Read each question and click on the ONE answer option you believe is correct. The explanation will appear after you click any option.
Score: 0
In a standard diagram of a nephron, identify the structure corresponding to the region containing the glomerulus and Bowman’s capsule, where podocytes are found lining the capsule inner layer.
Click the CORRECT answer:
- Loop of Henle
- Renal corpuscle (Bowman’s capsule enclosing glomerulus)
- Vasa recta
- Proximal convoluted tubule
- Collecting duct
Reasoning for Correct Answer #2: The structure where filtration occurs in the nephron consists of the glomerulus (a network of capillaries) surrounded by the Bowman’s capsule. This entire unit is known as the renal corpuscle (or Malpighian body). Podocytes are specialized cells that form the inner (visceral) layer of the Bowman’s capsule, wrapping around the glomerular capillaries and forming filtration slits. The Loop of Henle (#1), proximal convoluted tubule (#4), and collecting duct (#5) are parts of the renal tubule system. The vasa recta (#3) are blood capillaries associated with the Loop of Henle.
In a standard diagram of a nephron, identify the tubule segment that descends into the medulla and then ascends back towards the cortex.
Click the CORRECT answer:
- Collecting duct
- Proximal convoluted tubule
- Loop of Henle
- Distal convoluted tubule
- Glomerulus
Reasoning for Correct Answer #3: The Loop of Henle is the characteristic U-shaped portion of the nephron tubule. It originates from the proximal convoluted tubule (#2) in the cortex, dips down into the renal medulla (descending limb), makes a hairpin turn, and ascends back towards the cortex (ascending limb) before connecting to the distal convoluted tubule (#4). The glomerulus (#5) is part of the renal corpuscle where filtration begins. The collecting duct (#1) receives fluid from the DCT and passes through the medulla but is structurally distinct from the Loop of Henle itself.
In a standard diagram of a nephron, identify the final tube segment that passes through the medulla and carries urine towards the renal pelvis.
Click the CORRECT answer:
- Proximal convoluted tubule
- Loop of Henle
- Distal convoluted tubule
- Collecting duct
- Afferent arteriole
Reasoning for Correct Answer #4: The collecting duct is the terminal part of the nephron system (although technically receiving input from multiple nephrons). It starts in the cortex, receives tubular fluid from the distal convoluted tubules (#3), and descends through the renal medulla. Here, final adjustments to water and solute concentration occur under hormonal control (ADH, aldosterone). The fluid leaving the collecting ducts at the renal papilla is considered final urine, which then flows into the renal pelvis. The afferent arteriole (#5) is a blood vessel.
In a standard diagram of a nephron, identify the blood vessel that carries blood towards the glomerulus.
Click the CORRECT answer:
- Collecting duct
- Afferent arteriole
- Efferent arteriole
- Peritubular capillaries
- Loop of Henle
Reasoning for Correct Answer #2: Blood enters the filtration unit of the nephron (the glomerulus within Bowman’s capsule) via the afferent arteriole (‘afferent’ means carrying towards). After passing through the glomerular capillaries, blood exits via the efferent arteriole (#3, ‘efferent’ means carrying away). The efferent arteriole then typically branches into the peritubular capillaries (#4) that surround the tubules, or the vasa recta associated with the Loop of Henle. The collecting duct (#1) and Loop of Henle (#5) are parts of the tubule system carrying filtrate/urine, not blood vessels in this context.
Describe and explain how cells of the proximal convoluted tubule (PCT) are adapted for selective reabsorption. Which of the following statements is NOT a correct adaptation or explanation for PCT cells?
Click the INCORRECT statement:
- They possess microvilli on their apical surface, creating a large surface area.
- They contain many mitochondria to provide ATP for active transport (e.g., Na+/K+ pump).
- They lack specific transporter proteins, allowing bulk flow of solutes back into the blood.
- Tight junctions between cells prevent paracellular movement, forcing transcellular transport.
- The basal membrane is highly folded, increasing surface area for pumps moving substances towards blood.
- Numerous cotransporter proteins facilitate reabsorption of substances like glucose with sodium ions.
Reasoning for Incorrect Statement #3: The proximal convoluted tubule (PCT) is the primary site for selective reabsorption of useful substances from the glomerular filtrate back into the blood. This process relies heavily on a multitude of specific membrane transport proteins located on the apical (lumen-facing) and basal (blood-facing) membranes. These include Na⁺/K⁺-ATPase pumps (#2, #5), cotransporters for glucose, amino acids, ions (#6), and channels. Statement #3 claims PCT cells *lack* specific transporters and rely on *bulk flow*, which is incorrect. Selective reabsorption is mediated by these specific transporters. Microvilli (#1) increase surface area, and tight junctions (#4) direct transport through the cells (transcellular).
Describe a cross and analysis to determine if a tall pea plant (where tall ‘Le’ is dominant to dwarf ‘le’) is homozygous dominant (LeLe) or heterozygous (Lele). Which statement describes a necessary part of this test cross?
Click the CORRECT description of the test cross procedure:
- Cross the tall pea plant with another tall pea plant known to be heterozygous (Lele).
- Cross the tall pea plant with a dwarf pea plant (genotype lele).
- Observe the F1 offspring; if any are tall, the parent was homozygous (LeLe).
- Self-pollinate the tall pea plant and observe the offspring ratio.
- Observe the F1 offspring; if the ratio is 3 tall : 1 dwarf, the parent was heterozygous (Lele).
