Q1
The F2 phenotypic ratio of a dihybrid cross is:
A1:1:1:1
B3:1
C9:3:3:1
D1:2:1
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Answer: C (9:3:3:1). Independent assortment of two gene pairs gives 9:3:3:1 in F2.
Q2
The test-cross ratio of a dihybrid (RrYy Γ rryy) is:
A3:1
B9:3:3:1
C1:1:1:1
D1:2:1
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Answer: C (1:1:1:1). The heterozygote makes 4 gamete types equally; the recessive parent makes only ry.
Q3
How many types of gametes can be formed by genotype AaBbCc?
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Answer: B (8). 2^n where n = 3 heterozygous pairs β 8 gametes.
Q4
Linked genes:
AAre on different chromosomes
BTend to be inherited together
CAssort independently
DAlways give 9:3:3:1
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Answer: B (Tend to be inherited together). Genes on the same chromosome stay together unless separated by crossing over.
Q5
Crossing over takes place between:
ACentromeres
BNon-sister chromatids of homologous chromosomes
CNon-homologous chromosomes
DSister chromatids
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Answer: B (Non-sister chromatids of homologous chromosomes). It occurs between non-sister chromatids during prophase I of meiosis.
Q6
Crossing over occurs during:
AInterphase
BMetaphase I
CAnaphase II
DProphase I (pachytene)
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Answer: D (Prophase I (pachytene)). Chiasmata form and exchange occurs in pachytene of prophase I.
Q7
Haemophilia A is due to deficiency of clotting factor:
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Answer: B (VIII). Haemophilia A β factor VIII; haemophilia B (Christmas disease) β factor IX.
Q8
A carrier woman marries a normal man. The chance that their son will be haemophiliac is:
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Answer: C (50%). Sons receive either X^H or X^h from the mother: 50% affected.
Q9
A haemophiliac man marries a normal (non-carrier) woman. Their daughters will be:
AAll normal non-carriers
BAll haemophiliac
CAll carriers
D50% carriers
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Answer: C (All carriers). Every daughter receives the father's X^h and mother's X^H β all carriers.
Q10
Sex-linked recessive disorders are more common in males because males are:
AHemizygous for X-linked genes
BHomozygous
CPolyploid
DHeterozygous
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Answer: A (Hemizygous for X-linked genes). Males have one X; a single recessive allele is expressed.