GRE Biology Subject Test Practice Questions

Question 1:

Which of the following cellular components is primarily responsible for the synthesis of integral membrane proteins and proteins destined for secretion outside the cell?

A. Free ribosomes

B. Smooth endoplasmic reticulum

C. Golgi apparatus

D. Rough endoplasmic reticulum

Correct Answer: D. Rough endoplasmic reticulum

The rough endoplasmic reticulum (RER) is studded with ribosomes that synthesize proteins destined for secretion, insertion into membranes, or delivery to certain organelles. These proteins enter the RER lumen or membrane during synthesis for folding and modification. Free ribosomes, in contrast, synthesize proteins that remain in the cytosol. The smooth endoplasmic reticulum is involved in lipid synthesis and detoxification, while the Golgi apparatus modifies, sorts, and packages proteins and lipids.

Question 2:

During DNA replication, which enzyme is responsible for synthesizing short RNA primers that provide a starting point for DNA polymerase?

A. DNA ligase

B. Helicase

C. Primase

D. Topoisomerase

Correct Answer: C. Primase

Primase is a type of RNA polymerase that synthesizes short RNA primers, which are essential for DNA polymerase to initiate DNA synthesis as it can only add nucleotides to an existing 3′-OH group. DNA ligase joins Okazaki fragments on the lagging strand. Helicase unwinds the DNA double helix, and topoisomerase relieves supercoiling ahead of the replication fork, preventing tangling.

Question 3:

In a population of fruit flies, the allele for red eyes (R) is dominant over the allele for white eyes (r). If a population is in Hardy-Weinberg equilibrium and the frequency of the white-eyed allele (r) is 0.4, what is the expected frequency of heterozygous (Rr) individuals?

A. 0.16

B. 0.36

C. 0.48

D. 0.64

Correct Answer: C. 0.48

In Hardy-Weinberg equilibrium, p + q = 1 (allele frequencies) and p² + 2pq + q² = 1 (genotype frequencies). Given q (frequency of r) = 0.4, then p (frequency of R) = 1 – 0.4 = 0.6. The frequency of heterozygous individuals (Rr) is represented by 2pq. Therefore, 2 * 0.6 * 0.4 = 0.48. This demonstrates your understanding of population genetics principles.

Question 4:

Which of the following terms describes the process where a single ancestral species diversifies into multiple new species, each adapted to a specific ecological niche?

A. Genetic drift

B. Allopatric speciation

C. Adaptive radiation

D. Convergent evolution

Correct Answer: C. Adaptive radiation

Adaptive radiation is the rapid diversification of a single ancestral species into many new species, each filling a different ecological niche. This often occurs when new resources become available or new challenges arise. Genetic drift is random changes in allele frequencies. Allopatric speciation involves geographic isolation, and while it can lead to new species, it doesn’t specifically describe the broad diversification pattern. Convergent evolution is when unrelated species evolve similar traits due to similar environmental pressures.

Question 5:

A scientist observes that a particular insect species exhibits a Type III survivorship curve. What does this pattern typically indicate about the life history strategy of this species?

A. High mortality rates early in life, with few individuals surviving to old age.

B. Relatively constant mortality rates throughout the lifespan.

C. Low mortality rates early in life, with most individuals surviving to old age.

D. A balanced sex ratio with equal numbers of males and females surviving.

Correct Answer: A. High mortality rates early in life, with few individuals surviving to old age.

A Type III survivorship curve is characterized by very high mortality rates among the young, followed by much lower mortality rates for those few individuals that survive early life. This strategy is common in species that produce many offspring but provide little parental care. Type I curves show low early mortality and Type II show constant mortality. This question tests your knowledge of population ecology and life history strategies.

Question 6:

The adrenal medulla secretes hormones that primarily prepare the body for “fight or flight” responses. Which of the following hormones are the main secretions of the adrenal medulla?

A. Cortisol and aldosterone

B. Insulin and glucagon

C. Epinephrine and norepinephrine

D. Thyroxine and calcitonin

Correct Answer: C. Epinephrine and norepinephrine

The adrenal medulla secretes catecholamines, primarily epinephrine (adrenaline) and norepinephrine (noradrenaline), which are crucial for the body’s short-term stress response, often referred to as the “fight or flight” response. Cortisol and aldosterone are secreted by the adrenal cortex. Insulin and glucagon regulate blood sugar from the pancreas, while thyroxine and calcitonin are thyroid hormones. This question assesses your understanding of the endocrine system.

Question 7:

In C3 plants, the enzyme RuBisCO can bind with both CO2 and O2. When RuBisCO binds with O2 instead of CO2, it initiates a process known as photorespiration. Which of the following conditions would most likely favor photorespiration?

A. High CO2 concentration and low O2 concentration

B. Low CO2 concentration and high O2 concentration

C. Low light intensity and low temperatures

D. High water availability and high humidity

Correct Answer: B. Low CO2 concentration and high O2 concentration

Photorespiration occurs when RuBisCO oxygenates RuBP instead of carboxylating it, consuming ATP and releasing CO2 without producing sugar. This process is favored under conditions of high oxygen and low carbon dioxide concentrations, often occurring in hot, dry environments when stomata close to conserve water, limiting CO2 uptake while O2 builds up. High CO2 and low O2 would favor photosynthesis. The other options are less directly linked to RuBisCO’s substrate affinity.

Question 8:

During early animal development, the process of gastrulation results in the formation of:

A. A solid ball of cells called a morula.

B. A hollow ball of cells called a blastula.

C. Three distinct germ layers (ectoderm, mesoderm, endoderm).

D. The first differentiated organ systems.

Correct Answer: C. Three distinct germ layers (ectoderm, mesoderm, endoderm).

Gastrulation is a critical stage in embryonic development where the single-layered blastula is reorganized into a multilayered structure called the gastrula. This process establishes the three primary germ layers: the ectoderm, mesoderm, and endoderm, from which all tissues and organs will eventually develop. The morula is an earlier stage, and the blastula precedes gastrulation. Organ systems differentiate much later from these germ layers.

Question 9:

A bacterial cell acquires a plasmid containing genes for antibiotic resistance from another bacterium through a direct cell-to-cell contact mechanism. This process is known as:

A. Transformation

B. Transduction

C. Conjugation

D. Transposition

Correct Answer: C. Conjugation

Conjugation is a form of horizontal gene transfer in bacteria that involves the direct transfer of genetic material (often a plasmid) from one bacterial cell to another through physical contact, typically via a pilus. Transformation is the uptake of naked DNA from the environment, while transduction involves bacteriophages transferring DNA. Transposition refers to the movement of DNA segments within a genome. This question tests your knowledge of bacterial genetics.

Question 10:

Which type of enzyme inhibitor binds to the active site of an enzyme and directly competes with the substrate, often having a similar structure to the natural substrate?

A. Non-competitive inhibitor

B. Uncompetitive inhibitor

C. Competitive inhibitor

D. Allosteric inhibitor

Correct Answer: C. Competitive inhibitor

A competitive inhibitor is a molecule that structurally resembles the enzyme’s natural substrate and binds reversibly to the enzyme’s active site, thereby preventing the substrate from binding. This competition can be overcome by increasing the substrate concentration. Non-competitive inhibitors bind to a different site, altering enzyme conformation. Uncompetitive inhibitors bind only to the enzyme-substrate complex, and allosteric inhibitors bind to a regulatory site, which can include non-competitive inhibition.

You’re doing great! Each question you tackle builds your knowledge and confidence. Keep practicing and reviewing these concepts to solidify your understanding for the GRE Biology Subject Test!