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Georgia Unified Biology study materials
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Georgia Unified Biology practice questions

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Question 1Biological inquiry and interpretation of evidence

A researcher predicts that tomato plants given a nitrogen-enriched fertilizer will produce more fruit than plants given a standard fertilizer. In this study, the amount of fruit produced is best described as which of the following?

  1. The hypothesis
  2. A controlled variable
  3. The dependent variable Correct answer.
  4. The independent variable

Explanation

The dependent variable is the measured outcome that responds to the manipulation, which here is fruit yield. The fertilizer type is the factor the researcher deliberately changes, so it is the independent variable rather than the measured result.

Question 2Biological inquiry and interpretation of evidence

A student measures the mass of the same sample five times and obtains nearly identical values, but the balance was never zeroed and reads 2.0 g too high. How should these measurements be described?

  1. Both accurate and precise
  2. Precise but not accurate Correct answer.
  3. Accurate but not precise
  4. Neither accurate nor precise

Explanation

Precision refers to how closely repeated measurements agree with one another, while accuracy refers to how close they are to the true value. Because the readings cluster tightly but are systematically offset by the unzeroed balance, they are precise but inaccurate.

Question 3Biological inquiry and interpretation of evidence

Which of the following best distinguishes a scientific hypothesis from a scientific theory?

  1. A hypothesis is a testable proposed explanation, while a theory is a well-supported explanatory framework built from extensive evidence Correct answer.
  2. A hypothesis is always correct once tested, while a theory remains an untested guess
  3. A hypothesis is a broad framework, while a theory is a single testable prediction
  4. A hypothesis applies only to biology, while a theory applies only to physics

Explanation

In science, a hypothesis is a testable proposed explanation for a specific observation, whereas a theory is a broad, well-substantiated explanation supported by many lines of evidence. The common usage of theory as a mere guess is incorrect in scientific contexts.

Question 4Biological inquiry and interpretation of evidence

A scientist repeats an experiment several times and obtains consistent results. Which conclusion is most justified?

  1. The findings are reliable and reproducible under the tested conditions Correct answer.
  2. The results now apply to all organisms everywhere
  3. The hypothesis has been proven true beyond any doubt
  4. The experiment has no remaining sources of error

Explanation

Reproducible results indicate reliability, meaning the procedure yields consistent outcomes under the tested conditions. Science does not establish absolute proof, and replication does not eliminate error or guarantee universal applicability.

Question 5Biological inquiry and interpretation of evidence

A graph plots plant height on the y-axis against days of growth on the x-axis. Which statement correctly identifies the placement of these variables?

  1. Days of growth is the independent variable on the x-axis, and height is the dependent variable on the y-axis Correct answer.
  2. Both variables are dependent and should share the x-axis
  3. Days of growth is the dependent variable because it is measured over time
  4. Height is the independent variable on the y-axis, and days of growth is the dependent variable on the x-axis

Explanation

By convention, the independent variable is placed on the x-axis and the dependent variable on the y-axis. Height responds to elapsed time, so time is independent and height is dependent.

Question 6Biological inquiry and interpretation of evidence

A student notices that mold grows on bread left in a warm, humid place but not on bread stored in a dry, cool place. Which of the following is the most appropriate next step in the scientific process?

  1. Formulate a testable hypothesis and design a controlled experiment to test it Correct answer.
  2. Conclude immediately that warmth and moisture cause mold growth
  3. Publish the observation as a confirmed scientific law
  4. Discard the observation because it was not made in a laboratory

Explanation

An observation generates questions that should be framed as a testable hypothesis and examined through a controlled experiment. Reaching a firm causal conclusion from a single uncontrolled observation is premature.

Question 7Biochemistry

Which property of water is most directly responsible for its ability to moderate temperature in living organisms and large bodies of water?

  1. Its low specific heat
  2. Its inability to form hydrogen bonds
  3. Its nonpolar covalent bonds
  4. Its high specific heat Correct answer.

Explanation

Water has a high specific heat because hydrogen bonds must absorb or release substantial energy before water molecules change temperature. This allows water and water-rich organisms to resist large temperature swings, helping maintain homeostasis.

Question 8Cell structure, membranes and division

Which structure is found in a typical plant cell but NOT in a typical animal cell?

  1. Mitochondrion
  2. Nucleus
  3. Ribosome
  4. Cell wall made of cellulose Correct answer.

Explanation

Plant cells have a rigid cell wall composed primarily of cellulose that surrounds the plasma membrane, providing structural support. Animal cells lack a cell wall, relying on the cytoskeleton and extracellular matrix instead.

