Govt. Exams
Entrance Exams
Phenotypic plasticity is the ability of a single genotype to produce different phenotypes in response to environmental variation (e.g., hydrangea flower color).
Natural selection is the non-random differential reproduction of organisms with traits that increase fitness. Organisms with advantageous traits leave more offspring.
Silent mutation occurs in the third position of a codon and does not change the amino acid due to degeneracy of genetic code. No phenotypic effect.
Color blindness is an X-linked recessive trait. To find the percentage of color-blind sons, we need to perform a cross between a carrier mother and a normal father, then analyze the male offspring.
Step 1: Identify the genotypes
Mother (carrier): \(X^B X^b\) — has one normal allele (\(X^B\)) and one color-blind allele (\(X^b\))
Father (normal): \(X^B Y\) — has the normal allele on his X chromosome
Step 2: Set up the Punnett square
Cross all maternal alleles with all paternal alleles:
$$\begin{array}{c|cc}
& X^B & Y \\
\hline
X^B & X^B X^B & X^B Y \\
X^b & X^B X^b & X^b Y \\
\end{array}$$
Step 3: Identify offspring genotypes and phenotypes
The four possible offspring are:
- \(X^B X^B\) (daughter, normal)
- \(X^B Y\) (son, normal)
- \(X^B X^b\) (daughter, carrier)
- \(X^b Y\) (son, color blind)
Step 4: Calculate percentage of color-blind sons
Among all sons (male offspring only):
- \(X^B Y\) (normal): 1 out of 2 sons
- \(X^b Y\) (color blind): 1 out of 2 sons
Answer: 50% of their sons will be color blind (Option C)
Haemophilia A (clotting factor VIII deficiency) is X-linked recessive. Affected males (XᵃY) pass the allele to all daughters but no sons.
X-linked recessive traits show more affected males (XᵃY) than females because males need only one copy of recessive allele on their single X chromosome.
Recessive phenotype = 360/1000 = 0.36 = q². Therefore q = √0.36 = 0.60. This represents the frequency of recessive allele.
In Hardy-Weinberg equation, p and q represent frequencies of two alleles in a population. p² and q² are homozygous frequencies; 2pq is heterozygous frequency.
The correct hierarchical order is: Nucleotide (basic unit) → DNA (polymer of nucleotides) → Gene (segment of DNA) → Chromosome (condensed DNA with proteins).
The 1:1 ratio indicates a testcross result (Rr × rr). The round seed parent must be heterozygous (Rr) to produce wrinkled (rr) offspring.
