the fur color in a population of dogs is controlled by two alleles black b and tan b the black allele exhibits complete dominance if the phenotype of

HESI A2

Biology HESI A2 Practice Test

1. The fur color in a population of dogs is controlled by two alleles: black (B) and tan (b). The black allele exhibits complete dominance. If the phenotype of the dog is tan fur, what is its genotype?

A. BB
B. Bb
C. bb
D. Not enough information

Correct answer: C

Rationale: The correct answer is C: "bb." Since tan fur is the recessive trait controlled by the allele 'b,' the only possible genotype for a dog with tan fur is 'bb.' If a dog had the genotype 'BB' or 'Bb,' the dominant trait (black fur) would be expressed, not tan fur. Choice D, 'Not enough information,' is incorrect because based on the information provided, we can deduce the genotype of a dog with tan fur.

2. Which is true of the term Phenotype?

A. It is the genetic code
B. Phenotype determines genotype
C. Genotype determines phenotype
D. Genotype, epigenetic factors, and environmental factors determine phenotype

Correct answer: D

Rationale: The term "phenotype" refers to the observable characteristics or traits of an organism, which are determined by a combination of genotype (genetic makeup), epigenetic factors (modifications to gene expression), and environmental factors (external influences and experiences). Genotype, epigenetic factors, and environmental factors all play a role in determining an individual's phenotype, making option D the correct statement. Options A, B, and C are incorrect because the phenotype is not the genetic code, the phenotype does not determine the genotype, and it is the genotype that influences the phenotype, not the other way around.

3. What happens to glucose during glycolysis?

A. Its energy is entirely lost.
B. It splits into molecules of pyruvic acid.
C. It is stored in NADH.
D. It joins with molecules of citric acid.

Correct answer: B

Rationale: During glycolysis, glucose undergoes a series of enzymatic reactions in the cytoplasm of the cell, resulting in its breakdown into two molecules of pyruvic acid. This process also generates ATP and NADH as energy carriers. Choice A is incorrect because glucose is not entirely lost, but rather converted into other molecules. Choice C is incorrect because NADH is a product of glycolysis, not a storage form for glucose. Choice D is incorrect as glucose does not join with molecules of citric acid during glycolysis, but rather in subsequent stages of cellular respiration.

4. Which of the following is not found within a bacterial cell?

A. mitochondria
B. DNA
C. vesicles
D. ribosome

Correct answer: A

Rationale: The correct answer is A: mitochondria. Mitochondria are not found in bacterial cells. Bacterial cells lack membrane-bound organelles like mitochondria, which are commonly found in eukaryotic cells. Choices B, C, and D are all components that can be found within a bacterial cell. Bacterial cells contain DNA as their genetic material, vesicles for various cellular functions, and ribosomes for protein synthesis. Therefore, the presence of mitochondria is the distinguishing factor that is not found in bacterial cells, making option A the correct answer.

5. Huntington’s disease is carried on the dominant allele. In a situation where two heterozygous parents have the disease, what percentage of their offspring are predicted to be disease-free?

A. 0%
B. 25%
C. 50%
D. 100%

Correct answer: B

Rationale: In this scenario, both parents are heterozygous for Huntington's disease, meaning each carries one dominant allele (representing the disease) and one recessive allele (representing no disease). When they have offspring, there is a 25% chance that each child will inherit two recessive alleles, making them disease-free. The Punnett square for two heterozygous parents (Hh x Hh) yields a 25% probability of offspring being homozygous recessive (hh) and therefore disease-free. Choice A (0%) is incorrect because there is a possibility of disease-free offspring. Choice C (50%) is incorrect as it represents the likelihood of being a carrier. Choice D (100%) is incorrect as all offspring will not be disease-free in this scenario.