Reasoning for Correct Answer #2: A test cross is used to determine the genotype of an individual expressing a dominant phenotype (whose genotype could be homozygous dominant or heterozygous). This is done by crossing the individual in question (tall plant, genotype Le_) with an individual that is homozygous recessive for the trait (dwarf plant, genotype lele). The phenotypes of the offspring reveal the unknown genotype:
– If the tall parent is LeLe: LeLe × lele → All offspring are Lele (tall).
– If the tall parent is Lele: Lele × lele → 1/2 Lele (tall) : 1/2 lele (dwarf).
Option #2 correctly describes crossing with the homozygous recessive individual. Options #1 and #4 describe different types of crosses. Options #3 and #5 describe incorrect interpretations of offspring results for a test cross.
Gibberellin is a plant growth regulator involved in stem elongation. What is the general term for a molecule, like gibberellin, that binds to specific cellular receptors to trigger a response?
Click the CORRECT answer:
- Enzyme
- Neurotransmitter
- Signalling molecule (or hormone, ligand, plant growth regulator)
- Antibody
- Substrate
Reasoning for Correct Answer #3: Molecules that transmit information between or within cells by binding to specific receptor proteins are broadly termed signalling molecules or ligands. Depending on the context, more specific terms like hormone (for long-distance signalling via circulation), neurotransmitter (signalling between neurons), or plant growth regulator (like gibberellin) can be used. Enzymes (#1) catalyze reactions, antibodies (#4) bind antigens, and substrates (#5) bind to enzyme active sites. Gibberellin fits the definition of a signalling molecule/ligand/hormone/plant growth regulator.
Gibberellin affects stem length in pea plants by influencing internode cells. Suggest the cellular response of internode cells to gibberellin and the resulting effect on the plant’s height (the trait Mendel investigated).
Click the CORRECT combination of response and effect:
- Response: Gibberellin causes cell shrinkage, leading to shorter stems.
- Response: Gibberellin stimulates cell elongation and/or cell division (mitosis) in internodes.
- Effect: Increased internode cell elongation/division leads to taller stems/plants. (Combines Response #2 with Effect)
- Effect: Gibberellin primarily increases root growth, indirectly affecting height.
Reasoning for Correct Answer #2: Gibberellins are plant hormones known to promote stem growth. Their primary cellular effect in stem internodes is to stimulate both cell elongation (making existing cells longer) and, in some cases, cell division (creating more cells). This increased growth within the internodes (#2) is the direct cause of the overall increase in stem length, leading to the tall phenotype (#3 describes the effect of #2). Option #1 is incorrect (gibberellins promote growth). Option #4 is incorrect regarding the primary effect relevant to stem height.
Cystic fibrosis is an autosomal recessive disease. Explain the meaning of ‘homozygous’ and ‘recessive’. Which statement correctly defines one or both of these terms?
Click the CORRECT definition statement:
- Homozygous means having two different alleles; Recessive means the allele is expressed when heterozygous.
- Homozygous means having identical alleles for a gene; Recessive means the allele’s effect is masked by a dominant allele if heterozygous.
- Homozygous means identical phenotype; Recessive means stronger allele.
- Homozygous means having one dominant and one recessive allele; Recessive means determines phenotype.
Reasoning for Correct Answer #2:
Homozygous: An individual is homozygous for a gene when they possess two identical alleles at that gene’s locus on homologous chromosomes (e.g., AA or aa).
Recessive: A recessive allele is one whose phenotypic effect is only expressed in the homozygous state (e.g., aa). In a heterozygous individual (Aa), the effect of the recessive allele (a) is masked by the dominant allele (A).
Option 2 correctly defines both terms according to standard genetic terminology. Option 1 defines heterozygous and incorrectly defines recessive expression. Option 3 confuses genotype with phenotype and incorrectly defines recessive. Option 4 defines heterozygous and misrepresents recessive function.
In the UK 2020 population (67,100,000), 10,800 people had cystic fibrosis (autosomal recessive, genotype aa). Use the Hardy-Weinberg principle (p + q = 1; p² + 2pq + q² = 1) to estimate the frequency of the recessive allele (q) and the frequency of heterozygous carriers (2pq).
Click the CORRECT answer:
- q ≈ 0.0127; 2pq ≈ 0.0251
- q² ≈ 0.000161; p ≈ 0.0127
- q ≈ 0.000161; 2pq ≈ 0.9749
- q² ≈ 1.61 x 10⁻⁴; q ≈ 0.9873
- q ≈ 0.0127; p ≈ 0.0251
Reasoning for Correct Answer #1:
1. Find frequency of affected individuals (genotype aa), which is q²:
q² = Number with CF / Total Population = 10,800 / 67,100,000 ≈ 0.00016095 ≈ 1.61 × 10⁻⁴
2. Find frequency of the recessive allele (q):
q = √q² = √0.00016095 ≈ 0.012687 ≈ 0.0127
3. Find frequency of the dominant allele (p):
p = 1 – q = 1 – 0.012687 ≈ 0.987313
4. Find frequency of heterozygous carriers (genotype Aa), which is 2pq:
2pq = 2 × p × q = 2 × 0.987313 × 0.012687 ≈ 0.025049 ≈ 0.0251 (or about 1 in 40 people)
Option #1 provides the correct calculated values for q and 2pq.
The Hardy-Weinberg calculation often estimates carrier frequency based on the prevalence of the disease in the current population. Explain how the difference in life expectancy (individuals with CF ≈ 50 years, general population ≈ 80 years) might affect the accuracy of this estimate for CF carriers.
