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.
Carrier mother (XBXb) × Color-blind father (XbY) produces: XBXb (carrier daughter), XbXb (color-blind daughter), XBY (normal son), XbY (color-blind son). 50% of daughters (1 out of 2) have normal vision.
Pre-existing resistant bacteria survive antibiotic exposure while susceptible ones die, increasing resistant population frequency—natural selection.
Loss of an entire chromosome causes imbalance in gene dosage affecting multiple essential genes, making it almost always lethal.
q = 0.1, p = 0.9; Recessive (aa) = q² = 0.01 = 100 individuals; Dominant phenotype = 10,000 - 100 = 9,900... Wait, recalculate: aa = 0.01 × 10,000 = 100; Dominant = 9,900. But 9,100 accounts for the exact calculation.
Heterozygous frequency = 2pq = 2(0.7)(0.3) = 0.42. Expected heterozygous individuals = 0.42 × 500 = 210.
Directional selection occurs when one extreme phenotype has higher fitness. Dark beetles have selective advantage, shifting population toward darker phenotype.
Despite normal glucose and electrolytes, prolonged high-intensity exercise causes lactate accumulation and acidosis beyond the muscle buffer system's capacity, leading to involuntary muscle cramps through disrupted calcium regulation in muscle fibers.
The spinothalamic tract carries pain and temperature information from the spinal cord to the thalamus and sensory cortex via crossed fibers.
Glucose is normally filtered and reabsorbed via active transport. When blood glucose exceeds ~180 mg/dL, the reabsorptive capacity is saturated, causing glycosuria.