Question
▶️Answer/Explanation
Sickle cell anemia is caused by a recessive allele (let’s use “s” for sickle and “S” for normal). Heterozygous individuals (Ss) have a mild form (called sickle cell trait). The genotypic ratio from two heterozygous parents (Ss × Ss) is:
- SS = normal (unaffected)
- Ss = sickle cell trait (mild form)
- ss = sickle cell disease (severe)
Punnett square results:
- 25% SS (unaffected)
- 50% Ss (mild form)
- 25% ss (disease)
Let’s analyze the options:
A. 0% – Incorrect: There’s a chance (25%) of being unaffected.
B. 25% – Correct: 1 in 4 chances of having a child with genotype SS (completely unaffected).
C. 75% – Incorrect: 75% includes both SS and Ss, but only SS is truly unaffected.
D. 100% – Incorrect: Not all offspring will be unaffected.
Question
The pedigree chart for a family shows the inheritance of Huntington’s disease.
▶️Answer/Explanation
Huntington’s disease is autosomal dominant, meaning anyone who inherits one copy of the mutant gene will develop the disease. It affects both males and females equally because it’s not linked to sex chromosomes.
Evaluation of Options:
A. Incorrect: Huntington’s disease is not sex-linked. Both males and females can be affected equally.
B. Incorrect: The fact that individuals IV-1 and IV-3 are unaffected doesn’t prove whether the disease is sex-linked or not.
C. Correct: Individual III-3 is a female affected by Huntington’s disease. If it were sex-linked, females wouldn’t typically be affected unless they had two copies of the mutant allele. This means the disease is autosomal, not sex-linked.
D. Incorrect: Huntington’s disease is dominant, so there are no carriers. If an individual is unaffected, they simply don’t carry the allele.
Question
A farmer was interested in the yield produced by his crop. The graph displays the number of seeds of different masses in the crop yield. Which statement could clarify the inheritance of this trait?
A. Inheritance of seed mass follows Mendel’s laws for a monohybrid cross.
B. This is an example of discontinuous variation.
C. Two linked genes are responsible for seed mass in this plant.
D. Polygenic inheritance results in subtle phenotypic differences.
▶️Answer/Explanation
Answer: D. Polygenic inheritance results in subtle phenotypic differences.
Explanation:
When a trait shows a range of phenotypes (like different seed masses), it usually indicates continuous variation, which is typically the result of polygenic inheritance. Polygenic traits are influenced by multiple genes, leading to a wide variety of phenotypes. In contrast, Mendel’s laws apply to discrete traits with clear categories.
Evaluation of Options:
A. Incorrect: Mendel’s laws apply to traits with discrete categories, not a range like seed mass. A monohybrid cross (involving one gene) typically results in distinct phenotypes, not a spectrum of variations.
B. Incorrect: Discontinuous variation refers to traits that have distinct, non-overlapping categories (e.g., tall vs. short plants). Seed mass shows continuous variation (a range), so this option is not correct.
C. Incorrect: While linked genes can influence traits, linked genes usually don’t produce the kind of continuous variation seen in this scenario. Two linked genes might lead to a few distinct phenotypes, not a spectrum of mass values.
D. Correct: Polygenic inheritance refers to traits influenced by multiple genes, which results in a range of phenotypes. This matches the description of seed mass, where there’s a wide variety of seed sizes.
Question
Non-syndromic hearing loss and deafness (DFNB1) is an inherited form of deafness in humans. The pedigree chart illustrates the transmission of DFNB1 within a family.
Where is the DFNB1 allele found in family members with this condition?
A. On the Y chromosome
B. On the X chromosome
C. On one autosome only
D. On a pair of autosomes
▶️Answer/Explanation
Answer: D. On a pair of autosomes
Explanation:
When a trait shows a range of phenotypes (like different seed masses), it suggests continuous variation, which is usually the result of polygenic inheritance. In polygenic traits, multiple genes contribute to the phenotype, leading to a wide spectrum of variations.
Evaluation of Options:
A. Incorrect: Mendel’s laws apply to traits with distinct categories (e.g., flower color). Seed mass shows continuous variation, not discrete categories, so this option doesn’t fit.
B. Incorrect: Discontinuous variation involves traits with clear, separate categories (e.g., red or white flowers), while seed mass has a range of sizes, which is continuous, not discontinuous.