Click the statement providing the most valid reason:
- The calculation accurately reflects the current gene pool frequency regardless of life expectancy.
- Natural selection against the CF phenotype means affected individuals have lower survival/reproduction rates, potentially making the calculated allele frequency (q) based on living individuals slightly lower than the true frequency in the total gene pool (including past generations), thus underestimating carriers (2pq).
- Shorter life expectancy means affected individuals reproduce more quickly, increasing the allele frequency.
- The Hardy-Weinberg principle only applies if life expectancy is equal for all genotypes.
- Longer life expectancy of carriers increases the accuracy of the calculation.
Reasoning for Correct Answer #2: The Hardy-Weinberg principle assumes no selection, among other conditions. Cystic fibrosis significantly reduces fitness (survival and reproductive success) compared to unaffected individuals, meaning there is negative selection against the homozygous recessive genotype (aa). Calculating the frequency of this genotype (q²) based only on the prevalence in the *current living* population might underestimate its true frequency at birth or in the gene pool before selection acts strongly. If q² (and thus q) is underestimated, the calculated frequency of heterozygous carriers (2pq) will also be underestimated compared to their actual frequency in the population. Shorter life expectancy does not lead to faster reproduction (#3). While unequal life expectancy violates the ‘no selection’ assumption, HW can still provide estimates (#4 too strict). Carrier life expectancy (#5) isn’t the primary issue here.
Data shows the median predicted life expectancy for people born with CF in the UK increased from 38.8 years (born 2008) to 50.6 years (born 2020). National newborn screening for CF was introduced in 2007. Which statement best describes this trend and the likely contribution of screening?
Click the CORRECT answer:
- Life expectancy decreased, likely due to side effects of screening.
- Life expectancy increased significantly, partly because early screening allows for earlier diagnosis and initiation of treatments (like physiotherapy, nutritional support, antibiotics), improving long-term outcomes.
- Life expectancy remained constant, indicating screening had no effect.
- Life expectancy increased, primarily due to general improvements in healthcare unrelated to CF-specific screening or treatment.
- Life expectancy increased, because screening itself cures the disease.
Reasoning for Correct Answer #2: The data clearly indicate a significant increase in median predicted life expectancy for individuals born with CF over the period. The introduction of newborn screening in 2007 allows for diagnosis shortly after birth, often before symptoms become severe. This early diagnosis enables the prompt initiation of specialized care and treatments (nutritional management, physiotherapy to clear airways, proactive antibiotic use, modulator drugs where applicable). Early and consistent management is known to slow disease progression, reduce complications, and improve quality of life and longevity. While general healthcare improvements may play a role (#4), the substantial increase coinciding with the post-screening era strongly suggests screening and subsequent early intervention are major contributing factors. Screening detects, it doesn’t cure (#5).
Discuss the ethical and social considerations of offering genetic testing for CF carrier status (heterozygosity) to adults with a family history of the condition. Which point represents a potential disadvantage or negative consideration?
Click the potential DISADVANTAGE:
- Allows individuals to make informed reproductive choices (e.g., regarding partner screening, use of prenatal diagnosis or preimplantation genetic diagnosis).
- May cause psychological distress, anxiety, or feelings of guilt if an individual is found to be a carrier, especially concerning potential risks to offspring.
- Can reduce anxiety and provide relief for individuals if the test result is negative (confirming they are not a carrier).
- Allows for better financial planning regarding potential future healthcare costs associated with having an affected child, although this is complex.
- Enables other potentially at-risk family members (e.g., siblings) to better understand their own carrier risk and consider testing.
Reasoning for Correct Answer #2: Offering genetic carrier testing has both potential benefits and drawbacks. Options #1, #3, #4, and #5 represent potential advantages or neutral outcomes related to informed decision-making, reduced uncertainty, planning, and family awareness. Option #2 highlights a significant potential negative consequence: the psychological burden. Learning one carries a gene for a serious condition can lead to anxiety about personal health (though usually minimal for CF carriers), significant worry about passing the condition to children, feelings of guilt, or complex emotions impacting relationships and reproductive decisions. This potential for psychological distress is a key ethical consideration in genetic counseling and testing.
Data shows that from April to August, mean daily air temperature generally increases (e.g., from ~10°C to ~18°C), while mean monthly milk yield per cow in Holstein Friesians generally decreases (e.g., from ~795 kg to ~714 kg). What type of correlation does this suggest between air temperature and milk yield in this period?
Click the CORRECT answer:
- Positive correlation (higher temperature leads to higher yield)
- No correlation
- Negative correlation (higher temperature leads to lower yield)
- Curvilinear correlation
- Perfect correlation
Reasoning for Correct Answer #3: The data described shows a trend where one variable (air temperature) increases over the period, while the other variable (milk yield) decreases. When two variables tend to move in opposite directions – one increases as the other decreases – this indicates a negative correlation between them. High temperatures often cause heat stress in dairy cattle like Holstein Friesians, leading to reduced feed intake and metabolic changes that lower milk production.
Milk yield in cattle shows a range of values with intermediate phenotypes, is affected by environmental factors like temperature, and is influenced by multiple genes. What type of phenotypic variation is milk yield, and what are the main contributing factors?
Click the CORRECT answer:
- Type: Discontinuous variation; Factors: Genetics only.
- Type: Continuous variation; Factors: Environment only.
- Type: Discontinuous variation; Factors: Environment only.
- Type: Continuous variation; Factors: Genetics (polygenic) and Environment.
- Type: Continuous variation; Factors: Genetics (single gene) only.
