Entrance Exams
Govt. Exams
Recessive frequency = 400/50,000 = 0.008 = q². Thus q = √0.008 ≈ 0.089. Therefore p = 1 - 0.089 ≈ 0.911 ≈ 0.9 (Option should read 0.9, but closest match is A: 0.8 or recalculate: q² = 0.008, q ≈ 0.09, p ≈ 0.91). If recessive phenotype frequency is 0.008, then q = 0.089, p = 0.911.
UAA is a stop codon. A missense mutation creating a stop codon is called a nonsense mutation, resulting in premature protein termination.
Cross: XAXa (female) × XaY (male). Male offspring: 1/2 XAY (red) and 1/2 XaY (white). Therefore, 50% white-eyed males.
Microevolution involves allele frequency changes within a population over relatively short periods. Macroevolution is long-term divergence into new species.
A color-blind female must be homozygous recessive (XbXb) because she has only two X chromosomes. She has two copies of the recessive allele.
Ligers are viable hybrids but sterile, representing post-zygotic isolation where hybrids have reduced fitness or viability.
Segregation of alleles and recombination during meiosis create genetic variation in offspring, leading to phenotypic differences.
Antibiotic resistance emerges through natural selection. Resistant bacteria survive antibiotic treatment and reproduce, increasing resistance frequency in the population.
For AaBb × AaBb: probability of aa = 1/4, probability of bb = 1/4. Combined: 1/4 × 1/4 = 1/16
Variation provides the raw material for natural selection. Without variation, there would be no differential reproductive success.