Read the given passage and answer the following questions:
During post-industrialisation period, the tree trunks became dark due to industrial smoke and soots. Under this condition the white-winged moth did not survive due to predators, dark-winged or melanised moth survived. Before industrialisation set in, thick growth of almost white-coloured lichen covered the trees - in that background the white winged moth survived but the dark-coloured moth were picked out by predators. Do you know that lichens can be used as industrial pollution indicators? They will not grow in areas that are polluted. Hence, moths that were able to camouflage themselves, i.e., hide in the background, survived. This understanding is supported by the fact that in areas where industrialisation did not occur e.g., in rural areas, the count of melanic moths was low. This showed that in a mixed population, those that can better-adapt, survive and increase in population size. Remember that no variant is completely wiped out. Similarly, excess use of herbicides, pesticides, etc., has only resulted in selection of resistant varieties in a much lesser time scale. This is also true for microbes against which we employ antibiotics or drugs against eukaryotic organisms/cell. Hence, resistant organisms/cells are appearing in a time scale of months or years and not centuries. These are examples of evolution by anthropogenic action. This also tells us that evolution is not a directed process in the sense of determinism. It is a stochastic process based on chance events in nature and chance mutation in the organisms.
Predation changed the allele frequencies in the moth population described in the passage. Given below are two statements about such frequencies: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In a diploid population with allele frequencies p (for A) and q (for a), p2 + 2pq + q2 = 1.
Reason (R): In a diploid population with allele frequencies p (for A) and q (for a), the frequency of heterozygous (Aa) individuals is q2.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Assertion (A) is true but Reason (R) is false.
The Hardy-Weinberg principle deals with the frequencies of the alleles of a gene in a population. For a gene with two alleles, A and a, p is the frequency of A and q is the frequency of a, and p + q = 1.
Assertion (A): In a diploid population with allele frequencies p (for A) and q (for a), p2 + 2pq + q2 = 1. — True. The chance that an individual gets A on both chromosomes is p × p = p2 (AA). In the same way, aa is q2 and Aa is 2pq. These three genotypes make up the whole population, so p2 + 2pq + q2 = 1. This is the binomial expansion of (p + q)2.
Reason (R): In a diploid population with allele frequencies p (for A) and q (for a), the frequency of heterozygous (Aa) individuals is q2. — False. q2 is the frequency of the homozygous recessive (aa) individuals, who carry allele a on both chromosomes. A heterozygote can get A from one parent and a from the other, or a from one and A from the other, so its frequency is 2pq, not q2.
A is true and R is false, so Option (3) is the answer.
NCERT Extract:
Hardy-Weinberg Principle – “… In a diploid, p and q represent the frequency of allele A and allele a. The frequency of AA individuals in a population is simply p2. This is simply stated in another ways, i.e., the probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p2. Similarly of aa is q2, of Aa 2pq. Hence, p2+2pq+q2=1. This is a binomial expansion of (p+q)2.”