A teacher notes that coastal cities have smaller daily and seasonal temperature swings than inland cities at the same latitude. Which explanation of this pattern is most accurate at the molecular level?

Energy goes into breaking hydrogen bonds between water molecules before the molecules move faster.

Water's nonpolar molecules reflect most incoming solar radiation, so the ocean warms and cools very little.

Ionic bonds between dissolved salts and water absorb the heat that would otherwise warm the air over the coast.

Heat absorbed by water breaks the polar covalent bonds within water molecules, storing energy as free hydrogen and oxygen.

The compound 2,4-dinitrophenol (DNP) makes the inner mitochondrial membrane freely permeable to H⁺ ions. It was once sold as a weight-loss drug before being banned for causing dangerous overheating. In cells treated with DNP, which outcome is expected?

Electron transport and O₂ consumption continue or even increase, but ATP synthesis falls and more energy is released as heat.

ATP synthesis increases, because protons flow back into the matrix faster than normal.

Electron transport stops, so O₂ consumption falls to zero and no heat is produced.

Glycolysis stops, because the leaking protons enter the cytosol and denature glycolytic enzymes.

A guinea pig with black fur (dominant to white) is crossed with a white guinea pig. Their litters include 7 black and 6 white offspring. What can be concluded about the black parent?

It is heterozygous, since each white offspring got a recessive allele from it

It is homozygous recessive, because nearly half the offspring were white

Its genotype cannot be determined without crossing it to another black guinea pig

It is homozygous dominant, because most of the offspring had black fur

A microbiology class grows E. coli in a medium in which sulfate labeled with radioactive ³⁵S is the only sulfur source. After several generations, the cells are broken open and their macromolecules are separated. In which class of macromolecule would most of the ³⁵S be found?

Proteins, in cysteine and methionine side chains

Phospholipids, in the polar head groups of the membrane

Polysaccharides, in the glycosidic bonds of glycogen

Nucleic acids, in the sugar–phosphate backbone

Cultured cells are treated with a drug that prevents microtubules from attaching to kinetochores. The cells condense their chromosomes but then remain arrested with sister chromatids still joined, never entering anaphase. Which explanation best accounts for this arrest?

The spindle checkpoint keeps the anaphase-promoting complex off until all kinetochores attach, so cohesin stays intact.

The G1 checkpoint detects the missing microtubules and stops the cells before they can begin DNA replication.

The cells enter anaphase, but the chromatids cannot separate because cyclin levels drop too low to drive their separation.

The cells arrest in S phase, because spindle microtubules are required to separate the replicating DNA strands.

A student's food web for a meadow shows arrows from grass to grasshoppers, grass to rabbits, grasshoppers to shrews, rabbits to foxes, and shrews to foxes. Which question can this food web NOT answer?

What share of the fox's energy comes from rabbits versus shrews

In which direction energy moves from one organism to the next

Which organisms function as secondary consumers in this meadow

Which organism in the web is the producer that captures solar energy

OSAT Biological Sciences (110)

Free OSAT Biological Sciences (110) Practice Questions

10 sample questions for the OSAT Biological Sciences (110), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Structures and Processes of Cells and Organisms, Ecosystem Energy Flow, Interactions, and Dynamics, Heredity and Evolution.

