Question

In the cross AaBBcc

aaBbCc, what is the probability of an offspring that is aaBbcc?
(a) 1/2
(b) 1/4
(c) 1/8
(d) 1/16

Answer 1

Hint: In a trihybrid cross, each gamete gets one of the alleles from each parent. Hence, we perform such crosses in punnett squares. Trihybrid cross follows all the laws given by Mendel i.e. law of segregation of gametes, law of dominance and law of independent assortment.

Complete answer:
Here, let us break down the given problem into small parts. First, we will consider a cross between Aa and aa. We will look for priority of getting a homozygous individual (since we want the final genotype to be aaBbcc.)

	A	a
a	AaHeterozygote	aaHomozygote
a	AaHeterozygote	aaHomozygote

 So, the probability of getting a homozygote (aa) is 2/4 i.e. 1/2.
Similarly, let us look into the cross between BB and Bb.

	B	B
B	BBHomozygote	BBHomozygote
b	BbHeterozygote	BbHeterozygote

So, the probability of getting a heterozygote (Bb) is 2/4 i.e. 1/2.
Last part of the given genotype-aaBbcc is ‘cc’. Now, let us look into a cross between parental parts cc and Cc.

	C	c
c	CcHeterozygote	ccHomozygote
c	CcHeterozygote	ccHomozygote

So, the probability of getting a homozygote (cc) is 2/4 i.e. 1/2.
In order to get all of them happening together, we will now reduce the overall chance for each genotype. So, we will multiply all three probabilities as - $$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\text{ = }\frac{1}{8}$$
So, the correct answer is ‘1/8.’

Note: Traditional method of solving this problem can be by using the punnett square. So, if we look into the punnett square of a normal trihybrid cross, we can get the probability by looking into the genotype.