ESAT Worked Solutions · Biology
ESAT Paper 4 Biology Worked Solutions
Complete step-by-step worked solutions. Part of the ESAT preparation guide.
Question 1
Back to top ↑A particular species of fish is diploid and has five pairs of chromosomes in each of its body cells. During meiosis, how many ways are there in which these pairs of chromosomes can arrange themselves across the equator during metaphase I?
Key Idea (💡): During metaphase I of meiosis, homologous chromosome pairs line up randomly across the cell equator. This is known as independent assortment. Since each pair can orient in 2 different ways independently of the others, the total number of combinations is $2^n$, where $n$ is the number of pairs.
Reveal the answer & worked solution — commit to an option first
Correct Answer: C. 32
Fastest Approach (🚀):
Identify the number of pairs $n=5$. Calculate $2^5 = 32$.
Step-by-Step Breakdown:
1. Identify the number of homologous pairs
The question states that the fish has five pairs of chromosomes ($n = 5$).
2. Apply the rule of independent assortment
During metaphase I of meiosis, homologous pairs of chromosomes align along the equator of the cell. Each pair can arrange itself in one of two possible orientations (paternal chromosome on the left and maternal on the right, or vice versa).
Because the orientation of each pair is independent of the other pairs, we calculate the total number of possible arrangements by multiplying the possibilities for each pair: $2 \times 2 \times 2 \times 2 \times 2 = 2^5$.
3. Calculate the final value
There are 32 different ways the chromosomes can arrange themselves.
Common Mistake (⚠️):
Calculating $5 \times 2 = 10$ instead of $2^5$, or misinterpreting 'pairs of chromosomes' as individual chromosomes.
Question 2
Back to top ↑Which of the following statements best explains how oestrogen acts as a contraceptive in humans?
Key Idea (💡): In contraceptive pills, high levels of oestrogen exert negative feedback on the pituitary gland, inhibiting the secretion of FSH. Without FSH, follicles in the ovary do not mature, which ultimately prevents ovulation.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. It decreases the secretion of FSH
Fastest Approach (🚀):
Oestrogen inhibits FSH release. Option D directly states this.
Step-by-Step Breakdown:
1. Role of Oestrogen in Contraception
Combined oral contraceptives contain synthetic oestrogen (and progesterone). A continuous high level of oestrogen throughout the cycle causes negative feedback on the anterior pituitary gland.
2. Effect on FSH
This negative feedback specifically inhibits the release of Follicle Stimulating Hormone (FSH). Normally, FSH stimulates the maturation of a follicle in the ovaries. By decreasing FSH secretion, no follicle matures, and therefore no egg is released (ovulation is prevented).
3. Analyzing other options
A: Progesterone (not just oestrogen) maintains the endometrial lining.
B: Progesterone is responsible for thickening cervical mucus to prevent sperm entry.
- C and E: Oestrogen does not significantly alter vaginal acidity to kill sperm or harden the zona pellucida as its primary contraceptive mechanism.
Common Mistake (⚠️):
Confusing the roles of oestrogen and progesterone (e.g., assuming oestrogen thickens cervical mucus, which is actually progesterone's role).
Question 3
Back to top ↑A disease caused by a genetic disorder is associated with a family. A woman from this family does not have the disease and she has children with a man who also does not have the disease. They have three children, which include a healthy girl and two boys with the disease.
Which of the following statements could be true?
- The disease is recessive
- The disease is dominant
- The disease is caused by a mutation on the X-chromosome
- The mother is homozygous
Key Idea (💡): If two healthy parents have an affected child, they must both be carriers of a recessive allele (or the mother is a carrier of an X-linked recessive allele). The trait cannot be dominant, and the mother cannot be homozygous if she is a healthy carrier.
Reveal the answer & worked solution — commit to an option first
Correct Answer: C. 1 and 3
Fastest Approach (🚀):
Healthy parents -> affected child = Recessive (1 is true). It could be X-linked since only boys are affected (3 is true). It cannot be dominant (2 is false). Thus, 1 and 3.
Step-by-Step Breakdown:
1. Recessive vs Dominant
The parents do not have the disease, but they have children who do. This is the classic hallmark of a recessive disease. Both parents (or at least the mother, if X-linked) must be hiding the diseased allele. Therefore, Statement 1 is true, and Statement 2 is false.