Question 1 · Structures and Processes of Cells and Organisms
A teacher notes that coastal cities have smaller daily and seasonal temperature swings than inland cities at the same latitude. Which explanation of this pattern is most accurate at the molecular level?
- A.Water's nonpolar molecules reflect most incoming solar radiation, so the ocean warms and cools very little.
- B.Energy goes into breaking hydrogen bonds between water molecules before the molecules move faster.
- C.Ionic bonds between dissolved salts and water absorb the heat that would otherwise warm the air over the coast.
- D.Heat absorbed by water breaks the polar covalent bonds within water molecules, storing energy as free hydrogen and oxygen.
Show answer and explanation
Answer: B. Energy goes into breaking hydrogen bonds between water molecules before the molecules move faster.
Water's high specific heat comes from hydrogen bonding: energy must disrupt intermolecular hydrogen bonds before temperature rises, and that energy is released as bonds re-form during cooling. The covalent O–H bonds inside each molecule are not broken at ordinary temperatures. Salts are not the cause, since large freshwater lakes moderate climate too, and water is strongly polar, not nonpolar.
Question 2 · Structures and Processes of Cells and Organisms
The compound 2,4-dinitrophenol (DNP) makes the inner mitochondrial membrane freely permeable to H⁺ ions. It was once sold as a weight-loss drug before being banned for causing dangerous overheating. In cells treated with DNP, which outcome is expected?
- A.ATP synthesis increases, because protons flow back into the matrix faster than normal.
- B.Electron transport stops, so O₂ consumption falls to zero and no heat is produced.
- C.Electron transport and O₂ consumption continue or even increase, but ATP synthesis falls and more energy is released as heat.
- D.Glycolysis stops, because the leaking protons enter the cytosol and denature glycolytic enzymes.
Show answer and explanation
Answer: C. Electron transport and O₂ consumption continue or even increase, but ATP synthesis falls and more energy is released as heat.
DNP is an uncoupler: protons leak back into the matrix without passing through ATP synthase, so the gradient no longer drives ATP production. The electron transport chain keeps running, often faster, and the energy of the gradient is lost as heat, which explains the overheating. ATP synthesis requires protons to flow through ATP synthase, so faster leakage reduces, rather than increases, ATP output.
Question 3 · Ecosystem Energy Flow, Interactions, and Dynamics
In a field, hawks eat snakes, the snakes eat mice, and the mice eat seeds. Using the 10% rule, about how many kilograms of seeds are needed to produce 2 kg of new hawk biomass?
- A.2,000 kg
- B.20,000 kg
- C.20 kg
- D.200 kg
Show answer and explanation
Answer: A. 2,000 kg
Working backward, 2 kg of hawk requires about 20 kg of snake, which requires about 200 kg of mice, which requires about 2,000 kg of seeds, three tenfold steps in all. The values of 20 kg and 200 kg stop one or two steps short of the producers. The value of 20,000 kg adds an extra trophic step that is not in this food chain.
Question 4 · Ecosystem Energy Flow, Interactions, and Dynamics
A rabbit population grows logistically with an intrinsic rate of increase r = 0.4 per year and a carrying capacity K = 1,000. Using dN/dt = rN(K − N)/K, what is the population's growth rate when N = 250?
- A.75 rabbits per year
- B.300 rabbits per year
- C.100 rabbits per year
- D.25 rabbits per year
Show answer and explanation
Answer: A. 75 rabbits per year
dN/dt = 0.4 × 250 × (1,000 − 250)/1,000 = 100 × 0.75 = 75 rabbits per year. A value of 100 is the exponential rate rN, ignoring the limit set by K. A value of 25 incorrectly multiplies by N/K instead of (K − N)/K, and 300 multiplies r by the unused capacity (K − N) instead of by N.
Question 5 · Heredity and Evolution
A guinea pig with black fur (dominant to white) is crossed with a white guinea pig. Their litters include 7 black and 6 white offspring. What can be concluded about the black parent?
- A.It is homozygous recessive, because nearly half the offspring were white
- B.Its genotype cannot be determined without crossing it to another black guinea pig
- C.It is homozygous dominant, because most of the offspring had black fur
- D.It is heterozygous, since each white offspring got a recessive allele from it
Show answer and explanation
Answer: D. It is heterozygous, since each white offspring got a recessive allele from it
This is a test cross: the white parent (bb) can contribute only b, so any white offspring must have received b from the black parent, making it Bb, and the roughly 1:1 ratio fits Bb × bb. A homozygous dominant parent would produce only black offspring, and a homozygous recessive parent would be white. Crossing to the homozygous recessive is exactly what reveals the unknown genotype, whereas a cross with another black animal would be less informative.
Question 6 · Structures and Processes of Cells and Organisms