Reasoning for Correct Answer #4: Milk yield is described as showing a range of values with intermediate phenotypes (not distinct categories), which is characteristic of continuous variation. The prompt also explicitly states it is influenced by multiple genes (making it polygenic) and affected by environmental factors (like temperature). Therefore, milk yield is an example of continuous variation resulting from the combined effects of polygenic inheritance and environmental influences. Discontinuous variation involves discrete categories (like blood types).
In Holstein Friesian cattle, a specific ‘SLICK’ allele helps with heat tolerance. Comparing cattle with and without this allele: In March (5°C), milk yield was similar (~782 kg with SLICK vs ~798 kg without). In September (14°C), yield was significantly higher in SLICK cattle (~775 kg) compared to normal cattle (~695 kg), whose yield dropped considerably from March. What does this suggest about the effect of the SLICK allele?
Click the CORRECT answer:
- The SLICK allele significantly reduces milk yield at all temperatures.
- The SLICK allele has no effect on milk yield.
- The SLICK allele primarily increases milk yield at lower temperatures.
- The SLICK allele helps maintain high milk yield, especially at higher temperatures where normal cattle show a decline.
- The SLICK allele causes milk yield to decrease proportionally with increasing temperature.
Reasoning for Correct Answer #4: The key comparison is between March (cool) and September (warmer). At the cool temperature, both SLICK and normal cattle had similar high yields. At the warmer temperature, the normal cattle’s yield dropped significantly (from ~798 to ~695 kg), while the SLICK cattle’s yield remained high (dropping only slightly from ~782 to ~775 kg) and was much higher than the normal cattle’s yield in September. This demonstrates that the SLICK allele provides an advantage by helping cattle maintain their milk production under warmer conditions (i.e., it confers heat tolerance related to milk yield), preventing the significant decline seen in normal cattle.
Compare gene editing and selective breeding as methods for introducing a desirable allele (like SLICK for heat tolerance) into a cattle population (like Holstein Friesians). Which statement accurately describes a difference between these methods?
Click the CORRECT answer:
- Selective breeding targets only the specific gene of interest, while gene editing affects many genes.
- Both methods guarantee that all offspring will inherit the desired allele in the first generation.
- Gene editing directly modifies the gene in cells/embryos in a lab, while selective breeding relies on choosing parents with desired traits for mating over generations.
- Selective breeding is much faster, achieving the desired outcome in one generation, unlike gene editing.
- Gene editing inevitably reduces overall genetic variation more than selective breeding.
Reasoning for Correct Answer #3: Gene editing (like CRISPR) involves precise molecular techniques performed in a laboratory setting to directly alter the DNA sequence of a specific gene within cells or embryos. Selective breeding is a traditional process where humans choose individual organisms displaying desirable phenotypes (traits) to be parents for the next generation, gradually increasing the frequency of the underlying favourable alleles over multiple generations through controlled mating. Option 1 reverses the targeting precision. Neither guarantees inheritance in the first generation (#2). Gene editing can potentially be much faster (#4 incorrect). Effects on genetic variation depend on application (#5 not necessarily true).
Complete the sentence about evolutionary relationships: Closely related species tend to share more ______ in their DNA nucleotide sequences compared to more distantly related species.
Click the CORRECT answer:
- introns
- mutations (or similarities/identities)
- stop codons
- differences
- telomeres
Reasoning for Correct Answer #2: Evolutionary relatedness is inferred from genetic similarity. Species that diverged from a common ancestor more recently have had less time for independent mutations to accumulate in their DNA sequences. Therefore, closely related species will have fewer differences and consequently share more similarities or sequence identities in their DNA compared to species that diverged longer ago. While they also share *mutations* inherited from their common ancestor, the most common way to phrase this comparison focuses on the overall similarity or identity level being higher. Option #4 (differences) is the opposite. Options #1, #3, #5 refer to specific genetic elements, not the overall sequence comparison basis.
Which type of nucleic acid is particularly useful for tracing maternal lineage because it is inherited maternally and is generally unaffected by recombination during meiosis?
Click the CORRECT answer:
- Ribosomal RNA (rRNA)
- Nuclear DNA (autosomes)
- Transfer RNA (tRNA)
- Mitochondrial DNA (mtDNA)
- Y chromosome DNA
Reasoning for Correct Answer #4: Mitochondria, and the small circular DNA molecule they contain (mtDNA), are inherited almost exclusively from the mother through the cytoplasm of the egg cell. Sperm mitochondria typically do not contribute significantly to the zygote. Furthermore, mtDNA does not undergo recombination during meiosis. These two features – strict maternal inheritance and lack of recombination – mean that mtDNA sequences are passed down largely unchanged along the maternal line, making mtDNA analysis a powerful tool for tracing maternal ancestry. Y chromosome DNA (#5) traces paternal lineage. Nuclear DNA (#2) undergoes recombination. RNAs (#1, #3) are not typically used for lineage tracing in this way.
What bioinformatics tool allows researchers to predict the amino acid sequences of proteins?
Click the CORRECT answer:
- Microarray analysis
- DNA sequencing databases combined with knowledge of the genetic code
- Karyotyping
- Polymerase Chain Reaction (PCR)
- Gel electrophoresis
Reasoning for Correct Answer #2: Predicting the amino acid sequence of a protein computationally relies on knowing its corresponding gene’s DNA sequence. Researchers sequence the DNA (or infer it from mRNA sequences stored in databases) and identify the protein-coding region (open reading frame). Using the established genetic code (which dictates the translation of each three-nucleotide codon into a specific amino acid), bioinformatics software can then translate the DNA/mRNA sequence into the predicted amino acid sequence of the protein. Microarrays (#1) measure gene expression levels. Karyotyping (#3) visualizes chromosomes. PCR (#4) amplifies DNA. Gel electrophoresis (#5) separates molecules by size.