Question 9Genetics, inheritance and molecular information

Meiosis differs from mitosis in that meiosis produces

  1. Two genetically identical diploid cells
  2. One diploid cell with double the chromosomes
  3. Two haploid cells identical to the parent
  4. Four genetically varied haploid cells Correct answer.

Explanation

Meiosis consists of two divisions that produce four haploid gametes, each with half the chromosome number of the parent. Crossing over and independent assortment generate genetic variation, distinguishing meiosis from the identical-diploid products of mitosis.

Question 10Metabolism

An enzyme lowers the activation energy of a reaction. Which statement best describes how this affects the reaction?

  1. It increases the reaction rate without being consumed in the reaction Correct answer.
  2. It changes the overall free-energy difference between reactants and products
  3. It supplies the energy needed to make an endergonic reaction exergonic
  4. It permanently binds to the product to stop the reaction

Explanation

Enzymes are biological catalysts that lower activation energy, speeding up reactions without being consumed. They do not change the net free-energy difference (delta G) between reactants and products; they only make the reaction reach equilibrium faster.

Question 11Physiology

Which feature allows a steroid hormone to pass directly through the plasma membrane?

  1. Its lipid (hydrophobic) nature lets it dissolve through the phospholipid bilayer Correct answer.
  2. Its large size lets it push proteins aside
  3. Its negative charge attracts it to the membrane interior
  4. Its attachment to a carbohydrate tag opens a channel

Explanation

Steroid hormones are lipid-soluble (hydrophobic), so they can dissolve through the hydrophobic interior of the phospholipid bilayer without a transport protein. Once inside, they typically bind intracellular receptors that act on the DNA to regulate gene expression.

Question 12Ecology

In an energy pyramid, only about 10 percent of the energy at one trophic level is transferred to the next. Most of the remaining energy is

  1. Converted directly into new biomass at the next level
  2. Lost as heat, largely through cellular respiration and metabolic processes Correct answer.
  3. Stored permanently in the soil
  4. Recycled back to producers as light energy

Explanation

At each trophic level, organisms use most of the energy they consume for metabolism and lose much of it as heat through cellular respiration. Only about 10 percent is stored as biomass available to the next level, which limits food chain length.

Question 13Evolution

Which of the following is a necessary condition for natural selection to occur in a population?

  1. All individuals are genetically identical
  2. Heritable variation exists and affects survival or reproduction Correct answer.
  3. The environment never changes
  4. Every individual produces the same number of offspring

Explanation

Natural selection requires heritable variation among individuals and differences in reproductive success linked to that variation. Individuals with advantageous heritable traits leave more offspring, shifting the population's traits over generations.

Question 14Cell structure, membranes and division

Which type of chemical reaction joins two monosaccharide molecules to form a disaccharide such as maltose, with the simultaneous release of a water molecule?

  1. Condensation (dehydration synthesis) Correct answer.
  2. Hydrolytic cleavage
  3. Decarboxylation
  4. Oxidative phosphorylation

Explanation

A condensation reaction (dehydration synthesis) combines two monosaccharide monomers by removing a hydroxyl group from one and a hydrogen atom from the other, forming a glycosidic bond and releasing water. In contrast, hydrolysis breaks this bond by adding water back into the structure. This fundamental reaction builds complex carbohydrates like starch and glycogen.

Question 15Cell structure, membranes and division

Why do phospholipids naturally assemble into a bilayer when placed in an aqueous biological environment?

  1. Their hydrophilic phosphate heads face the water while their hydrophobic fatty acid tails face inward. Correct answer.
  2. Their hydrophobic glycerol heads face outward while their hydrophilic hydrocarbon tails form the core.
  3. Covalent disulfide bonds link adjacent phospholipid molecules to stabilize the membrane surface.
  4. Electrostatic repulsion between phosphate groups forces the molecules to stack end-to-end linearly.

Explanation

Phospholipids are amphipathic molecules possessing hydrophilic (polar) phosphate-containing heads and hydrophobic (nonpolar) fatty acid tails. In water, hydrophobic interactions cause the fatty acid tails to cluster away from water molecules, while the charged phosphate heads interact favorably with the polar aqueous interior and exterior.

Question 16Cell structure, membranes and division

Hemoglobin consists of two alpha and two beta polypeptide chains associated with iron-containing heme prosthetic groups. Which level of protein structural organization does this arrangement represent?

  1. Primary structure
  2. Secondary structure
  3. Tertiary structure
  4. Quaternary structure Correct answer.

Explanation

Quaternary structure refers to the assembly of multiple distinct polypeptide subunits into a single functional oligomeric protein complex. Hemoglobin is a classic tetramer composed of four polypeptide subunits (two alpha and two beta chains) non-covalently bound together. Single polypeptides that fold completely achieve tertiary structure, but multi-subunit assembly is required for quaternary classification.

