Numerical Problems/ISC/Class 12
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
BioNotesBioNotesCreated by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Online
Numerical Problems - Evolution
Welcome to the advanced numerical problems section on Evolution. In population genetics, the Hardy-Weinberg Principle is the fundamental null model for evaluating whether a population is evolving.
By utilizing the principles of population genetics, Hardy-Weinberg equilibrium (HWE), selection coefficients, and effective population size, you can mathematically track microevolution.
Steps for Hardy-Weinberg Calculations:
In a population of 1000 field mice, coat color is determined by a single locus with two alleles (B for brown, b for white). The population consists of:
Calculate the exact allele frequencies of B and b in this population. Is the population in HWE?
1. Calculate total number of alleles:
2. Count the 'B' alleles (p):
3. Count the 'b' alleles (q):
4. Check for Hardy-Weinberg Equilibrium:
Since observed values precisely match expected values, yes, the population is in Hardy-Weinberg equilibrium.
Color blindness is an X-linked recessive trait. In a human population at Hardy-Weinberg equilibrium, 9% of the males are color-blind. Calculate the expected percentage of color-blind females and carrier females in this population.
1. Understand X-linked traits in HWE:
2. Identify given information:
3. Calculate expected frequencies in females:
Answer: 0.81% of females will be color-blind, and 16.38% will be carriers.
In a population in HWE, the frequencies of the ABO blood group alleles are:
Calculate the percentage of individuals in the population with Blood Type A and Blood Type O.
For three alleles, the Hardy-Weinberg equation expands to: (p + q + r)² = p² + 2pq + q² + 2pr + 2qr + r² = 1
1. Calculate Blood Type O frequency:
2. Calculate Blood Type A frequency:
In a population of endangered cheetahs, there are 50 breeding males and 10 breeding females. Calculate the effective population size (Ne) and explain the implications for genetic drift.
The formula for effective population size based on unequal sex ratios is: Ne = (4 × Nm × Nf) / (Nm + Nf) Where Nm = number of breeding males, and Nf = number of breeding females.
1. Plug in the values:
Answer: Even though there are 60 breeding adults in the population, the effective population size is only ~33 individuals.
Implications: A highly skewed sex ratio dramatically reduces the effective population size. This small Ne means the cheetah population will suffer from rapid genetic drift, leading to a severe loss of heterozygosity and an increased risk of inbreeding depression.
These questions test if you are blindly applying formulas or genuinely understanding the biological concepts behind them.
Question: In a population of butterflies, there are 400 blue (dominant) and 600 red (recessive) individuals. Calculate the exact frequency of heterozygous individuals.
The Pitfall: Students immediately assume the population is in HWE and calculate q = √0.6 ≈ 0.77, then p = 0.23, and 2pq = 2(0.77)(0.23) = 0.35.
The Reality: The problem does not state the population is in Hardy-Weinberg Equilibrium! If a population is not in HWE, you cannot use √q² to find q, because the frequency of the recessive phenotype doesn't necessarily equal q². You can only calculate allele frequencies directly if you are given the exact number of heterozygotes vs homozygotes. In this scenario, it is impossible to calculate without more information.
Question: In a human population, the frequency of an X-linked recessive allele is 0.2. What is the frequency of males who are heterozygous for this trait?
The Pitfall: Students quickly calculate 2pq = 2 × 0.8 × 0.2 = 0.32.
The Reality: Males are hemizygous (XY) for X-linked traits. They cannot be heterozygous! The frequency of heterozygous males is exactly 0.
Genotype Distribution for X-linked trait in Males (q=0.2)
Question: The frequency of Cystic Fibrosis (a recessive disease) in a population is 1 in 2500. What is the frequency of the recessive allele?
The Pitfall: Students often answer 1/2500 (0.0004).
The Reality: The disease frequency is the frequency of the phenotype or genotype (q²), not the allele (q). You must take the square root! q = √(1/2500) = 1/50 = 0.02. The allele frequency is 2%.
In a population of peppered moths, the relative fitness (w) of the dark morph (melanic) is 1.0, while the relative fitness of the light morph is 0.6. Calculate the Selection Coefficient (s) against the light morph.
1. Understand the terms:
2. Use the formula:
Answer: The selection coefficient against the light morph is 0.4. This means the light morph has a 40% disadvantage in survival/reproduction compared to the dark morph.