ Aa x Aa

F1) 1/4 AA, 2/4 Aa, 1/4 aa

The probability of having aa individuals is P(aa) = 1/4

Gene B ⇒ Bb x Bb

F1) 1/4 BB, 2/4 Bb, 1/4 bb

The probability of having bb individuals is P(bb) = 1/4

Gene C ⇒ Cc x Cc

F1) 1/4 Cc, 2/4 Cc, 1/4 cc

The probability of having cc individuals is P(cc) = 1/4

The probability of having a h0m0zyg0us recessive individual for the three traits is

P(aa) x P(bb) x P(cc) = 1/4 x 1/4 x 1/4 = 1/64 aabbcc

The fraction of the offspring expected to be h0m0zyg0us recessive for the three traits is Option D) 1/64

You will learn more about independent genes at

https://brainly.com/question/6460312