Evolution Set-5

Q1. A large randomly mating population is initially at Hardy–Weinberg with allele frequencies $p=0.7$ for $A$ and $q=0.3$ for $a$. Viability selection acts on zygotes with fitnesses $W_{AA}=1,\;W_{Aa}=1,\;W_{aa}=0.64$. Calculate the frequency of allele $a$ among survivors after one generation (round to three decimal places).

Q2. Populations P1 and P2 have allele $A$ frequencies $p_1=0.8$ and $p_2=0.2$. Each generation a fraction $m=0.1$ of individuals in P1 are migrants from P2; after migration, random mating occurs in P1. Compute the allele frequency $p_1'$ in P1 after migration and the expected heterozygote frequency $2p_1'(1-p_1')$ (round to three decimal places).

Q3. A breeding population has $N_m=12$ males and $N_f=48$ females. Using $N_e=\dfrac{4N_mN_f}{N_m+N_f}$, compute the effective population size $N_e$ (to one decimal) and then compute the fraction of heterozygosity retained after one generation $H_{t+1}/H_t=1-\dfrac{1}{2N_e}$ (give the retention factor to three decimals).

Q4. Assertion (A): Under neutral theory, the rate of molecular evolution (substitutions per site per generation) equals the neutral mutation rate and is independent of effective population size $N_e$.
Reason (R): Although larger populations produce more new neutral mutations per generation, the fixation probability of a neutral mutation is lower in larger populations in such a way that the product (new neutral mutations × fixation probability) equals the neutral mutation rate, making substitution rate independent of $N_e$.

Q5. Characters (three binary sites) are scored for four taxa as follows:
A: 0 0 0
B: 0 1 0
C: 1 1 1
D: 1 0 1
Using parsimony (minimum number of character changes), which unrooted tree topology is most parsimonious for these data?

Q6. In a large randomly mating population under Hardy–Weinberg equilibrium, 160 out of 1000 individuals exhibit a recessive phenotype. If $q^2=\dfrac{160}{1000}=0.16$, what is the expected number of heterozygotes in the population? (Use heterozygote frequency $2pq$)

Q7. In a population with allele frequencies $p=0.7$ (A) and $q=0.3$ (a), relative fitnesses are $w_{AA}=1.0$, $w_{Aa}=0.9$, $w_{aa}=0.6$. Calculate the allele frequency $p'$ of A after one generation of selection using

Q8. A newly arisen beneficial allele has selection coefficient $s=0.02$ and the effective population size is $N_e=100$. Using the rule-of-thumb that selection is effective when $2N_es\gg1$, which statement best describes the expected evolutionary dynamics for this allele? (Compute $2N_es$.)

Q9. Assertion (A): The rate of neutral molecular evolution (neutral substitution rate) equals the neutral mutation rate $\mu$ per generation.
Reason (R): Populations with larger effective size $N_e$ have higher neutral substitution rates because they produce more new neutral mutations per generation (about $2N_e\mu$).

Q10. A deleterious recessive allele is maintained by mutation–selection balance with mutation rate $\mu=2\times10^{-6}$ and selection coefficient against homozygote $s=0.02$. Using $q\approx\sqrt{\mu/s}$, the equilibrium frequency is $q_0\approx\sqrt{\dfrac{2\times10^{-6}}{0.02}}$. If either (i) $\mu$ is halved or (ii) $s$ is doubled, which management action produces the larger decrease in the equilibrium frequency?

Q11. In a large randomly mating population at Hardy–Weinberg equilibrium the observed frequency of heterozygotes for a diallelic locus is $0.42$. What are the allele frequencies $p$ and $q$?

Q12. A mainland population has allele $A$ with frequency $p=0.6$. Four diploid founders are randomly sampled to establish a new island population (so a sample of 8 alleles). What is the probability that the frequency of allele $A$ among the founders is at least $0.75$? (Compute $P(k\ge6)$ for $k\sim\text{Binomial}(n=8,p=0.6)$ and give answer to three significant figures.)

Q13. Assertion (A): When gene trees are incongruent with the species tree, it always indicates horizontal gene transfer.
Reason (R): Incomplete lineage sorting (ILS) occurs when ancestral polymorphism persists through rapid successive speciation events and different alleles become fixed in different descendant lineages, producing discordant gene trees.

Q14. A 3000 bp gene shows 12 nucleotide differences between two species. Assuming a neutral substitution rate $\mu=1\times10^{-9}$ substitutions per site per year per lineage, estimate the divergence time $t$ (years) between the species using $t=\dfrac{D}{2\mu}$ where $D$ = proportion of differing sites.

Q15. Assertion (A): If the heterozygote has the highest fitness (overdominance), both alleles are maintained at a stable internal equilibrium and fixation of either allele is prevented.
Reason (R): Sympatric speciation can occur without geographic isolation, for example via polyploidy in plants or via disruptive selection combined with assortative mating.