In ABO blood group genetics, if both parents have AB blood group, what blood groups are IMPOSSIBLE in their children?
Answer: A
AB parents (IAIB × IAIB) produce IAIA, IAIB, and IBIB only. O group (ii) requires two recessive alleles, which neither parent possesses.
Q.22Medium
A gene has four alleles in a population. If three alleles have frequencies of 0.4, 0.3, and 0.2, what is the frequency of the fourth allele?
Answer: A
According to Hardy-Weinberg principle, total allele frequencies must equal 1. Therefore, 1 - (0.4 + 0.3 + 0.2) = 0.1
Q.23Medium
In a population at Hardy-Weinberg equilibrium, if the frequency of recessive allele (q) = 0.3, what is the frequency of heterozygous individuals?
Answer: A
If q = 0.3, then p = 0.7. Heterozygous frequency = 2pq = 2(0.7)(0.3) = 0.42
Q.24Medium
A tall plant (genotype unknown) is crossed with a dwarf plant. All offspring are tall. What is the most likely genotype of the tall parent?
Answer: A
All offspring being tall when crossed with dwarf (tt) indicates the tall parent is TT. If it were Tt, approximately half offspring would be dwarf.
Q.25Medium
Which of the following best represents anagenesis in evolution?
Answer: B
Anagenesis refers to phyletic evolution or linear evolution where one species gradually transforms into another without branching.
Advertisement
Q.26Medium
A person is heterozygous for cystic fibrosis (Ff). Their partner is homozygous normal (FF). What is the probability of having an affected child?
Answer: A
Cross: Ff × FF produces FF and Ff offspring. All are phenotypically normal. Probability of affected (ff) child = 0%
Q.27Medium
In Darwin's theory of natural selection, 'variation' is important because it:
Answer: A
Variation provides the raw material for natural selection. Without variation, there would be no differential reproductive success.
Q.28Medium
A dihybrid cross (AaBb × AaBb) is performed. What is the probability of getting the recessive phenotype for both traits?
Answer: A
For AaBb × AaBb: probability of aa = 41, probability of bb = 41. Combined: 41 × 41 = 161
Q.29Medium
Antibiotic resistance in bacteria develops rapidly due to:
Answer: B
Antibiotic resistance emerges through natural selection. Resistant bacteria survive antibiotic treatment and reproduce, increasing resistance frequency in the population.
Q.30Medium
Which genetic concept explains why children of the same parents can have different traits despite inheriting genes from both?
Answer: B
Segregation of alleles and recombination during meiosis create genetic variation in offspring, leading to phenotypic differences.
Q.31Medium
A new species can be defined by reproductive isolation from other species. Which reproductive barrier is exemplified by lions and tigers producing sterile offspring (ligers)?
Answer: D
Ligers are viable hybrids but sterile, representing post-zygotic isolation where hybrids have reduced fitness or viability.
Q.32Medium
In human genetics, color blindness (red-green) is X-linked recessive. A color-blind woman has how many copies of the color-blindness allele?
Answer: B
A color-blind female must be homozygous recessive (XbXb) because she has only two X chromosomes. She has two copies of the recessive allele.
Q.33Medium
Microevolution differs from macroevolution primarily in that microevolution:
Answer: B
Microevolution involves allele frequency changes within a population over relatively short periods. Macroevolution is long-term divergence into new species.
Q.34Medium
A heterozygous red-eyed Drosophila female (XAXa) is crossed with a white-eyed male (XaY). What percentage of male offspring will have white eyes?
Answer: B
Cross: XAXa (female) × XaY (male). Male offspring: 21 XAY (red) and 21 XaY (white). Therefore, 50% white-eyed males.
Q.35Medium
A point mutation in a gene's coding sequence results in UAA instead of CAA. This type of mutation is called:
Answer: C
UAA is a stop codon. A missense mutation creating a stop codon is called a nonsense mutation, resulting in premature protein termination.
Q.36Medium
In a test cross with a heterozygous individual (Aa), if 40% of offspring show the dominant phenotype, what might this indicate?
Answer: C
In a normal test cross (Aa × aa), expected ratio is 1:1 (50% each). 40% dominant suggests some homozygous dominant (AA) offspring are dying, indicating lethal homozygous dominant allele.
Q.37Medium
Which of the following represents the correct sequence of speciation in peripatric speciation?
Answer: A
Peripatric speciation occurs when a small population is geographically isolated, genetic drift acts strongly, leading to reproductive isolation and speciation.
Q.38Medium
A recessive genetic disorder appears in 1% of a population. What is the frequency of the recessive allele (q)?
Answer: B
If q² = 0.01, then q = √0.01 = 0.1. This is the frequency of the recessive allele.
Q.39Medium
Which mechanism of evolution is most likely responsible for the loss of eyes in blind cave fish?
Answer: B
In cave environments, eyes provide no selective advantage. Neutral mutations affecting eye development accumulate through genetic drift, resulting in eye loss.
Q.40Medium
In Drosophila, if the coefficient of inbreeding (F) increases, what is the expected change in heterozygosity?
Answer: A
Inbreeding increases homozygosity and decreases heterozygosity. The relationship is: Ht = H0(1-F), where heterozygosity decreases with increasing F.
