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Biology Set-1 MCQ Online Practice Test | Class 12 | Quiz Tweek

Biology Set-1

Practice Important MCQs for Evolution — Biology, Class 12.

Evolution is a core Class 12 Biology chapter because it explains how genetic variation arises and how populations change over time through mechanisms like Hardy–Weinberg equilibrium, natural selection, genetic drift, gene flow, and molecular clock reasoning. These ideas are frequently tested in both board and competitive exams, especially through numerical problems (allele/genotype frequencies, mutation–selection balance) and conceptual questions (speciation, assortative mating/inbreeding effects, overdominance, drift vs selection).

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Q1. In a large, randomly mating population, 4% of individuals show the phenotype of a completely recessive autosomal trait (genotype $aa$ ). Assuming Hardy–Weinberg equilibrium ( $p^2+2pq+q^2=1$ ), calculate allele frequencies $p$ and $q$ and the expected percentage of heterozygous carriers ( $Aa$ ).

(A)

$p=0.9,\ q=0.1,\ \text{carriers}=18\%$

(B)

$p=0.6,\ q=0.4,\ \text{carriers}=48\%$

(C)

$p=0.8,\ q=0.2,\ \text{carriers}=32\%$

(D)

$p=0.2,\ q=0.8,\ \text{carriers}=64\%$

Q2. In a large, randomly mating population with alleles $A$ and $a$ , relative fitnesses are $w_{AA}=1-s,\ w_{Aa}=1,\ w_{aa}=1-t$ with $s=0.2$ and $t=0.5$ . Assuming selection is the only evolutionary force and a stable polymorphism exists due to heterozygote advantage, what is the equilibrium frequency $\hat{p}$ of allele $A$ ?

(A)

$\hat{p}=\dfrac{t}{s+t}=\dfrac{0.5}{0.7}\approx 0.714$

(B)

$\hat{p}=\dfrac{s}{s+t}\approx 0.286$

(C)

$\hat{p}=0.5$

(D)

$\hat{p}=0.2$

Q3. Two isolated populations show average sequence divergence $d=0.12$ substitutions per site in a presumed neutral gene. If the neutral substitution rate per lineage is $\mu=2\times10^{-8}$ substitutions/site/year and the molecular clock applies, estimate the time since their divergence using $t=\dfrac{d}{2\mu}$ .

(A)

$6\times 10^{6}\ \text{years}$

(B)

$0.75\times 10^{6}\ \text{years}$

(C)

$30\times 10^{6}\ \text{years}$

(D)

$3\times 10^{6}\ \text{years}$

Q4. Assertion (A): Genetic drift can cause rapid fixation of neutral alleles and loss of genetic variation in small isolated populations.
Reason (R): Gene flow between populations always inhibits speciation because it homogenizes allele frequencies and prevents divergence.

(A)

Both A and R are true and R is the correct explanation of A.

(B)

Both A and R are true but R is NOT the correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q5. At a locus with alleles $A$ and $a$ , relative fitnesses are $w_{AA}=1,\ w_{Aa}=0.6,\ w_{aa}=0.8$ . For general two-allele selection the internal equilibrium frequency of $A$ is $p^*=\dfrac{w_{aa}-w_{Aa}}{w_{AA}-2w_{Aa}+w_{aa}}$ . Calculate $p^*$ and determine the evolutionary fate of allele $A$ if its initial frequency is $p_0=0.25$ .

(A)

$p^*=\dfrac{1}{3}\approx 0.333$ ; allele $A$ will go to fixation ( $p\to 1$ ).

(B)

$p^*=\dfrac{1}{3}\approx 0.333$ ; allele $A$ will be lost ( $p\to 0$ ).

(C)

$p^*=\dfrac{2}{3}\approx 0.667$ ; allele $A$ will be lost ( $p\to 0$ ).

(D)

$p^*=0.5$ ; allele $A$ will be lost ( $p\to 0$ ).

Q6. In a large randomly mating population at Hardy–Weinberg equilibrium the frequency of the recessive phenotype (homozygous recessive) is $0.16$ . What is the expected frequency of heterozygous carriers? (Use $p^2+2pq+q^2=1$ and heterozygote frequency $2pq$ .)

(A)

$0.36$

(B)

$0.48$

(C)

$0.16$

(D)

$0.64$

Q7. Consider a large randomly mating population with allele $A$ frequency $p=0.6$ and $a$ frequency $q=0.4$ . Genotype relative fitnesses are $w_{AA}=1.0,\; w_{Aa}=0.9,\; w_{aa}=0.7$ . After one generation of viability selection, the allele frequency of $A$ becomes $p'$ . Using $p'=\dfrac{p^2 w_{AA}+ p q w_{Aa}}{\bar w}$ (where $\bar w$ is mean fitness), what is $p'$ approximately?

(A)

$0.600$

(B)

$0.664$

(C)

$0.720$

(D)

$0.637$

Q8. Two species differ by 12 nucleotide substitutions in a 1000 bp neutral gene. Assuming these differences are neutral substitutions and the neutral mutation rate is $\mu=1\times10^{-9}$ substitutions per site per year, estimate their divergence time $t$ in million years using $t=\dfrac{d}{2\mu}$ where $d$ is substitutions per site.