2. Is it X-linked?
Could the disease be caused by a mutation on the X-chromosome? If it is an X-linked recessive trait, the mother could be a heterozygous carrier ($X^D X^d$). The father is healthy ($X^D Y$). They can produce affected sons ($X^d Y$) and healthy daughters ($X^D X^d$ or $X^D X^D$). This perfectly matches the children described. Therefore, Statement 3 could be true.
3. Is the mother homozygous?
The mother does not have the disease. If the trait is recessive, she must have at least one dominant healthy allele. Since she passed a diseased allele to her sons, she must also have the recessive diseased allele. Thus, she must be heterozygous, not homozygous. Statement 4 is false.
Statements 1 and 3 could be true.
Common Mistake (⚠️):
Assuming that because it's recessive, both parents must be carriers of an autosomal trait and completely forgetting the possibility of X-linked inheritance.
Question 4
Back to top ↑Integrase is an enzyme produced by HIV. The function of integrase is to cut open a section of the host cell's DNA and insert the virus' DNA and then to close the DNA.
Which of the following enzymes can catalyse the formation of phosphodiester bonds?
- Integrase
- Restriction enzyme
- Ligase
Key Idea (💡): Phosphodiester bonds link the sugar of one nucleotide to the phosphate of the next. Forming these bonds connects DNA strands together.
Reveal the answer & worked solution — commit to an option first
Correct Answer: E. 1 and 3
Fastest Approach (🚀):
Integrase 'closes' DNA (forms bonds). Ligase joins DNA fragments (forms bonds). Restriction enzymes only cut (break bonds). Thus 1 and 3.
Step-by-Step Breakdown:
1. Understanding Phosphodiester Bonds
Phosphodiester bonds are the strong covalent bonds that form the sugar-phosphate backbone of a DNA molecule. 'Forming' these bonds means joining pieces of DNA together.
2. Evaluating the Enzymes
- Integrase: The prompt states integrase cuts open DNA, inserts viral DNA, and then closes* the DNA. Closing the DNA requires sealing the sugar-phosphate backbone, meaning it must catalyse the formation of phosphodiester bonds.
- 2. Restriction enzyme: Restriction endonucleases function strictly as 'molecular scissors'. They cut DNA by breaking phosphodiester bonds. They do not form them.
- 3. Ligase: DNA ligase is the classic enzyme used in cells to seal nicks in the sugar-phosphate backbone, directly catalysing the formation of phosphodiester bonds to join Okazaki fragments or recombinant DNA.
Therefore, Integrase and Ligase (1 and 3) can catalyse the formation of these bonds.
Common Mistake (⚠️):
Believing Integrase only cuts DNA, ignoring the 'and then to close the DNA' part of the prompt.
Question 5
Back to top ↑Which of the following describes the series of events following a person becoming dehydrated?
Key Idea (💡): When a person is dehydrated, their blood water potential falls. This triggers the release of more ADH, which makes the kidney collecting ducts more permeable to water, allowing more water to be reabsorbed into the blood.
Reveal the answer & worked solution — commit to an option first
Correct Answer: B. Release of ADH increases, causing urine to decrease in volume and increase in concentration
Fastest Approach (🚀):
Dehydration means the body needs to save water. So ADH increases. This produces a small volume of highly concentrated urine.
Step-by-Step Breakdown:
1. Initial stimulus
Dehydration means there is a low water concentration (low water potential) in the blood.
2. Hormone release
Osmoreceptors in the hypothalamus detect this and stimulate the posterior pituitary gland to release more Antidiuretic Hormone (ADH) into the bloodstream.
3. Effect on Kidneys
ADH travels to the kidneys and binds to receptors on the collecting ducts, increasing their permeability to water. This causes more water to be reabsorbed by osmosis back into the blood.
4. Result on Urine
Because more water has been removed from the filtrate and kept in the body, the resulting urine has a smaller volume and is more highly concentrated (containing more urea and salts per unit volume).
Common Mistake (⚠️):
Confusing the effects of ADH, thinking it increases urine volume (diuresis) when its name is anti-diuretic (stops diuresis).
Question 6
Back to top ↑Which of the following processes is not/are not directly involved in the carbon cycle?
- Combustion
- Dissolving of limestone
- Denitrification
Key Idea (💡): The carbon cycle tracks the movement of carbon atoms. Combustion (burning carbon-based fossil fuels) releases CO2. Dissolving of limestone (calcium carbonate) releases carbon into water. Denitrification is part of the nitrogen cycle.