Question 17Cell structure, membranes and division

According to the induced-fit model of enzyme action, how does an enzyme lower the activation energy of a biochemical reaction?

  1. The substrate permanently alters its chemical identity prior to contacting the enzyme surface.
  2. The enzyme active site changes shape slightly upon substrate binding to strain specific chemical bonds. Correct answer.
  3. The enzyme increases the ambient kinetic temperature inside the active site pocket.
  4. The enzyme undergoes irreversible covalent breakdown to supply free energy to the reaction products.

Explanation

The induced-fit model postulates that initial binding of the substrate induces a conformational change in the enzyme's catalytic active site. This shape adjustment positions catalytic residues precisely and places mechanical strain on specific substrate bonds, stabilizing the transition state and lowering activation energy. Enzymes remain structurally intact and chemically unchanged at the conclusion of the reaction.

Question 18Cell structure, membranes and division

What primary molecular event causes an enzyme to lose its catalytic activity when exposed to temperatures significantly above its optimum?

  1. Peptide bonds between amino acids are rapidly cleaved by thermal hydrolysis.
  2. Non-covalent hydrogen and ionic bonds stabilizing the tertiary structure are disrupted. Correct answer.
  3. Ribosomes re-absorb the enzyme to recycle its constituent mRNA codons.
  4. Substrate molecules are converted into competitive inhibitors by atmospheric oxidation.

Explanation

Excess thermal kinetic energy breaks weak non-covalent bonds, including hydrogen bonds, hydrophobic interactions, and ionic bridges, that maintain the enzyme's specific three-dimensional tertiary conformation. Once the native geometry of the active site is lost, the substrate can no longer bind, causing denaturation. High physiological temperatures typically do not hydrolyze strong covalent peptide bonds directly.

Question 19Cell structure, membranes and division

In a double-stranded DNA molecule, if thymine accounts for 32% of the total nitrogenous bases, what percentage of the bases is expected to be guanine?

  1. 18% Correct answer.
  2. 32%
  3. 36%
  4. 64%

Explanation

According to Chargaff's rules, double-stranded DNA exhibits 1:1 base pairing where adenine equals thymine (A = T) and guanine equals cytosine (G = C). If thymine is 32%, adenine is also 32%, making the A + T total 64%. The remaining 36% consists equally of guanine and cytosine, meaning guanine represents 36% / 2 = 18%.

Question 20Cell structure, membranes and division

Which RNA molecule possesses an anticodon loop and carries a specific amino acid to the ribosome during protein synthesis?

  1. Transfer RNA (tRNA) Correct answer.
  2. Messenger RNA (mRNA)
  3. Ribosomal RNA (rRNA)
  4. Small nuclear RNA (snRNA)

Explanation

Transfer RNA (tRNA) acts as the physical adapter molecule between genetic code and polypeptide synthesis. Each tRNA molecule has a specific three-base anticodon at one end that base-pairs with a complementary mRNA codon, and an amino acid attachment site at its 3' CCA terminal end. This ensures the correct sequential incorporation of amino acids into the growing polypeptide chain.

Question 21Cell structure, membranes and division

Which chemical bonds within an ATP molecule release a large amount of free energy upon hydrolysis to power cellular work?

  1. Phosphoanhydride bonds linking adjacent phosphate groups Correct answer.
  2. Phosphodiester bonds connecting ribose to the alpha phosphate
  3. Glycosidic bonds connecting adenine to the ribose sugar
  4. Hydrogen bonds established between complementary purine rings

Explanation

ATP contains two high-energy phosphoanhydride bonds linking its three phosphate groups (beta-to-alpha and gamma-to-beta). Hydrolysis of the terminal phosphoanhydride bond yields ADP, inorganic phosphate, and approximately 30.5 kJ/mol of standard free energy. This energy release occurs because hydrolysis relieves significant electrostatic repulsion between the adjacent negatively charged oxygen atoms.

Question 22Cell structure, membranes and division

What is the physiological purpose of the extensive inward foldings (cristae) of the inner mitochondrial membrane?

  1. To segregate glycolysis enzymes from the pyruvate dehydrogenase complex
  2. To increase the total surface area available for electron transport chains and ATP synthase complexes Correct answer.
  3. To store calcium carbonate crystals for osmotic pressure regulation
  4. To allow ribosomes direct physical passage into the intermembrane space

Explanation

Cristae are deep invaginations of the inner mitochondrial membrane that dramatically expand its total membrane surface area. This expanded surface accommodates thousands of copies of electron transport chain complexes (I-IV) and ATP synthase molecules per mitochondrion. A higher density of these respiratory complexes directly maximizes the rate of cellular oxidative phosphorylation.

