### Shortcut Methods

**JEE Mains and Advanced**

**Numerical 1:**

- Let (p) be the frequency of the dominant gene.
- (p^2) is the frequency of affected individuals.
- So, (p^2 = 0.01 \Rightarrow p = 0.1)
- Frequency of carriers (= 2pq = 2 \times 0.1 \times 0.9 = 0.18 ) or (18%).

**Numerical 2:**

- Let (μ = 10) and (σ = 2).
- Sample size (= n = 100).
- Probability of sample mean between 9.5 mm and 10.5 mm (= P(9.5 < \overline{X} < 10.5)).
- Using CLT, (P \left (\frac{9.5-10}{0.2} < \frac{\overline{X} - \mu}{σ/\sqrt{n}} \frac{10.5-10}{0.2} \right )).
- (P(-2.5 < Z < 2.5) = 0.99) (using standard normal tables).

**CBSE Board Exams**

**Numerical 1:**

- Tall allele: (T), Short allele: (t), Dominant Tall: (TT), Recessive Short: (tt), Heterozygous: (Tt).
- In Hardy-Weinberg equilibrium:
- (p^2) is the frequency of homozygous dominant type,

- (q^2) is the frequency of recessive homozygous type,
- (2pq) is the frequency of heterozygous type, and (p+q=1).
- Given that 75% are tall, so frequency of dominant allele (p^2 = 0.75), so (p=0.87) and (q=1-p=0.13).
- Expected tall plants: (p^2 + 2pq = (0.87)^2 + 2(0.87)(0.13) = 0.76 + 0.224 = 0.984), which is roughly (76).
- Expected short plants: (q^2 = (0.13)^2 = 0.0169) or roughly (2).

**Numerical 2:**

- Similar to Numerical 1, we get the following;
- (p^2 =0.75 \Rightarrow p=0.87)
- (q^2 = 0.25 \Rightarrow q=0.5).
- So expected number of butterflies with brown wings (= 0.87^2 + 0.87 \times 0.5 \times 2= 0.76 + 0.87 = 0.933) i.e., roughly (93).
- Similarly, expected number of butterflies with white wings (= 0.5^2 = 0.25 ) i.e., roughly (25).