Reveal the answer & worked solution — commit to an option first
Correct Answer: C. 3 only
Fastest Approach (🚀):
Identify denitrification as part of the nitrogen cycle. Combustion and limestone both involve carbon.
Step-by-Step Breakdown:
1. Evaluate Combustion
Combustion involves burning organic material (like fossil fuels or wood). This reacts carbon-based compounds with oxygen to produce carbon dioxide (CO$_2$), directly releasing carbon into the atmosphere. This is a key part of the carbon cycle.
2. Evaluate Dissolving of limestone
Limestone is primarily composed of calcium carbonate (CaCO$_3$). When it dissolves, it releases carbonate and bicarbonate ions into the water, which can eventually release CO$_2$. This is part of the long-term geological carbon cycle.
3. Evaluate Denitrification
Denitrification is the process by which soil bacteria convert nitrates (NO$_3^-$) into nitrogen gas (N$_2$). This is a fundamental step in the nitrogen cycle, not the carbon cycle.
Therefore, only Denitrification (3) is not directly involved in the carbon cycle.
Common Mistake (⚠️):
Confusing denitrification with a carbon-releasing process, or assuming limestone isn't part of the carbon cycle because it's a rock.
Question 7
Back to top ↑Between 1954 and 1971 a mine in Northern Australia was releasing excess copper into the nearby Finnis River. Copper ions are poisonous and most of the types of fish living in the river died. The river remains polluted with copper, but scientists have discovered one type of rainbow fish that is able to survive and live in the river.
A student wrote the following statements to explain this information.
- One type of rainbow fish did not die out in the river because this type was able to adapt to the changing environment
- None of the other types of fish showed any type of genetic variation.
- The presence of copper ions acted as a selective pressure
Which of the following statements could be correct?
Key Idea (💡): The copper ions created a strong selective pressure. The rainbow fish population had pre-existing genetic variation that included a resistance allele, allowing them to survive, reproduce, and adapt as a population. Other fish populations died because they lacked that specific resistance allele, but they still had genetic variation in other traits.
Reveal the answer & worked solution — commit to an option first
Correct Answer: C. 1 and 3 only
Fastest Approach (🚀):
Statement 2 is clearly false (all species show genetic variation). Eliminate options B, D, and E. Statement 3 is definitely true (copper is the selective pressure). Therefore, it must be 1 and 3.
Step-by-Step Breakdown:
1. Evaluating Statement 1
"One type of rainbow fish... was able to adapt..." In biology, populations adapt to changing environments over generations via natural selection. Because they survived while others died, it is correct to say the species (type) adapted to the new environment.
2. Evaluating Statement 2
"None of the other types of fish showed any type of genetic variation." This is false. All sexually reproducing populations show genetic variation. The other types of fish simply didn't possess the specific genetic variation (allele) required to survive copper toxicity.
3. Evaluating Statement 3
"The presence of copper ions acted as a selective pressure." This is true. A selective pressure is any environmental factor that alters the reproductive success of organisms in a population. The copper killed susceptible fish, selecting for resistant ones.
Therefore, statements 1 and 3 could be correct.
Common Mistake (⚠️):
Believing statement 2 is true because the other fish died out, confusing a lack of one specific trait with a lack of any genetic variation.
Question 8
Back to top ↑The diagram shows a plant cell, where arrow 1 points into the cell and arrow 2 points out of the cell across the cell surface membrane, while arrow 3 points into the vacuole and arrow 4 points out of the vacuole across the tonoplast.
Which of the arrows show the net movement of water molecules, by osmosis, when the cell is surrounded by a solution that is more concentrated than the solution in the cytoplasm?
Key Idea (💡): Osmosis is the net movement of water from a higher water potential to a lower water potential. A 'more concentrated' external solution has a lower water potential than the cell.
Reveal the answer & worked solution — commit to an option first
Correct Answer: F. 2 and 4 only
Fastest Approach (🚀):
More concentrated outside = water leaves cell (Arrow 2). Vacuole then loses water to cytoplasm (Arrow 4). Thus, 2 and 4.
Step-by-Step Breakdown:
1. Movement across the cell surface membrane
The cell is placed in a solution that is more concentrated (hypertonic) than the cytoplasm. This means the external solution has a lower water potential.
Water always moves down its water potential gradient, so water will leave the cytoplasm and exit the cell. This is represented by Arrow 2 (pointing outward).