Question 23Cell structure, membranes and division

Where do the light-independent reactions (Calvin cycle) of photosynthesis take place within a plant cell?

  1. Across the thylakoid lumen
  2. Inside the stroma of the chloroplast Correct answer.
  3. Within the intermembrane space of the chloroplast
  4. Along the outer envelope membrane of the chloroplast

Explanation

The Calvin cycle (dark or light-independent reactions) occurs in the stroma, the fluid-filled space surrounding the thylakoid stacks within chloroplasts. The stroma contains soluble enzymes, including ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO), along with dissolved CO2, ATP, and NADPH produced by the light reactions. In contrast, light-dependent photophosphorylation takes place directly in and across the thylakoid membranes.

Question 24Cell structure, membranes and division

Which organelle is primarily responsible for synthesizing lipids, metabolizing carbohydrates, and detoxifying drugs and poisons in human liver cells?

  1. Rough endoplasmic reticulum
  2. Smooth endoplasmic reticulum Correct answer.
  3. Lysosome
  4. Nucleolus

Explanation

The smooth endoplasmic reticulum (SER) lacks surface-bound ribosomes and specializes in lipid and steroid hormone synthesis, carbohydrate metabolism, and drug detoxification via cytochrome P450 enzymes. In muscle cells, a specialized SER called the sarcoplasmic reticulum also regulates intracellular calcium storage. In contrast, the rough ER focuses on synthesizing proteins destined for secretion or membrane insertion.

Question 25Cell structure, membranes and division

A newly synthesized secretory protein moves from the rough endoplasmic reticulum to the cis face of the Golgi apparatus. What primary modification and sorting step occurs within the Golgi cisternae?

  1. Transcription of genomic exons into introns
  2. Post-translational glycosylation and packaging into transport vesicles Correct answer.
  3. Synthesis of ribosomal RNA for export through nuclear pores
  4. Uncoupling of electron transport from ATP synthesis

Explanation

The Golgi apparatus acts as the cell's post-translational processing, sorting, and dispatch center. As proteins progress from the cis to the trans cisternae, oligosaccharide chains are sequentially trimmed or added (glycosylation), and the proteins are sorted and packaged into secretory vesicles or lysosomes. These vesicles bud from the trans-Golgi network to reach their specific cellular destinations.

Question 26Cell structure, membranes and division

Which sedimentation coefficient combination correctly characterizes intact ribosomes found in prokaryotic bacteria compared to those in eukaryotic cytosol?

  1. 70S in prokaryotes (composed of 50S and 30S subunits); 80S in eukaryotes (composed of 60S and 40S subunits) Correct answer.
  2. 80S in prokaryotes (composed of 50S and 30S subunits); 70S in eukaryotes (composed of 60S and 40S subunits)
  3. 60S in prokaryotes (composed of 40S and 20S subunits); 90S in eukaryotes (composed of 60S and 30S subunits)
  4. 50S in prokaryotes (composed of 30S and 20S subunits); 100S in eukaryotes (composed of 70S and 30S subunits)

Explanation

Bacterial (prokaryotic) ribosomes are 70S particles composed of a large 50S subunit and a small 30S subunit. Eukaryotic cytosolic ribosomes are 80S particles composed of a large 60S subunit and a small 40S subunit. This structural difference provides the biological basis for selective antibiotic action, where drugs like tetracycline or chloramphenicol inhibit 70S bacterial ribosomes without affecting host 80S cytosolic ribosomes.

Question 27Cell structure, membranes and division

What is the primary function of the nucleolus located inside the eukaryotic nucleus?

  1. To synthesize ribosomal RNA (rRNA) and assemble precursor ribosomal subunits Correct answer.
  2. To translate messenger RNA into functional cytoplasmic enzymes
  3. To replicate mitochondrial circular plasmid DNA during G1 phase
  4. To break down damaged nuclear envelope proteins via acidic proteolysis

Explanation

The nucleolus is a dense non-membrane-bound subnuclear structure organized around chromosomal nucleolar organizer regions (NORs). It is the site where RNA polymerase I transcribes ribosomal DNA into precursor rRNA transcripts, which are then processed and assembled with imported ribosomal proteins to form 40S and 60S ribosomal subunit precursors before export to the cytoplasm.

Question 28Cell structure, membranes and division

Which cytoskeletal component consists of hollow cylinders assembled from alpha- and beta-tubulin dimers, forming the mitotic spindle and the structural core of cilia and flagella?