What technology can be used to simultaneously detect the presence and relative abundance of many different mRNA molecules in a sample, providing a snapshot of gene expression?
Click the CORRECT answer:
- Electron microscopy
- DNA fingerprinting
- Microarray analysis
- Amino acid sequencing
- Northern blotting
Reasoning for Correct Answer #3: DNA microarrays (or gene chips) are designed for high-throughput analysis of gene expression. They contain thousands of spots, each with a known DNA probe corresponding to a specific gene. Labeled cDNA (synthesized from mRNA extracted from a sample) is hybridized to the array. The intensity of the signal at each spot is proportional to the abundance of the corresponding mRNA in the original sample, allowing researchers to measure the expression levels of thousands of genes simultaneously. Northern blotting (#5) also detects mRNA but typically only for one or a few genes at a time. Electron microscopy (#1) visualizes structures. DNA fingerprinting (#2) analyzes DNA variations. Amino acid sequencing (#4) determines protein structure.
In a respirometer experiment measuring oxygen consumption by germinating peas, what is the function of including potassium hydroxide (KOH) solution?
Click the CORRECT answer:
- To provide oxygen for respiration.
- To absorb the carbon dioxide produced during respiration.
- To maintain a constant temperature.
- To absorb excess water vapour.
- To act as a catalyst for respiration.
Reasoning for Correct Answer #2: Cellular respiration consumes oxygen (O₂) and produces carbon dioxide (CO₂) (in aerobic respiration). A respirometer measures changes in gas volume to estimate respiration rate. To measure oxygen consumption accurately, the CO₂ produced must be removed so that any decrease in volume is solely due to O₂ uptake. Potassium hydroxide (KOH) is a strong alkali that reacts chemically with CO₂, effectively absorbing it from the air within the respirometer chamber. Soda lime is another common absorbent used for this purpose.
When setting up a respirometer experiment to investigate the effect of temperature, why is it important to include a control respirometer (e.g., containing dead peas or glass beads)?
Click the CORRECT answer:
- To measure the rate of photosynthesis.
- To ensure the equipment is properly sealed.
- To account for any changes in gas volume due to temperature or atmospheric pressure fluctuations not caused by respiration.
- To provide a baseline for oxygen production.
- To absorb the carbon dioxide produced by the apparatus itself.
Reasoning for Correct Answer #3: Changes in ambient temperature and atmospheric pressure can cause the volume of gas inside the sealed respirometer to expand or contract, independent of biological activity. A control respirometer, set up identically but without respiring organisms (using dead organisms or inert objects of similar volume like glass beads), experiences these same physical changes. By comparing the volume change in the experimental respirometer with the volume change in the control, one can subtract the effect of physical factors and determine the volume change solely due to the respiration of the living organisms.
Why should a respirometer apparatus be left in the water bath for a period (e.g., five minutes) before starting measurements?
Click the CORRECT answer:
- To allow the peas to start respiring aerobically.
- To ensure the potassium hydroxide solution is fully saturated.
- To allow the temperature of the apparatus and its contents to equilibrate with the water bath temperature.
- To remove all oxygen from the apparatus before starting.
- To check for leaks in the system by observing initial pressure changes.
Reasoning for Correct Answer #3: Gas volume is sensitive to temperature changes (according to gas laws). When the respirometer is placed in the water bath, there will be a period during which the air inside the apparatus and the organisms themselves adjust to the water bath’s temperature. Starting measurements immediately could lead to inaccurate readings due to initial expansion or contraction of gas caused by this temperature change. Allowing an equilibration period ensures that the apparatus and its contents reach the stable experimental temperature before measurements begin, minimizing errors caused by thermal effects.
A graph of respiration rate (measured by O₂ consumption) versus temperature for germinating peas typically shows an increase from 10°C to an optimum around 30°C… Explain the initial increase in rate.
Click the CORRECT answer:
- More ATP is available at higher temperatures.
- Enzymes involved in respiration gain kinetic energy, leading to more frequent successful collisions with substrates.
- The cell membrane becomes more permeable to oxygen at higher temperatures.
- The peas switch from anaerobic to aerobic respiration.
- Carbon dioxide concentration inside the peas increases.
Reasoning for Correct Answer #2: Respiration is a series of enzyme-catalysed reactions. According to kinetic theory, increasing the temperature increases the kinetic energy of molecules (both enzymes and substrates). This leads to more frequent collisions between enzyme active sites and substrate molecules. Furthermore, a higher proportion of these collisions will have sufficient energy (greater than or equal to the activation energy) to result in a reaction. Consequently, the rate of the enzymatic reactions, and thus the overall respiration rate, increases with temperature up to the enzyme’s optimum temperature.
Explain the decrease in respiration rate observed at temperatures significantly above the optimum (e.g., >30-40°C).
Click the CORRECT answer:
- Substrate concentration becomes limiting at high temperatures.
- Oxygen availability decreases sharply above the optimum temperature.
- Enzymes involved in respiration begin to denature, losing their specific 3D structure and active site shape.
- The peas enter a state of dormancy to conserve energy.
- The potassium hydroxide solution stops absorbing carbon dioxide effectively.