2. Movement across the tonoplast (vacuole membrane)
As water leaves the cytoplasm, the cytoplasm becomes more concentrated, lowering its water potential. The cytoplasm now has a lower water potential than the vacuole.
Water moves from the higher water potential in the vacuole to the lower water potential in the cytoplasm. This is represented by Arrow 4 (pointing out of the vacuole).
Therefore, arrows 2 and 4 correctly show the net movement of water.
Common Mistake (⚠️):
Thinking the cell absorbs water when in a concentrated solution, or forgetting the vacuole's interaction with the cytoplasm.
Question 9
Back to top ↑Albinism is a recessive genetic condition that results in the absence of the pigment melanin in the skin, hair and eyes. In a population of 580,000 people there were 29 albinos and 81,200 symptomless carriers. One living cheek cell was collected from every individual in the population.
What is the number of albinism alleles in these cells?
Key Idea (💡): A cheek cell is a somatic (diploid) cell. Therefore, an albino (homozygous recessive, aa) has 2 albinism alleles per cell. A carrier (heterozygous, Aa) has 1 albinism allele per cell. Normal individuals have 0.
Reveal the answer & worked solution — commit to an option first
Correct Answer: E. 81 258
Fastest Approach (🚀):
(29 albinos * 2) + (81200 carriers * 1) = 58 + 81200 = 81258.
Step-by-Step Breakdown:
1. Understand cell ploidy
The question specifies one living cheek cell was collected per person. Cheek cells are diploid somatic cells, meaning they contain two alleles for every gene.
2. Calculate alleles from albinos
Albinism is a recessive condition. To be an albino, an individual must be homozygous recessive (aa). Therefore, every cheek cell from an albino contains 2 albinism alleles.
Alleles from albinos = 29 individuals $\times$ 2 alleles/cell = 58 alleles.
3. Calculate alleles from carriers
Carriers are symptomless, meaning they are heterozygous (Aa). Every cheek cell from a carrier contains exactly 1 albinism allele.
Alleles from carriers = 81,200 individuals $\times$ 1 allele/cell = 81,200 alleles.
4. Total alleles
Individuals who are completely unaffected and not carriers (AA) contribute 0 albinism alleles.
Total albinism alleles = 58 + 81,200 = 81,258.
Common Mistake (⚠️):
Forgetting to multiply the number of albinos by 2, leading to the incorrect answer of 81,229.
Question 10
Back to top ↑A student set up the following apparatus at a temperature of 25°C and at pH 7: a test tube containing a cloudy protein solution had 10cm³ of human enzyme solution added to it, and 15 minutes later the solution had become clear.
What could the student change so that it would take less than 15 minutes for the solution to become clear?
Key Idea (💡): To decrease the time taken (increase the rate of reaction), we must increase the frequency of successful collisions between enzyme and substrate. Stirring achieves this. Drastic changes to human enzymes (70°C, pH 13) cause denaturation.
Reveal the answer & worked solution — commit to an option first
Correct Answer: B. Carry out the experiment, stirring the mixture once every 30 seconds.
Fastest Approach (🚀):
70°C and pH 13 denature human enzymes. Changing total volume without changing concentration keeps rate identical. Stirring increases collision frequency, speeding up the reaction.
Step-by-Step Breakdown:
1. Analyze extreme conditions (A and C)
The enzyme is a human enzyme. Human enzymes typically have an optimum temperature around 37°C and function near neutral pH (except for stomach enzymes).
- A: Increasing temperature to 70°C will quickly denature the enzyme by breaking the bonds maintaining its tertiary structure. The active site changes shape, and the reaction will stop, taking longer (or never finishing).
- C: A pH of 13 is extremely alkaline. This will also denature most human enzymes, stopping the reaction.
2. Analyze volume changes (D and E)
D and E: Doubling or halving the volumes of both* solutions keeps their relative concentrations exactly the same. The frequency of collisions per unit volume remains identical. Therefore, the time taken for the solution to clear will remain approximately the same (15 minutes).
3. Analyze stirring (B)
- B: Stirring the mixture adds kinetic energy and physically mixes the molecules, preventing the local depletion of substrate near enzymes (maintaining a steep concentration gradient). This increases the frequency of successful enzyme-substrate collisions, increasing the rate of reaction. Consequently, it will take less time to clear.
Common Mistake (⚠️):
Thinking that raising the temperature always increases the rate, forgetting that 70°C is well past the denaturation point for human enzymes.
