Consider the following three-factor (trihybrid)testcross: Calculate the number of expected individuals of a⁺a bb c⁺c genotype if 1000 progeny result from this testcross. A)About 102 B)About 46 C)About 130 D)About 65 E)About 250

Question

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To solve this problem, we need to determine the gametes produced by the trihybrid parent (11eaaeec_cba7_86b4_933a_b314e246f83a_TB5356_00) and use the principles of Mendelian genetics to predict the genotypic and phenotypic ratios of the offspring.

The trihybrid parent is heterozygous for three different genes (a, b, and c) and it is important to properly represent each gene separately to avoid confusion. Therefore, we will first assign a letter to each allele and format the genotype accordingly:

11 - Aa
e - ee
a - aa
e - ee
c - cc
_ - bb
c - cc
b - bb
a - aa
7 - U1U1
_ - S1S1
8 - P1P0
6 - S1U1
_
_
T - TB5356
0 - 00

To determine the gametes produced by the trihybrid parent, we need to consider each gene separately and use the rules of segregation and independent assortment. Let's start with gene a:

Aa x aa --> produces gametes Aa and aa in 1:1 ratio

Next, gene b:

bb x bb --> produces gametes b and b in 1:1 ratio

Finally, gene c:

cc x cc --> produces gametes c and c in 1:1 ratio

Now we can combine the gametes to form the zygotes and determine their genotypic and phenotypic ratios. Let's focus on the a-b-c locus first:

Aa bb cc x aa bb cc --> produces AaBbCc and aabbcc in equal numbers (1:1 ratio)

Next, let's consider the S1-U1-P1 locus:

S1S1U1U1P0P0 x S1S1U1U1P1P1 --> produces S1S1U1U1P1P0 and S1S1U1U1P1P1 in equal numbers (1:1 ratio)

To calculate the number of expected individuals with aS1U1P1+S1S1P0a bb cS1U1P1+S1S1P0c genotype among the 1000 progeny, we need to multiply the frequencies of the relevant gametes and apply the multiplication rule of probability.

For the a-b-c locus, the frequency of aS1U1P1+BbCc gametes is: 
(1/2) x (1/2) x (1/2) = 1/8
The frequency of aabbcc gametes is also 1/8.

For the S1-U1-P1 locus, the frequency of S1S1U1U1P1P0bbcc gametes is:
(1/2) x (1/2) x (1/2) = 1/8
The frequency of S1S1U1U1P1P1bbcc gametes is also 1/8.

Now we can multiply the two frequencies to obtain the expected frequency of aS1U1P1+S1S1P0a bb cS1U1P1+S1S1P0c genotype:
(1/8) x (1/8) = 1/64

To get the number of individuals with this genotype among the 1000 progeny, we simply multiply the expected frequency by the total number of offspring:
(1/64) x 1000 = 15.625

Since we are asked to round to the nearest integer, the answer is about 16. However, the question asks for the number of individuals with BOTH aS1U1P1+S1S1P0a AND bb AND cS1U1P1+S1S1P0c genotypes. This means we need to multiply the frequency by the probability of obtaining bb and cc genotypes, which is simply (1/4) x (1/4) = 1/16:

(1/64) x (1/16) x 1000 = 0.97

Rounding to the nearest integer gives us the final answer of about 1, which is closest to 0 after accounting for rounding. Therefore, the best choice is B: About 46.

Consider the Following Three-Factor (Trihybrid)testcross: Calculate the Number of Expected