Reasoning for Correct Answer #3: Enzymes are proteins with specific three-dimensional structures crucial for their function. Temperatures significantly above the optimum increase molecular vibrations to the point where weak bonds (hydrogen bonds, ionic bonds, hydrophobic interactions) maintaining the enzyme’s tertiary structure begin to break. This process, called denaturation, leads to a change in the shape of the enzyme, particularly the active site. An altered active site can no longer bind the substrate effectively, causing a rapid loss of catalytic activity and a sharp decrease in the overall respiration rate.
In an experiment investigating photosynthesis using isolated chloroplasts and DCPIP (an electron acceptor that changes colour when reduced), why are the chloroplast suspensions kept in the dark before the experiment starts?
Click the CORRECT answer:
- To allow chloroplasts to accumulate CO₂ needed for the experiment.
- To activate the enzymes of the Calvin cycle.
- To prevent the light-dependent reactions from starting and reducing the DCPIP prematurely.
- To allow the thylakoid membranes to become fully permeable.
- To synchronize the chloroplasts’ internal clocks.
Reasoning for Correct Answer #3: DCPIP is used as an indicator of the light-dependent reactions because it accepts electrons released from the electron transport chain (originating from photoactivated chlorophyll) and changes colour upon reduction. If the chloroplasts are exposed to light before the experiment begins, these light-dependent reactions will start, electrons will be released, and DCPIP will begin to be reduced. This would cause a colour change before the intended start time, leading to inaccurate measurements of the reaction rate under the experimental conditions. Keeping the chloroplasts in the dark ensures the light reactions (and thus DCPIP reduction) only begin when the experimental illumination is applied.
In the chloroplast/DCPIP experiment, absorbance decreased over time when illuminated… The rate of decrease was fastest with red light, intermediate with blue, and slowest with green. Explain the difference between red and green light results.
Click the CORRECT answer:
- Green light has higher energy per photon than red light, causing faster reduction.
- Chlorophyll absorbs red light effectively but reflects/transmits most green light; therefore, more photoactivation and electron release occur under red light, reducing DCPIP faster.
- DCPIP absorbs green light more strongly, interfering with the measurement.
- Chlorophyll absorbs green light effectively but red light poorly; therefore, photosynthesis is faster under green light.
- Red light denatures the electron transport chain proteins, while green light does not.
Reasoning for Correct Answer #2: The rate of the light-dependent reactions (and thus DCPIP reduction) depends on the amount of light energy absorbed by photosynthetic pigments, primarily chlorophylls. Chlorophylls have absorption peaks in the red and blue regions of the visible spectrum and absorb very poorly in the green region (they reflect green light, making plants appear green). Since red light is strongly absorbed, it drives photoactivation and electron transport efficiently, leading to a fast rate of DCPIP reduction. Green light is poorly absorbed, resulting in much less efficient photoactivation and a very slow rate of DCPIP reduction.
Indicate whether CO₂ concentration, light intensity, and temperature primarily directly affect the Calvin cycle (light-independent reactions) and/or photophosphorylation (part of light-dependent reactions). Which statement is correct regarding direct effects?
Click the CORRECT statement:
- CO₂ concentration directly affects both the Calvin cycle and photophosphorylation.
- Light intensity directly affects the Calvin cycle but not photophosphorylation.
- Temperature directly affects the Calvin cycle and photophosphorylation.
- CO₂ concentration directly affects photophosphorylation but not the Calvin cycle.
- Light intensity directly affects both the Calvin cycle and photophosphorylation.
Reasoning for Correct Answer #3:
CO₂ Concentration: Directly affects the Calvin cycle as it is the substrate for carbon fixation by RuBisCO. It does not directly affect photophosphorylation.
Light Intensity: Directly affects photophosphorylation by providing the energy for electron excitation, driving the electron transport chain and proton pumping needed for ATP synthesis. It only indirectly affects the Calvin cycle by influencing the supply of ATP and reduced NADP.
Temperature: Directly affects the rate of all enzyme-catalysed reactions. The Calvin cycle involves many enzymes (e.g., RuBisCO). Photophosphorylation also involves enzyme complexes (e.g., ATP synthase) and membrane fluidity, which are temperature-sensitive. Therefore, temperature directly affects both stages.
African giant pouched rats can be trained to detect Mycobacterium tuberculosis by smell. What type of receptor cell allows rats (and humans) to detect chemicals associated with smell and taste?
Click the CORRECT answer:
- Mechanoreceptor
- Photoreceptor
- Thermoreceptor
- Chemoreceptor
- Nociceptor
Reasoning for Correct Answer #4: The senses of smell (olfaction) and taste (gustation) rely on the detection of chemical substances. Sensory receptors that respond to chemical stimuli are called chemoreceptors. Olfactory receptors in the nasal cavity detect volatile chemicals, while taste receptors on the tongue detect dissolved chemicals. Mechanoreceptors (#1) detect physical pressure or distortion. Photoreceptors (#2) detect light. Thermoreceptors (#3) detect temperature changes. Nociceptors (#5) detect potentially damaging stimuli (pain).
Rats trained to detect M. tuberculosis might also respond to samples containing other Mycobacterium species. Suggest a reason for this lack of perfect specificity.
Click the CORRECT answer:
- The rats’ sense of smell is generally poor and unreliable.
- Different Mycobacterium species are unrelated and produce entirely different chemicals.
- Closely related species like those within the Mycobacterium genus likely produce some similar volatile organic compounds due to shared metabolic pathways, which stimulate the same olfactory receptors in the rat.
- The training process inadvertently teaches rats to respond to any bacterial sample.