Question 11
Back to top ↑In an investigation, a molecule of DNA was extracted and separated into its single strands 1 and 2. The percentage of each base present in each strand was found.
The table shows some of the results in strand 1:
| DNA Sample | Adenine (A) | Cytosine (C) | Guanine (G) | Thymine (T) |
|---|---|---|---|---|
| Strand 1 | 26 | ? | 28 | 14 |
P, Q, R and S are the percentages of each base in the complementary strand 2.
| DNA Sample | Adenine (A) | Cytosine (C) | Guanine (G) | Thymine (T) |
|---|---|---|---|---|
| Strand 2 | P | Q | R | S |
A student calculates the following percentages for P, Q, R and S:
P = 14%
Q = 28%
R = 26%
S = 28%
What percentage is/are correct?
Key Idea (💡): The total percentage of bases in a single strand must be 100%. Because the strands are complementary, the percentage of Adenine in Strand 1 is equal to the percentage of Thymine in Strand 2, and so on.
Reveal the answer & worked solution — commit to an option first
Correct Answer: E. P and Q only
Fastest Approach (🚀):
Strand 1 Cytosine = 100 - (26+28+14) = 32%. Strand 2: P(A)=T1=14%. Q(C)=G1=28%. R(G)=C1=32%. S(T)=A1=26%. Compare with student's values: P(14) is correct. Q(28) is correct. R(26) is wrong. S(28) is wrong. Thus P and Q only.
Step-by-Step Breakdown:
1. Complete the data for Strand 1
The total percentage of bases in Strand 1 must equal 100%.
%C = 100% - (%A + %G + %T)
%C = 100% - (26% + 28% + 14%) = 100% - 68% = 32%.
So for Strand 1: A = 26%, C = 32%, G = 28%, T = 14%.
2. Use complementary base pairing for Strand 2
Strand 2 is complementary to Strand 1, meaning Adenine pairs with Thymine, and Cytosine pairs with Guanine.
- P (Adenine in Strand 2) = Thymine in Strand 1 = 14%
- Q (Cytosine in Strand 2) = Guanine in Strand 1 = 28%
- R (Guanine in Strand 2) = Cytosine in Strand 1 = 32%
- S (Thymine in Strand 2) = Adenine in Strand 1 = 26%
3. Check the student's calculations
- Student's P = 14%. (Correct)
- Student's Q = 28%. (Correct)
- Student's R = 26%. (Incorrect, should be 32%)
- Student's S = 28%. (Incorrect, should be 26%)
Only P and Q are correct.
Common Mistake (⚠️):
Applying Chargaff's rule (%A = %T) to a single strand, rather than across complementary strands.
Question 12
Back to top ↑The sex of species Q is controlled by two chromosomes X and Y.
The sex of females of species Q is controlled by inheriting the same combination of sex chromosomes as healthy male humans. The sex of males in species Q is inherited in the same way as healthy female humans.
The family tree for one population of species Q is shown below (circles represent females, squares represent males). At the top, one male and one female have four children: two males and two females. Each of the two male children marries a female from outside the family. One of these couples has three children, two males and one female, and the other couple has three children, all female.
What is the ratio in its simplest form of males to females and the total number of Y chromosomes in this family tree?
Key Idea (💡): In species Q, females are XY (heterogametic) and males are XX (homogametic). To find the ratio, count squares (males) and circles (females). To find total Y chromosomes, simply count the number of females, as each female has exactly one Y chromosome and males have none.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. Ratio 5:9, Total Y 9
Fastest Approach (🚀):
Count shapes: 5 Squares (males), 9 Circles (females). Ratio males:females = 5:9. Total Y chromosomes = number of females = 9. Option D.
Step-by-Step Breakdown:
1. Determine Sex Chromosomes
The prompt states females of species Q have the same chromosomes as human males (XY). Males of species Q have the same as human females (XX).
Females (Circles): XY (contains 1 Y chromosome)
Males (Squares): XX (contains 0 Y chromosomes)
2. Count the Individuals in the Pedigree
Carefully trace the family tree to count all squares and circles:
Top generation: 1 Circle, 1 Square.
Middle generation: The original couple has 4 children (Square, Circle, Circle, Square). The two Square children marry outsider Circles. So this generation has: 2 outsider Circles, 2 child Circles, 2 child Squares.