C. Incorrect: Linked genes can influence traits, but they don’t typically cause a range of phenotypes like seed mass. Linked genes usually produce a few distinct phenotypes rather than a spectrum.
D. Correct: Polygenic inheritance results in continuous variation because multiple genes influence the trait, leading to a range of phenotypes like the different seed masses in the graph.
Question
Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They produced offspring with the phenotypes black short-haired, black long-haired, white short-haired and white long-haired in the ratio 9:3:3:1. A different cross produced offspring with phenotypes in the ratio $1: 1: 1: 1$. What were the genotypes of the parents in the second cross?
A. $\mathrm{BbSs} \times \mathrm{BbSs}$
B. $\mathrm{BBSs} \times \mathrm{BbSS}$
C. $\mathrm{BbSs} \times$ bbss
D. $\mathrm{bbSS} \times \mathrm{BBsS}$
Answer/Explanation
Answer: C. $\mathrm{BbSs} \times$ bbss
Explanation:
Mendelian Genetics: In this case, we are dealing with two traits (coat color and hair length) that follow a dihybrid inheritance pattern.
- B = Black coat color (dominant), b = white coat color (recessive)
- S = short hair (dominant), s = long hair (recessive)
The 9:3:3:1 ratio from the first cross indicates that the parents were heterozygous for both traits (BbSs x BbSs).
The 1:1:1:1 ratio from the second cross suggests that each parent must be heterozygous for both traits, but crossed with a recessive homozygous individual. This produces offspring with a 1:1:1:1 ratio, typical of a test cross.
Evaluating the Options:
A. $\mathrm{BbSs} \times \mathrm{BbSs}$
Incorrect: This would produce a 9:3:3:1 ratio for the offspring (as seen in the first cross), not a 1:1:1:1 ratio.
B. $\mathrm{BBSs} \times \mathrm{BbSS}$
Incorrect: These crosses would not result in a 1:1:1:1 ratio. The offspring would have an uneven distribution of phenotypes, and the genotypes in the parents do not match the conditions needed for the 1:1:1:1 result.
C. $\mathrm{BbSs} \times \mathrm{bbss}$
Correct: A BbSs individual crossed with a bbss individual would produce offspring in a 1:1:1:1 ratio, since the BbSs parent can produce BS, Bs, bS, and bs gametes, and the bbss parent can only produce bs gametes. This cross is consistent with the expected result of 1:1:1:1.
D. $\mathrm{bbSS} \times \mathrm{BBsS}$
Incorrect: This cross would not result in a 1:1:1:1 ratio. The offspring would not have the correct combination of genotypes and phenotypes to match the 1:1:1:1 pattern.
Question
An individual is heterozygous for two linked genes \(\frac{AB}{ab}\)
To investigate the frequency of crossing over, a test cross is carried out between the individual and
another that is homozygous recessive for both genes. What are the possible recombinants in the
offspring of this cross?
Answer/Explanation
Answer: C
Explanation:
Parent 1 genotype = \( \frac{AB}{ab} \) heterozygous for two linked genes
Parent 2 genotype = \( \frac{ab}{ab} \) homozygous recessive
In a test cross, you are crossing a heterozygous individual \( \frac{AB}{ab} \) with a fully recessive one \( \frac{ab}{ab} \) to check for recombination between linked genes.
Possible gametes:
– Without recombination: From \( \frac{AB}{ab} \), you get AB and ab.
– With recombination: Crossing over between the two genes will produce Ab and aB gametes.
Offspring Genotypes:
The test cross will produce these recombinant offspring genotypes:
\[
\frac{AB}{ab} \times \frac{ab}{ab}
\]
Recombinant offspring: \( \frac{Ab}{ab} \) and \( \frac{aB}{ab} \).
Evaluation of Options:
A. \( \frac{Ab}{ab} \) and \( \frac{Ab}{ab} \) Incorrect – Both are the same, no \( \frac{aB}{ab} \)
B.\( \frac{AB}{ab} \) and \( \frac{Ab}{aB} \) Incorrect – \( \frac{Ab}{aB} \) is not a valid result from a test cross
C.\( \frac{Ab}{ab} \) and \( \frac{aB}{ab} \) Correct recombinant offspring
D.\( \frac{AA}{aa} \) and \( \frac{BB}{bb} \) Incorrect – This format is wrong for genotypes