Gene Inheritance Probability

Solution & Explanation

Understanding the Problem:

• Genetics of Appearance: An animal appears normal if it has either two normal genes (NN) or one normal and one mutated gene (NM or MN). An animal appears mutated if it has two mutated genes (MM).
• Given Information:
  • Animals A and B both have one normal (N) and one mutated (M) gene, i.e., NM.
  • Animals C and D are both normal, meaning their genotypes could be NN or NM.
  • We need to find the probability that D is NM given that E is normal.

Genetic Inheritance:

• Each parent contributes one of their two genes to their offspring with a 50% probability.

Possible Genotypes of D:

• D could be NN (normal-normal).
• D could be NM (normal-mutated).

Using Bayes’ Theorem:

We want to calculate $P(D = NM \mid E \text{ appears normal})$.

Bayes’ Theorem tells us: $P(D = NM \mid E \text{ appears normal}) = \frac{P(E \text{ appears normal} \mid D = NM) \cdot P(D = NM)}{P(E \text{ appears normal})}$

Calculate Prior Probabilities:

• $P(D = NM)$:
  • Since C and D have parents A and B who are both NM:
    • D inherits: N from A and M from B (probability $\frac{1}{4}$)
    • M from A and N from B (probability $\frac{1}{4}$)
  • Thus, $P(D = NM) = \frac{1}{2}$.

Calculate Likelihood:

• $P(E \text{ appears normal} \mid D = NM)$:
  • If D is NM and C is NM:
    • E could be:
      • NN (probability $\frac{1}{4}$)
      • NM (probability $\frac{1}{2}$)
      • MM (probability $\frac{1}{4}$)
  • So, the probability that E is normal (either NN or NM) given that D is NM is: $P(E \text{ appears normal} \mid D = NM) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

Calculate Total Probability of E Appearing Normal:

• $P(E \text{ appears normal})$ is calculated considering both cases (whether D is NN or NM).
• $P(E \text{ appears normal} \mid D = NN)$:
  • If D = NN, then E is definitely normal regardless of C’s genotype.
  • So, $P(E \text{ appears normal} \mid D = NN) = 1$.

Now, we combine both scenarios: $P(E \text{ appears normal}) = P(E \text{ appears normal} \mid D = NN) \cdot P(D = NN) + P(E \text{ appears normal} \mid D = NM) \cdot P(D = NM)$ $P(E \text{ appears normal}) = (1 \times \frac{1}{2}) + \left(\frac{3}{4} \times \frac{1}{2}\right) = \frac{1}{2} + \frac{3}{8} = \frac{7}{8}$

Apply Bayes’ Theorem:

Finally: $P(D = NM \mid E \text{ appears normal}) = \frac{\frac{3}{4} \times \frac{1}{2}}{\frac{7}{8}} = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}$

Conclusion:

The probability that D has one normal and one mutated gene given that E appears normal is $\frac{3}{7}$.