- Bottom generation: The left couple has 3 children (Square, Circle, Square). The right couple has 3 children (Circle, Circle, Circle).
Total Squares (Males) = 1 (top) + 2 (middle) + 2 (bottom left) + 0 (bottom right) = 5 Males.
Total Circles (Females) = 1 (top) + 4 (middle) + 1 (bottom left) + 3 (bottom right) = 9 Females.
3. Calculate Ratio and Y Chromosomes
Ratio of males to females: 5 : 9
Total Y chromosomes: Only females have Y chromosomes, and they each have exactly 1. Therefore, 9 females = 9 Y chromosomes.
This matches Option D.
Common Mistake (⚠️):
Miscounting the shapes in the pedigree, or forgetting that in this species, females are XY.
Question 13
Back to top ↑A graph shows how a number of factors — oxygen concentration, bacteria numbers, algae numbers, and fish numbers — vary with distance down a river, after a source of pollution flowed in.
(Assume the oxygen concentration is changing only based on the species present in the river)
Reading downstream from the pollution source: bacteria numbers spike first and oxygen begins to fall (points 1 and 2); further downstream, algae numbers start to rise while oxygen is still low (point 3); oxygen remains low a little further on (point 4); and further downstream still, oxygen has recovered and fish numbers increase (point 5).
Which one of the statements below can be correctly concluded from the graph?
Key Idea (💡): The graph depicts organic pollution (like sewage). Bacteria multiply rapidly to feed on it, using up dissolved oxygen through aerobic respiration. This kills fish. Later, minerals released allow algae to bloom and produce oxygen, allowing the ecosystem to recover.
Reveal the answer & worked solution — commit to an option first
Correct Answer: C. At point 3, the oxygen concentration is decreasing because bacteria are using up more oxygen than the algae are producing
Fastest Approach (🚀):
Check C: At point 3, oxygen is near its lowest, while algae are rising. The only way oxygen can still be low/decreasing while algae produce it is if the bacteria are still consuming it faster than it's produced. This is correct.
Step-by-Step Breakdown:
1. Analyze the initial drop in oxygen (Options A and B)
A: At point 1, oxygen decreases because bacteria are carrying out aerobic respiration, not anaerobic. Anaerobic respiration does not use oxygen.
B: At point 2, oxygen is decreasing because of bacteria, not algae. If high numbers of algae were photosynthesising, oxygen would increase.
2. Analyze the recovery phase (Options C and E)
C: At point 3, the algae population is increasing (producing oxygen), but the oxygen level in the water is still very low or decreasing. This indicates a net loss of oxygen, meaning the massive bacterial population is still consuming oxygen (via respiration) faster than the algae can produce it. This is a correct conclusion.
E: At point 5, fish numbers increase because the oxygen levels have recovered, not because of 'less competition with algae'. Algae are primarily oxygen producers during the day.
3. Analyze indicator species (Option D)
- D: Bloodworms and sludge worms are adapted to survive in highly polluted, low-oxygen environments. At point 4, where oxygen is low, their numbers would be at their highest, not lowest.
Therefore, statement C is the only correct conclusion.
Common Mistake (⚠️):
Choosing A because bacteria are associated with pollution, overlooking the word 'anaerobic' which means without oxygen.
Question 14
Back to top ↑The graph shows the effect of increasing the substrate concentration on an enzyme-controlled reaction when all the other variables were kept constant: a graph of the y-variable against substrate concentration is a horizontal straight line at a constant positive value, starting from substrate concentration = 0.
Which of the following labels, if any, could be correct for the y-axis?
- Rate of substrate loss /mg min⁻¹
- Rate of enzyme-substrate complex formation/ number of complexes s⁻¹
- Rate of product formed per enzyme molecule/ mg min⁻¹
Key Idea (💡): The graph shows a completely flat, horizontal line starting from the y-axis. The y-intercept is a positive, non-zero value. This means when the substrate concentration is exactly 0, the y-variable is already at a high constant value.
Reveal the answer & worked solution — commit to an option first
Correct Answer: A. None of them
Fastest Approach (🚀):
If [Substrate] = 0, reaction rate = 0. The graph has a non-zero y-intercept. None of the rate options can be correct.
Step-by-Step Breakdown:
1. Analyze the graph's features
The x-axis is 'concentration of substrate'. The graph features a perfectly horizontal line that intersects the y-axis at a positive value. This indicates that whatever is on the y-axis has a constant, non-zero value even when the substrate concentration is zero.