  1. Microfilaments (actin filaments)
  2. Microtubules Correct answer.
  3. Intermediate filaments (keratin)
  4. Collagen fibrils

Explanation

Microtubules are the thickest cytoskeletal fibers (approximately 25 nm in diameter) composed of alpha- and beta-tubulin heterodimers polymerizing into hollow tubes. They play critical roles in maintaining cell shape, providing tracks for organelle motor proteins (kinesin and dynein), segregating chromosomes as mitotic spindle fibers, and driving the 9+2 axoneme beating of eukaryotic cilia and flagella.

Question 29Cell structure, membranes and division

How does cholesterol modulate the physical state and fluidity of animal cell membranes across varying temperatures?

  1. It acts as a bidirectional fluidity buffer, preventing excessive fluidity at warm temperatures and preventing solidification at low temperatures. Correct answer.
  2. It dissolves transmembrane proteins to accelerate lateral lipid migration during heat shock.
  3. It permanently solidifies the outer leaflet to make animal cells impervious to osmotic water movement.
  4. It binds peripheral carbohydrates to replace the phospholipid bilayer with a rigid peptidoglycan wall.

Explanation

Cholesterol functions as a membrane fluidity buffer due to its steroid ring structure intercalated between phospholipid hydrocarbon tails. At high temperatures (such as 37°C), cholesterol restricts the movement of phospholipid fatty acid chains, reducing fluidity. At low temperatures, it prevents fatty acid tails from packing tightly together, inhibiting crystallization and maintaining membrane flexibility.

Question 30Cell structure, membranes and division

Which of the following substances crosses an intact animal cell membrane primarily via simple diffusion without the assistance of transport proteins?

  1. Oxygen gas (O2) Correct answer.
  2. Glucose (C6H12O6)
  3. Sodium ions (Na+)
  4. Calcium ions (Ca2+)

Explanation

Small, nonpolar molecules such as oxygen (O2), carbon dioxide (CO2), and nitrogen can readily dissolve in the hydrophobic core of the phospholipid bilayer and cross via simple diffusion down their concentration gradients. In contrast, large uncharged polar molecules (glucose) and charged inorganic ions (Na+, Ca2+) are repelled by the hydrophobic core and require specific channel or carrier proteins.

Question 31Cell structure, membranes and division

What happens when human red blood cells (erythrocytes) and green plant mesophyll cells are placed into pure distilled water (a hypotonic environment)?

  1. Red blood cells swell and undergo osmotic lysis (hemolysis), while plant cells become turgid due to their rigid cellulose cell wall. Correct answer.
  2. Both cell types shrink and undergo irreversible plasmolysis as water is drawn rapidly into the surrounding medium.
  3. Plant cells swell until they burst, whereas red blood cells maintain their shape by active sodium excretion.
  4. Both red blood cells and plant cells remain in dynamic equilibrium with zero net movement of water molecules.

Explanation

Distilled water has a higher water potential (lower solute concentration) than cell cytoplasm, causing water to enter both cells by osmosis. Lacking a rigid cell wall, red blood cells swell until their plasma membrane ruptures, an event termed osmotic lysis or hemolysis. Plant cells also absorb water, but the influx generates internal turgor pressure that presses against the rigid cellulose cell wall, reaching an equilibrium that makes the cell turgid without bursting.

Question 32Cell structure, membranes and division

During each catalytic cycle of the primary active transport sodium-potassium pump (Na+/K+-ATPase), what is the stoichiometry and direction of ion movement across the plasma membrane?

  1. 3 Na+ are pumped out of the cell and 2 K+ are pumped into the cell, consuming 1 ATP molecule. Correct answer.
  2. 2 Na+ are pumped out of the cell and 3 K+ are pumped into the cell, consuming 2 ATP molecules.
  3. 3 Na+ and 2 K+ are simultaneously pumped out of the cell, generating 1 ATP molecule.
  4. 1 Na+ is pumped out in exchange for 1 K+ entering, driven entirely by proton-motive force without ATP.

Explanation

The Na+/K+-ATPase is an electrogenic primary active transport pump. Hydrolysis of one molecule of ATP phosphorylates the pump, driving a conformational change that exports 3 Na+ ions out of the cytoplasm and imports 2 K+ ions into the cell, both against their electrochemical gradients. This net export of one positive charge per cycle contributes to maintaining the negative resting membrane potential and cellular osmotic balance.

Question 33Cell structure, membranes and division

Which cellular transport process involves the engulfment of large solid particulate matter, such as bacteria or cellular debris, via pseudopodial extensions of the plasma membrane?

  1. Pinocytosis
  2. Phagocytosis Correct answer.
  3. Receptor-mediated endocytosis
  4. Exocytosis

Explanation

Phagocytosis ('cell eating') is a form of endocytosis in which specialized cells, such as macrophages and neutrophils, extend pseudopodia around large particulate matter or whole microorganisms. The engulfed material is internalized within a large membrane-bound vesicle called a phagosome, which subsequently fuses with a primary lysosome to form a phagolysosome for enzymatic digestion.

