Govt. Exams
Entrance Exams
F1 is Rr (all round). F2: 1 RR : 2 Rr : 1 rr. Among round seeds (3 total), only 1 is homozygous RR, so 1/3.
Frequency of Aa = 2pq where p = 0.6 and q = 0.4; 2(0.6)(0.4) = 0.48
Silent mutations occur in the third position of codons (wobble base), changing the DNA sequence but not the amino acid due to codon degeneracy.
Female XwXw × Male XWY produces: XWXw (red female), XwXw (white female), XWY (red male), XwY (white male). 50% of total offspring are white-eyed.
Multiple lines of evidence support evolution: homologous structures indicate common ancestry, vestigial organs show evolutionary remnants, and fossil records show gradual changes over time.
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.