(A)

$6\ \text{million years}$

(B)

$3\ \text{million years}$

(C)

$12\ \text{million years}$

(D)

$0.006\ \text{million years}$

Q9. A single copy of a new dominant beneficial mutation arises in a diploid population with effective size $N_e=500$ and selection coefficient $s=0.001$ . Using approximations: fixation probability of a beneficial allele $\approx 2s$ and neutral fixation probability $\approx 1/(2N_e)$ , which statement is most accurate?

(A)

Selection will almost certainly fix the allele because $2N_e s \gg 1$ .

(B)

Fixation probability of the beneficial allele is essentially equal to the neutral expectation (both $\approx 1/(2N_e)$ ), so selection is negligible.

(C)

Fixation probability of the beneficial allele $\approx 0.002$ (about twice neutral), but genetic drift still plays a major role because $2N_e s \approx 1$ .

(D)

The new allele will be lost with certainty because selection is too weak to matter.

Q10. In a large randomly mating diploid population allele $A$ has initial frequency $p=0.10$ . Genotype fitnesses are additive: $w_{AA}=1+s,\; w_{Aa}=1+\tfrac{s}{2},\; w_{aa}=1$ . After selection the observed allele frequency is $p'=0.12$ . Using $p'=\dfrac{p^2 w_{AA}+ p q w_{Aa}}{\bar w}$ , estimate the selection coefficient $s$ (to two significant figures).

(A)

$0.465$

(B)

$0.0465$

(C)

$0.23$

(D)

$0.02$

Q11. In a large randomly mating population of 1000 individuals the genotype counts are AA = 490, Aa = 360 and aa = 150. Calculate the allele frequency of A and decide whether the population is in Hardy–Weinberg equilibrium. Which statement is correct?

(A)

$p=\dfrac{2\times490+360}{2000}=0.67$ , expected heterozygote frequency $2pq\approx0.44$ , and the population IS in Hardy–Weinberg equilibrium.

(B)

$p=0.67$ , expected heterozygote $2pq\approx0.44$ , the population is NOT in Hardy–Weinberg equilibrium; observed heterozygote frequency ( $360/1000=0.36$ ) is lower — likely indicating directional selection favoring homozygotes.

(C)

$p=0.67$ , expected heterozygote $2pq\approx0.44$ , the population is NOT in Hardy–Weinberg equilibrium; observed heterozygote frequency ( $0.36$ ) shows heterozygote deficiency consistent with inbreeding or assortative mating.

(D)

$p=\dfrac{2\times490+360}{2000}=0.67$ , expected heterozygote $2pq\approx0.44$ , observed heterozygote $0.36$ so the population shows excess heterozygotes relative to Hardy–Weinberg.

Q12. A large randomly mating population has allele frequency $q=0.20$ for a recessive allele $a$ . Genotype fitnesses are $w_{AA}=1,\;w_{Aa}=1,\;w_{aa}=0.80$ . Assuming Hardy–Weinberg proportions before selection, what is the allele frequency of $a$ after one generation of viability selection (approximate to three decimal places)?

(A)

$q' \approx 0.194$

(B)

$q' = 0.200$ (no change)

(C)

$q' \approx 0.206$ (allele increases)

(D)

$q' \approx 0.190$

Q13. Nucleotide divergence between two species is $D=0.12$ substitutions per site. Using the molecular clock formula $T=\dfrac{D}{2r}$ and assuming a neutral substitution rate $r=0.02$ substitutions per site per million years per lineage, the estimated time since their divergence is:

(A)

$1.5\ \text{million years}$

(B)

$6\ \text{million years}$

(C)

$0.6\ \text{million years}$

(D)

$3\ \text{million years}$

Q14. Assertion (A): In a diploid population with alleles $A$ and $a$ , if the heterozygote $Aa$ has higher fitness than both homozygotes (overdominance), both alleles will be maintained at a stable polymorphic equilibrium. Reason (R): If selection coefficients against $AA$ and $aa$ are $s_1$ and $s_2$ respectively (so $w_{AA}=1-s_1,\;w_{Aa}=1,\;w_{aa}=1-s_2$ ), the equilibrium frequency of allele $A$ is $\hat p=\dfrac{s_2}{s_1+s_2}$ .

(A)

Both A and R are true, but R is NOT the correct explanation of A.

(B)

Both A and R are true, and R correctly explains why the polymorphism is maintained (i.e. gives the stable equilibrium $\hat p=\dfrac{s_2}{s_1+s_2}$ ).

(C)

A is true but R is false.

(D)

A is false but R is true.

Q15. A deleterious recessive allele $a$ arises at mutation rate $\mu=10^{-5}$ per generation and homozygous fitness reduction is $s=0.02$ . Under mutation–selection balance for a fully recessive allele, $\hat q\approx\sqrt{\mu/s}$ ; for a fully dominant deleterious allele the equilibrium is $\hat q\approx\mu/s$ . Which pair gives the approximate equilibrium frequency of the allele in (recessive, dominant) cases?

(A)

(recessive $\approx 0.022$ , dominant $\approx 0.0005$ )

(B)

(recessive $\approx 0.0005$ , dominant $\approx 0.022$ )

(C)

(recessive $\approx 0.10$ , dominant $\approx 0.05$ )

(D)

(recessive $\approx 0.0022$ , dominant $\approx 0.005$ )

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