Question 34Cell structure, membranes and division

A diploid human cell contains 46 chromosomes (2n = 46) and an initial DNA content of 2C during the G1 phase of interphase. What is the chromosome count and DNA content in this same cell during the G2 phase, immediately prior to mitosis?

  1. 46 chromosomes and 4C DNA content Correct answer.
  2. 92 chromosomes and 4C DNA content
  3. 23 chromosomes and 2C DNA content
  4. 46 chromosomes and 2C DNA content

Explanation

During the S (synthesis) phase of interphase, each chromosome replicates its DNA, producing two identical sister chromatids joined at the centromere. Because sister chromatids remain physically attached at a single centromere, the chromosome count remains 46, but the total amount of DNA doubles from 2C to 4C. Chromatids are counted as individual chromosomes only after they separate at anaphase.

Question 35Cell structure, membranes and division

During which phase of mitosis do sister chromatids synchronously separate and move toward opposite poles of the dividing cell?

  1. Prophase
  2. Metaphase
  3. Anaphase Correct answer.
  4. Telophase

Explanation

Anaphase begins when the enzyme separase cleaves cohesin protein complexes holding sister chromatids together at the centromere. The kinetochore microtubules rapidly shorten, pulling the newly separated daughter chromosomes toward opposite centrosome poles. Simultaneously, non-kinetochore microtubules elongate to push the cell poles further apart.

Question 36Cell structure, membranes and division

Which critical cytological event occurs during prophase I of meiosis to generate novel combinations of maternal and paternal alleles on homologous chromosomes?

  1. Crossing over (chiasma formation) between non-sister chromatids of homologous pairs Correct answer.
  2. Independent segregation of sister chromatids to opposite poles
  3. Direct endoreduplication of mitochondrial plasmids
  4. Splicing of nuclear pre-mRNAs by cytoplasmic kinetochores

Explanation

During prophase I of meiosis (specifically pachytene), homologous chromosomes pair closely through synapsis to form bivalents (tetrads). Non-sister chromatids break and exchange reciprocal genetic segments at points called chiasmata in a process called crossing over. This recombination breaks parental linkage groups and generates new allele combinations, significantly increasing genetic diversity in gametes.

Question 37Cell structure, membranes and division

During aerobic cellular respiration, what is the immediate terminal electron acceptor at the end of the mitochondrial electron transport chain, and what molecule is formed?

  1. Molecular oxygen (O2), which is reduced to form water (H2O) Correct answer.
  2. NAD+, which is reduced to form NADH
  3. Pyruvate, which is reduced to form lactate
  4. Inorganic phosphate, which binds ADP to form ATP

Explanation

At complex IV (cytochrome c oxidase) of the mitochondrial inner membrane, electrons that have traveled down the respiratory chain are transferred to molecular oxygen (O2), the terminal electron acceptor. Oxygen combines with four electrons and four protons from the matrix to form two molecules of water (H2O). Without oxygen, the electron transport chain backs up, stopping the proton gradient and aerobic ATP synthesis.

Question 38Genetics, inheritance and molecular information

In pea plants, tall stem height (T) is completely dominant over dwarf stem height (t). If two heterozygous tall plants (Tt) are crossed, what is the expected phenotypic ratio among their offspring according to Mendel's law of segregation?

  1. 3 tall : 1 dwarf Correct answer.
  2. 1 tall : 2 intermediate : 1 dwarf
  3. 9 tall : 3 dwarf : 3 dwarf : 1 dwarf
  4. All tall plants

Explanation

According to Mendel's law of segregation, each heterozygous parent (Tt) produces gametes carrying either the T or t allele in equal proportions (1/2 T and 1/2 t). A monohybrid cross yields a genotypic ratio of 1 TT : 2 Tt : 1 tt. Because T is completely dominant over t, both TT and Tt individuals express the tall phenotype, resulting in a phenotypic ratio of 3 tall to 1 dwarf (75% tall, 25% dwarf).

Question 39Genetics, inheritance and molecular information

In a dihybrid cross between two individuals heterozygous for two unlinked autosomal traits (AaBb x AaBb), what fraction of the offspring is expected to display both dominant phenotypes?

  1. 9/16 Correct answer.
  2. 3/16
  3. 1/16
  4. 1/4

Explanation

According to Mendel's law of independent assortment, the segregation of alleles for gene A is completely independent of gene B. The probability of an offspring showing dominant trait A (AA or Aa) is 3/4, and the probability of showing dominant trait B (BB or Bb) is 3/4. Multiplying these independent probabilities yields (3/4) * (3/4) = 9/16 for individuals expressing both dominant phenotypes.