A microbiology class grows E. coli in a medium in which sulfate labeled with radioactive ³⁵S is the only sulfur source. After several generations, the cells are broken open and their macromolecules are separated. In which class of macromolecule would most of the ³⁵S be found?
- A.Proteins, in cysteine and methionine side chains
- B.Phospholipids, in the polar head groups of the membrane
- C.Polysaccharides, in the glycosidic bonds of glycogen
- D.Nucleic acids, in the sugar–phosphate backbone
Show answer and explanation
Answer: A. Proteins, in cysteine and methionine side chains
Sulfur enters proteins as part of the amino acids cysteine and methionine, which is why ³⁵S was historically used to label protein. Nucleic acid backbones contain phosphorus rather than sulfur, phospholipid head groups contain phosphate (and sometimes nitrogen), and glycogen contains only carbon, hydrogen, and oxygen.
Question 7 · Structures and Processes of Cells and Organisms
Cultured cells are treated with a drug that prevents microtubules from attaching to kinetochores. The cells condense their chromosomes but then remain arrested with sister chromatids still joined, never entering anaphase. Which explanation best accounts for this arrest?
- A.The spindle checkpoint keeps the anaphase-promoting complex off until all kinetochores attach, so cohesin stays intact.
- B.The G1 checkpoint detects the missing microtubules and stops the cells before they can begin DNA replication.
- C.The cells enter anaphase, but the chromatids cannot separate because cyclin levels drop too low to drive their separation.
- D.The cells arrest in S phase, because spindle microtubules are required to separate the replicating DNA strands.
Show answer and explanation
Answer: A. The spindle checkpoint keeps the anaphase-promoting complex off until all kinetochores attach, so cohesin stays intact.
Unattached kinetochores send a signal that keeps the anaphase-promoting complex inactive, so separase stays inhibited and cohesin remains intact; this is the M-phase (spindle) checkpoint. The cells have already condensed chromosomes, so they passed G1 and S. Falling cyclin is a result of anaphase-promoting complex activity, which is exactly what the checkpoint prevents.
Question 8 · Ecosystem Energy Flow, Interactions, and Dynamics
A student's food web for a meadow shows arrows from grass to grasshoppers, grass to rabbits, grasshoppers to shrews, rabbits to foxes, and shrews to foxes. Which question can this food web NOT answer?
- A.In which direction energy moves from one organism to the next
- B.What share of the fox's energy comes from rabbits versus shrews
- C.Which organisms function as secondary consumers in this meadow
- D.Which organism in the web is the producer that captures solar energy
Show answer and explanation
Answer: B. What share of the fox's energy comes from rabbits versus shrews
A typical food web shows only who eats whom; it does not show how much energy or biomass flows along each link, so the relative importance of rabbits versus shrews in the fox's diet cannot be determined. The web does show the producer (grass), identifies shrews and foxes as consumers of herbivores, and uses arrows to indicate the direction of energy flow.
Question 9 · Ecosystem Energy Flow, Interactions, and Dynamics
A population of 2,000 field mice has a per capita birth rate of 0.05 per individual per month and a per capita death rate of 0.02 per individual per month. Assuming exponential growth and no migration, how many individuals will the population gain in the next month?
- A.60
- B.140
- C.40
- D.100
Show answer and explanation
Answer: A. 60
The intrinsic rate of increase is r = b − d = 0.05 − 0.02 = 0.03 per month, and dN/dt = rN = 0.03 × 2,000 = 60 individuals. Using the birth rate alone gives 100, adding the two rates gives 140, and using only the death rate gives 40, none of which reflect net growth.
Question 10 · Heredity and Evolution
Plant height is controlled by three unlinked genes with additive effects. A plant with no height-increasing alleles is 30 cm tall, and each height-increasing allele adds 4 cm. Two plants heterozygous at all three genes are crossed. What fraction of the offspring are expected to be 42 cm tall?
- A.3/32
- B.15/64
- C.5/16
- D.1/64
Show answer and explanation
Answer: C. 5/16
A 42 cm plant has (42 − 30)/4 = 3 height-increasing alleles. Each offspring receives 6 alleles at these loci, each with a 1/2 chance of being height-increasing, so the probability of exactly 3 is 20/64 = 5/16. The fraction 1/64 is the chance of either extreme phenotype, 6/64 = 3/32 corresponds to exactly 1 or 5 alleles, and 15/64 corresponds to exactly 2 or 4 alleles.

Keep practicing for the OSAT Biological Sciences (110)

TTP Prep has 200 OSAT Biological Sciences (110) practice questions with explanations, timed and mock-exam modes, 61 flashcards, interactive lessons, a printable cheat sheet and a readiness score. Start free, no card needed.

Take a free Quick 10 See everything in the OSAT Biological Sciences (110) prep