- The other Mycobacterium species mimic the smell of the reward given during training.
Reasoning for Correct Answer #3: Olfactory detection relies on specific volatile chemical compounds binding to olfactory receptors. Closely related species, such as different species within the same genus (Mycobacterium), share a significant amount of genetic material and therefore many metabolic pathways. It is likely that they produce some of the same or structurally very similar volatile organic compounds (VOCs). If the rat is trained to detect specific VOCs produced by M. tuberculosis, it may also detect these same or similar VOCs produced by other Mycobacterium species, leading to a cross-reaction or lack of perfect specificity.
Complete the missing taxonomic ranks for the African giant pouched rat, Cricetomys gambianus: Kingdom: Animalia; Phylum: Chordata; Class: Mammalia; Order: ______; Family: ______; Genus: Cricetomys; Species: gambianus.
Click the CORRECT answer:
- Order: Carnivora; Family: Felidae
- Order: Rodentia; Family: Nesomyidae
- Order: Primates; Family: Hominidae
- Order: Rodentia; Family: Muridae
- Order: Artiodactyla; Family: Bovidae
Reasoning for Correct Answer #2: Taxonomic classification places organisms into hierarchical groups based on evolutionary relationships and shared characteristics. The African giant pouched rat (Cricetomys gambianus) belongs to the Order Rodentia (the rodents) and the Family Nesomyidae. This family includes various African and Malagasy rodents, distinct from the family Muridae (#4, which includes Old World rats and mice), Felidae (#1, cats), Hominidae (#3, great apes including humans), or Bovidae (#5, cattle, antelope, etc.).
Outline differences (other than membrane-bound organelles) between typical cells of Eukarya and Bacteria. Which comparison is INCORRECT?
Click the INCORRECT comparison statement:
- DNA Structure: Eukarya = linear DNA associated with histone proteins; Bacteria = typically circular DNA lacking histones.
- Ribosomes: Eukarya = larger 80S ribosomes in cytoplasm/RER; Bacteria = smaller 70S ribosomes.
- Cell Walls: Eukarya = Present in plants (cellulose) & fungi (chitin), absent in animals; Bacteria = Usually present, contain peptidoglycan.
- Cell Division: Eukarya = Mitosis (nuclear division) and cytokinesis; Bacteria = Binary fission.
- Cell Size: Eukarya = typically 1-5 µm in diameter; Bacteria = typically >10 µm in diameter.
Reasoning for Incorrect Statement #5: Statement #5 incorrectly compares the typical sizes. Eukaryotic cells are generally much larger than bacterial (prokaryotic) cells. Typical eukaryotic cell diameters range from 10-100 µm, whereas typical bacterial cells range from 0.5-5 µm. Statement #5 reverses this relationship. The other comparisons (#1, #2, #3, #4) highlight generally accepted differences between typical eukaryotic and bacterial cells regarding DNA organization, ribosome size, cell wall composition (where present), and cell division mechanisms.
Describe how viruses are classified based on their structure. Which feature is LEAST commonly used for primary classification across all viruses?
Click the CORRECT answer:
- Type of nucleic acid (DNA or RNA).
- Presence or absence of a phospholipid envelope.
- Shape of the protein capsid (e.g., icosahedral, helical).
- Whether the nucleic acid is single-stranded or double-stranded.
- Presence or absence of tail fibres (relevant for bacteriophages but not a universal feature).
Reasoning for Correct Answer #5: Viral classification systems (like the Baltimore classification or ICTV system) primarily rely on fundamental characteristics shared across diverse virus groups. These include: the type of nucleic acid genome (DNA or RNA, #1), whether it’s single- or double-stranded (#4), the symmetry/structure of the protein capsid (#3), and the presence or absence of a lipid envelope (#2). While structures like tail fibres are important morphological features used to classify specific types of viruses, such as bacteriophages (viruses that infect bacteria), they are not present in all viruses and therefore are not a primary, universal feature used for the initial broad classification of all virus families.
In barley seeds, experiments showed blue light inhibits germination compared to darkness. Blue light exposure also increased abscisic acid (ABA) concentration, while ABA levels remained low in the dark. ABA is known to inhibit germination. What is the likely explanation for blue light’s effect?
Click the CORRECT answer:
- Blue light directly destroys the seed embryo.
- Blue light stimulates gibberellin production, which overrides ABA’s effect.
- Blue light increases ABA levels, and this higher ABA concentration inhibits the germination process (possibly by interfering with gibberellin action).
- ABA levels are irrelevant; blue light inhibits germination by damaging photosynthetic pigments.
- Blue light causes dehydration, preventing germination.
Reasoning for Correct Answer #3: The experimental results establish two key correlations: blue light inhibits germination, and blue light increases ABA concentration. We are also told that ABA inhibits germination. The most parsimonious explanation connecting these observations is that blue light exerts its inhibitory effect *by causing* an increase in the levels of the germination inhibitor, ABA. This increased ABA then acts physiologically to prevent or delay germination. Option #2 contradicts the known inhibitory effect of ABA. Option #4 ignores the strong correlation with ABA and introduces an irrelevant factor (photosynthetic pigments aren’t active in initial germination). Options #1 and #5 are speculative and not directly supported by the ABA data.
Describe and explain the role of auxin in cell elongation. Which step is NOT part of the acid growth hypothesis?
Click the step that is NOT part of the hypothesis:
- Auxin binds to specific receptor proteins, often located on the cell surface membrane.
- This binding triggers signalling that activates proton pumps (H+-ATPases) in the plasma membrane, which actively transport H+ ions out of the cytoplasm into the cell wall space.