Question 40Genetics, inheritance and molecular information

What cross should a geneticist perform to determine whether an organism expressing a dominant phenotype is homozygous dominant or heterozygous?

  1. A test cross with a homozygous recessive individual Correct answer.
  2. A cross with an individual known to be homozygous dominant
  3. A self-fertilization cross with an identical dominant phenotype organism
  4. A cross with a wild-type individual of unknown genotype

Explanation

A test cross involves breeding the individual showing the dominant phenotype (genotype A?) with a known homozygous recessive individual (aa). If the mystery organism is homozygous dominant (AA), all offspring will express the dominant phenotype (Aa). If the organism is heterozygous (Aa), approximately half the offspring will display the recessive phenotype (aa), definitively revealing the parent's genotype.

Question 41Genetics, inheritance and molecular information

When a true-breeding red-flowered snapdragon (CRCR) is crossed with a true-breeding white-flowered snapdragon (CWCW), all F1 offspring produce pink flowers (CRCW). What type of inheritance is demonstrated?

  1. Incomplete dominance Correct answer.
  2. Complete dominance
  3. Codominance
  4. Epistasis

Explanation

Incomplete dominance occurs when neither allele is completely dominant over the other, resulting in a heterozygous phenotype that is an intermediate blend between the two homozygous parental phenotypes. In snapdragons, the CR allele produces red pigment while CW produces no pigment. The heterozygote CRCW produces half as much red pigment as the CRCR homozygote, creating a pink appearance.

Question 42Genetics, inheritance and molecular information

A mother with blood type A (genotype IAi) and a father with blood type B (genotype IBi) have children. Which ABO blood phenotypes are possible among their biological offspring?

  1. Only blood type AB
  2. Only blood types A and B
  3. Types A, B, AB, and O are all possible, each with a 25% probability. Correct answer.
  4. Only blood types AB and O

Explanation

The mother can donate either the IA or i allele, and the father can donate either the IB or i allele. Constructing a Punnett square yields four equally probable genotypic combinations: IAIB (type AB), IAi (type A), IBi (type B), and ii (type O). Each of these four blood groups has an expected frequency of 1/4 (25%).

Question 43Genetics, inheritance and molecular information

In human Rh blood group genetics, the Rh-positive allele (D) is dominant over the Rh-negative allele (d). Under what parental condition can hemolytic disease of the newborn (erythroblastosis fetalis) develop during a second pregnancy?

  1. An Rh-negative mother carries a second Rh-positive fetus after being sensitized by a previous Rh-positive pregnancy. Correct answer.
  2. An Rh-positive mother carries an Rh-negative fetus whose antibodies cross the placenta into maternal circulation.
  3. Both parents are Rh-negative, producing an Rh-positive child via spontaneous somatic mutation.
  4. An Rh-negative father produces sperm antibodies that attack the maternal endometrium prior to blastocyst implantation.

Explanation

Erythroblastosis fetalis occurs when an Rh-negative woman (dd) carries an Rh-positive fetus (Dd, inherited from an Rh-positive father). During the birth of the first Rh-positive child, fetal Rh antigens enter maternal circulation, stimulating the mother's immune system to produce anti-Rh IgG antibodies. In a subsequent Rh-positive pregnancy, these maternal anti-Rh antibodies cross the placenta and destroy fetal red blood cells.

Question 44Genetics, inheritance and molecular information

Red-green color blindness is an X-linked recessive disorder in humans. If a color-blind man marries a woman who has normal vision but whose father was color-blind, what is the probability that their first son will be color-blind?

  1. 0%
  2. 25%
  3. 50% Correct answer.
  4. 100%

Explanation

The woman's father was color-blind (XbY), meaning he passed his Xb chromosome to her, making her a heterozygous carrier with genotype XBXb. When assessing only their male offspring (sons), each son receives a Y chromosome from the father and either the XB allele (normal vision) or the Xb allele (color-blind) from the mother with equal probability. Therefore, the probability that a son is color-blind is 1/2, or 50%.

Question 45Genetics, inheritance and molecular information

In birds and some reptiles, chromosomal sex determination operates under the ZW system. Which individual is the heterogametic sex that determines the genetic sex of the offspring?

  1. The female (ZW) Correct answer.
  2. The male (ZZ)
  3. Both sexes are homogametic (ZZ in males and XX in females)
  4. Sex is determined strictly by the male supplying either a Z or W sperm chromosome

Explanation

In the ZW sex-determination system found in birds, butterflies, and certain reptiles, females are heterogametic possessing two distinct sex chromosomes (ZW), whereas males are homogametic possessing two identical sex chromosomes (ZZ). Because male sperm all carry a Z chromosome, it is the maternal ovum (carrying either Z or W) that determines whether the fertilized zygote develops into a male (ZZ) or female (ZW).

