29. What will be the probability of obtaining a plant with AaBBCc genotype from trihybrid (AaBbCc) parents?
1. 4 out of 64
2. 1 out of 64
3. 8 out of 64
4. 0 out of 64

Calculating the Probability of Obtaining AaBBCc from AaBbCc x AaBbCc Cross

Genetics can seem daunting when multiple traits are involved, but with the right approach, it becomes a logical puzzle. Let’s break down the probability of obtaining a specific genotype—AaBBCc—from a trihybrid cross: AaBbCc × AaBbCc.

Step-by-Step Breakdown

This trihybrid cross involves three independent gene pairs:

• A/a

• B/b

• C/c

We want to find the probability of getting:

• Aa from Aa × Aa

• BB from Bb × Bb

• Cc from Cc × Cc

Since these genes assort independently, we can calculate each probability separately and multiply them at the end.

1. Probability of Aa from Aa × Aa

→ So, P(Aa) = 1/2

2. Probability of BB from Bb × Bb

→ So, P(BB) = 1/4

3. Probability of Cc from Cc × Cc

→ So, P(Cc) = 1/2

Total Probability

Now, multiply all three:

P(AaBBCc)=P(Aa)×P(BB)×P(Cc)=12×14×12=116P(AaBBCc) = P(Aa) \times P(BB) \times P(Cc) \\ = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{2} = \frac{1}{16}P(AaBBCc)=P(Aa)×P(BB)×P(Cc)=21​×41​×21​=161​

But the total number of genotypic combinations from a trihybrid cross is 64 (since each locus has 3 outcomes and 33=273^3 = 2733=27, but in genotypic combinations, for specific genotypes, we go with $4\times 4\times 4=64$).

So,

116=464\frac{1}{16} = \frac{4}{64}161​=644​

✅ Correct Answer:

(1) 4 out of 64

Why This Matters

Understanding how to calculate genotype probabilities is essential in:

• Plant breeding

• Genetic research

• Predicting trait inheritance

The principles of Mendelian inheritance continue to be the foundation of classical genetics, and questions like this demonstrate the power of independent assortment and probability in predicting genetic outcomes.