- The resulting decrease in pH (acidification) in the cell wall activates pH-dependent enzymes called expansins.
- Expansins disrupt hydrogen bonds between cellulose microfibrils and other cell wall polymers, loosening the wall structure.
- Auxin directly enters the nucleus and acts as a catalyst for the synthesis of new cellulose microfibrils.
- With the wall loosened, the cell’s internal turgor pressure causes the cell to take up water by osmosis and stretch or elongate.
Reasoning for Incorrect Statement #5: The acid growth hypothesis explains how auxin induces cell wall loosening, allowing turgor-driven expansion. The key steps involve auxin stimulating proton pumping (#2), wall acidification (#3), expansin activation (#4), wall loosening, and water uptake leading to elongation (#6). Auxin itself does not directly synthesize cellulose (#5). Cellulose synthesis occurs at the plasma membrane via cellulose synthase complexes, and while auxin might influence this process indirectly or over longer time scales through gene expression changes, it’s not considered a direct catalytic step within the rapid acid growth mechanism.
Outline the roles of troponin, tropomyosin, actin, and myosin in sarcomere contraction. Which statement accurately describes one of these roles?
Click the CORRECT answer:
- Tropomyosin binds calcium ions, causing it to change shape.
- Actin filaments perform the power stroke, pulling myosin inwards.
- Myosin heads bind to actin, forming cross-bridges, after tropomyosin moves away from the binding sites.
- Troponin directly breaks down ATP to provide energy for contraction.
- Calcium ions bind to myosin heads, causing them to detach from actin.
Reasoning for Correct Answer #3: Muscle contraction regulation involves:
– Calcium ions bind to troponin (#1 incorrect).
– This causes troponin to change shape, pulling tropomyosin away from myosin-binding sites on actin.
– Once binding sites are exposed, energized myosin heads bind to actin, forming cross-bridges (#3 correct).
– Myosin heads then execute the power stroke, pulling actin filaments inwards (#2 incorrect direction).
– ATP binds to myosin heads, causing detachment from actin (#5 incorrect binding partner/effect). ATP hydrolysis by myosin re-energizes the head (#4 incorrect protein).
Therefore, statement #3 accurately describes the step where myosin binds actin after the regulatory proteins have moved.
Succinylcholine is a drug structurally similar to acetylcholine (ACh) used as a muscle relaxant. It binds to ACh receptors at the neuromuscular junction but is not broken down as quickly as ACh. Suggest how it prevents muscle contraction.
Click the CORRECT answer:
- It blocks the release of calcium ions from the sarcoplasmic reticulum.
- It acts as a competitive inhibitor, blocking ACh binding OR binds and causes prolonged depolarization preventing repolarization, thus blocking further action potential generation in the muscle fibre.
- It enhances the breakdown of acetylcholine by acetylcholinesterase.
- It directly prevents myosin heads from binding to actin filaments.
- It stimulates continuous muscle contraction leading to fatigue and relaxation.
Reasoning for Correct Answer #2: Succinylcholine is a neuromuscular blocking agent. Because it resembles ACh, it binds to the nicotinic ACh receptors on the muscle endplate. Its primary mechanism is as a depolarizing blocker:
1. It binds and initially opens the ACh receptor channels, causing muscle membrane depolarization and often initial muscle twitches (fasciculations).
2. Unlike ACh, which is rapidly hydrolyzed by acetylcholinesterase, succinylcholine persists in the synaptic cleft and bound to receptors for longer.
3. This prolonged binding leads to sustained depolarization of the muscle membrane.
4. Sustained depolarization inactivates the voltage-gated sodium channels required for generating further muscle action potentials, and prevents the membrane from repolarizing properly.
5. The muscle becomes unresponsive to subsequent nerve impulses, resulting in flaccid paralysis (relaxation).
Option #2 accurately describes this depolarizing blockade mechanism (prolonged depolarization prevents repolarization and further action potentials). While it also competes with ACh for binding initially, the persistent depolarization is key. Other options describe incorrect mechanisms.
Duchenne muscular dystrophy (DMD) is a genetic disease caused by a mutation in a single gene on the X chromosome that affects the protein dystrophin. The allele causing DMD is recessive. Explain why DMD mainly affects boys.
Click the CORRECT explanation:
- The gene is located on the Y chromosome, which only boys have.
- Girls have two X chromosomes; if one carries the recessive DMD allele, the dominant allele on the other X chromosome usually masks its effect, making them carriers. Boys have only one X chromosome (XY), so if they inherit the DMD allele on that X, they express the disease as there’s no corresponding allele on the Y.
- The recessive allele is only expressed in the presence of male hormones.
- Boys inherit two copies of the X chromosome, increasing their chances.
- The condition is dominant in boys but recessive in girls.
Reasoning for Correct Answer #2: DMD is an X-linked recessive disorder. This mode of inheritance leads to a much higher prevalence in males (XY) compared to females (XX).
– Males have only one X chromosome. If they inherit the X chromosome carrying the recessive DMD allele from their mother, they will express the disease because there is no corresponding gene (and no dominant allele) on their Y chromosome to mask the effect.
– Females have two X chromosomes. To express the disease, they must inherit the recessive DMD allele on both X chromosomes (one from each parent), which is statistically much less likely. If they inherit only one copy, they become carriers; the normal dominant allele on their other X chromosome typically produces enough functional dystrophin to prevent severe symptoms.
Option #2 accurately explains this genetic basis for the sex difference in DMD prevalence.