Question 46Genetics, inheritance and molecular information

Two gene loci on the same chromosome in fruit flies exhibit a recombination frequency of 14% in a test cross. What does this percentage indicate regarding the genetic distance between these two genes?

  1. The genes are separated by 14 centimorgans (map units) and assort non-independently due to linkage. Correct answer.
  2. The genes reside on different non-homologous chromosomes and follow Mendel's law of independent assortment.
  3. The genes undergo crossing over in 100% of meiotic divisions, but only 14% of zygotes survive.
  4. The genes are located at the exact same nucleotide locus and represent codominant alleles.

Explanation

By definition, one percent of recombination frequency between two linked loci corresponds to one map unit (1 cM or centimorgan) on a genetic linkage map. A recombination frequency of 14% means the genes are 14 cM apart on the same chromosome. Because 14% is significantly lower than the 50% recombination expected for independently assorting unlinked genes, these genes are linked and inherited together unless crossing over occurs.

Question 47Genetics, inheritance and molecular information

During DNA replication in Escherichia coli, why must the lagging strand be synthesized discontinuously as Okazaki fragments?

  1. DNA polymerases can only synthesize new DNA strands in the 5' to 3' direction along an antiparallel template. Correct answer.
  2. DNA ligase inhibits continuous strand elongation on both replication forks simultaneously.
  3. RNA primers can only bind to the 3' terminal hydroxyl groups of deoxyribose.
  4. Helicase unwinds the double helix in discontinuous bursts regulated by topoisomerase.

Explanation

DNA polymerases can only add free deoxyribonucleotides to the 3'-OH group of a pre-existing strand, meaning elongation proceeds exclusively in the 5' to 3' direction. Because the two parental DNA strands are antiparallel, as the replication fork opens, the lagging strand template runs in the 5' to 3' direction toward the fork. The replication machinery must therefore synthesize new DNA away from the fork in short segments known as Okazaki fragments.

Question 48Genetics, inheritance and molecular information

If a template strand of DNA has the sequence 3'-TAC GGC TTA CTG-5', what will be the corresponding sequence of the transcribed mRNA transcript?

  1. 5'-AUG CCG AAU GAC-3' Correct answer.
  2. 5'-ATG CCG AAT GAC-3'
  3. 5'-UAC GGC UUA CUG-3'
  4. 5'-CAG UAA GCC GUA-3'

Explanation

During transcription, RNA polymerase synthesizes an antiparallel, complementary mRNA strand in the 5' to 3' direction using the DNA template strand running 3' to 5'. Adenine pairs with Uracil (T->A, A->U) and Cytosine pairs with Guanine (C->G, G->C). Complementary pairing of 3'-TAC GGC TTA CTG-5' yields 5'-AUG CCG AAU GAC-3'.

Question 49Genetics, inheritance and molecular information

Which of the following modifications occurs during the post-transcriptional processing of eukaryotic pre-mRNA before it exits the nucleus?

  1. Addition of a 7-methylguanosine cap to the 5' end, a poly-A tail to the 3' end, and splicing out of non-coding introns Correct answer.
  2. Direct translation of introns into structural histone proteins within the nucleoplasm
  3. Removal of all exons followed by the ligation of intervening introns into circular rings
  4. Phosphorylation of nitrogenous bases to prevent ribosome attachment in the cytoplasm

Explanation

Eukaryotic primary transcripts (pre-mRNA) undergo extensive nuclear processing: a 5' 7-methylguanosine cap is added for ribosome recognition and ribonuclease protection, a polyadenylation signal directs cleavage and addition of a 3' poly-A tail for stability and nuclear export, and spliceosomes excise non-coding introns while joining coding exons together to form mature mRNA.

Question 50Genetics, inheritance and molecular information

What is meant by the statement that the universal genetic code is 'degenerate' (or redundant)?

  1. More than one distinct triplet codon can specify the same amino acid. Correct answer.
  2. A single codon can simultaneously code for several different amino acids in the same cell.
  3. Codons overlap with one another so that adjacent bases are read twice.
  4. The genetic code degrades and changes its codon assignments as an organism ages.

Explanation

The genetic code consists of 64 triplet codons, with 61 coding for 20 standard amino acids (and 3 acting as stop signals). Because there are far more codons than amino acids, multiple distinct codons can code for the same amino acid (for instance, leucine and arginine are each specified by six different codons). This degeneracy provides a buffer against deleterious mutations, especially at the third codon position.