- A plant heterozygous for both flower color (Rr) and seed shape (Ss) undergoes self-pollination. What is the probability of obtaining offspring that are homozygous for both traits?
a) 1/16
b) 2/16
c) 4/16
d) 9/16
Answer: (a) 1/16
Explanation: In a dihybrid cross (RrSs x RrSs), the probability of getting offspring that are homozygous for both traits (RRSS, RRss, rrSS, or rrss) is 1/16 for each specific combination. There are four possible homozygous combinations (RRSS, RRss, rrSS, rrss), and each has a probability of 1/16.
- In humans, the gene for red-green color blindness is located on the X chromosome. A man with normal vision marries a woman who is a carrier for the condition. What is the chance that their daughter will be colorblind?
a) 0%
b) 25%
c) 50%
d) 100%
Answer: (a) 0%
Explanation: Since the daughter inherits an X chromosome from both her parents, she would need to inherit two mutated X chromosomes (one from each parent) to be colorblind. Since the father has normal vision (XY), he can only pass a normal X chromosome. Therefore, their daughters cannot be colorblind, though they could be carriers.
- Which of the following patterns of inheritance is characterized by one gene masking the expression of another gene?
a) Pleiotropy
b) Codominance
c) Epistasis
d) Polygenic inheritance
Answer: (c) Epistasis
Explanation: Epistasis occurs when the expression of one gene is affected or completely masked by another gene. An example is coat color in mice, where one gene determines whether pigment is produced, while another gene determines the color of the pigment. If the pigment gene is inactive, the color gene has no effect.
- If two carriers of sickle cell anemia (Ss) have children, what is the probability that their offspring will have sickle cell disease?
a) 0%
b) 25%
c) 50%
d) 75%
Answer: (b) 25%
Explanation: Sickle cell anemia is caused by a recessive allele (s). If both parents are carriers (Ss), there is a 25% chance that the child will inherit two copies of the recessive allele (ss) and thus have the disease, a 50% chance the child will be a carrier (Ss), and a 25% chance the child will be normal (SS).
- Which of the following diseases is caused by a single-gene mutation?
a) Turner syndrome
b) Klinefelter syndrome
c) Huntington’s disease
d) Down syndrome
Answer: (c) Huntington’s disease
Explanation: Huntington’s disease is caused by a mutation in a single gene located on chromosome 4. It is an autosomal dominant disorder, meaning that only one copy of the mutated gene is necessary to cause the disease. The other conditions listed are chromosomal disorders caused by abnormalities in chromosome number, not by single-gene mutations.
- A woman with type A blood (IAIA) marries a man with type B blood (IBIB). What will be the blood type of their offspring?
a) A
b) B
c) AB
d) O
Answer: (c) AB
Explanation: Since the woman is homozygous for blood group A (IAIA) and the man is homozygous for blood group B (IBIB), all of their offspring will inherit one IA allele from the mother and one IB allele from the father, resulting in blood type AB.
- Which of the following is an example of a sex-limited trait?
a) Male-pattern baldness
b) Hemophilia
c) Lactation in females
d) Red-green color blindness
Answer: (c) Lactation in females
Explanation: A sex-limited trait is one that is expressed in only one sex, even though the gene for the trait may be present in both sexes. Lactation is a sex-limited trait because only females produce milk, even though both males and females have the genes for it. Male-pattern baldness is sex-influenced, and hemophilia and color blindness are sex-linked traits.
- What is the probability of producing a homozygous recessive offspring in a cross between two heterozygous individuals (Aa x Aa)?
a) 0%
b) 25%
c) 50%
d) 75%
Answer: (b) 25%
Explanation: In a cross between two heterozygous individuals (Aa x Aa), the offspring can inherit either the dominant allele (A) or the recessive allele (a) from each parent. The genotypic ratio of the offspring is 1:2:1 — one homozygous dominant (AA), two heterozygous (Aa), and one homozygous recessive (aa). Therefore, there is a 25% chance of producing a homozygous recessive offspring (aa).
- In a cross between a red-flowered plant and a white-flowered plant, the F1 generation has pink flowers. This is an example of:
a) Complete dominance
b) Codominance
c) Incomplete dominance
d) Epistasis
Answer: (c) Incomplete dominance
Explanation: In incomplete dominance, the heterozygous phenotype is an intermediate between the two homozygous phenotypes. In this case, the pink flowers in the F1 generation are a result of neither the red nor white alleles being completely dominant, so the resulting phenotype is a blend of the two.
- If a man with blood group AB marries a woman with blood group O, which blood groups are possible in their children?
a) A, B, AB, O
b) A, B, O
c) A, B
d) A, B, AB
Answer: (c) A, B
Explanation: The man has blood group AB (IAIB), and the woman has blood group O (ii). The children will inherit either the IA or IB allele from the father and the i allele from the mother, resulting in either blood group A (IAi) or blood group B (IBi). Blood groups AB and O are not possible in their children.
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