2. Evaluate the proposed labels
- 1. Rate of substrate loss: If there is no substrate present ([S] = 0), no substrate can be lost. The rate must be 0. The graph shows a non-zero rate at [S] = 0, so this label is incorrect.
- 2. Rate of enzyme-substrate complex formation: Without substrate, no complexes can form. The rate must be 0 at [S] = 0.
- 3. Rate of product formed: Without substrate, no product can be formed. The rate must be 0 at [S] = 0.
3. Conclusion
Because all three proposed labels represent reaction rates, they must all pass through the origin (0,0) when plotted against substrate concentration (they would increase and eventually plateau). A horizontal line with a positive y-intercept is impossible for these variables.
Therefore, none of the labels are correct.
Common Mistake (⚠️):
Focusing on the flat part of a typical enzyme kinetics graph (Vmax) and ignoring that the flat line shown starts at a substrate concentration of zero.
Question 15
Back to top ↑Which of the following will always lead to an increase in the size of a population?
- Removal of intraspecific competitors
- Supplying more of the nutrient that is at the lowest concentration
- Change in the organism's rate of reproduction.
Key Idea (💡): No single environmental change will always increase a population, because a population's size is controlled by a complex web of biotic and abiotic factors. If you fix one limiting factor, another immediately becomes the new limit.
Reveal the answer & worked solution — commit to an option first
Correct Answer: A. None of them
Fastest Approach (🚀):
Removal of your own species = population decreases initially. Nutrient at lowest concentration might not be the limiting nutrient (trace elements). Rate of reproduction could decrease. None 'always' increase it.
Step-by-Step Breakdown:
1. Evaluate Statement 1: Removal of intraspecific competitors
Intraspecific competitors are members of the same species. Removing them actually immediately decreases the population size. While it frees up resources, it does not guarantee the population will end up larger than it originally was, especially if another factor (like predation) is limiting it.
2. Evaluate Statement 2: Supplying more of the nutrient at lowest concentration
The nutrient at the 'lowest concentration' is not necessarily the limiting nutrient. An organism might require a macronutrient in huge quantities (making it limiting even if present in moderate amounts) but only require a micronutrient in trace amounts (so its very low concentration is perfectly sufficient). Even if it was the limiting nutrient, supplying more of it won't always increase the population because another factor (space, temperature, disease) might immediately take over as the limiting factor.
3. Evaluate Statement 3: Change in the rate of reproduction
A 'change' could be a decrease. A decrease in the rate of reproduction would lead to a population decline, not an increase.
Therefore, none of them will always lead to an increase.
Common Mistake (⚠️):
Assuming 'lowest concentration' equals 'limiting factor' and forgetting that a 'change' in reproduction could be negative.
Question 16
Back to top ↑A desert food chain is shown below
cactus -> rat -> rattlesnake -> hawk
Assume that 10% of the energy from each stage in the food chain is passed on.
If 150,000 units of energy are contained in the producer, how much energy will be lost in the transfer between the primary and secondary consumers?
Key Idea (💡): Energy is transferred at 10% efficiency. The rest (90%) is 'lost' to the environment (e.g., through heat from respiration). We need to calculate the energy in the primary consumer, then find 90% of that value to determine what is lost when transferring to the secondary consumer.
Reveal the answer & worked solution — commit to an option first
Correct Answer: B. 13,500
Fastest Approach (🚀):
Producer = 150,000. Primary consumer (rat) = 15,000. Secondary consumer (snake) = 1,500. Lost = 15,000 - 1,500 = 13,500.
Step-by-Step Breakdown:
1. Identify the trophic levels
Producer: Cactus
Primary Consumer: Rat
- Secondary Consumer: Rattlesnake
2. Calculate energy at each relevant level
Energy in Producer (Cactus): 150,000 units
Energy in Primary Consumer (Rat): 10% of 150,000 = 15,000 units
- Energy transferred to Secondary Consumer (Rattlesnake): 10% of 15,000 = 1,500 units
3. Calculate the energy lost
The question asks for the energy lost in the transfer between the primary and secondary consumers.
Energy lost = Energy available in primary consumer - Energy successfully transferred to secondary consumer
Energy lost = 15,000 - 1,500 = 13,500 units.
*(Alternatively, calculate 90% of 15,000 = 13,500).*
Matches Option B.
Common Mistake (⚠️):
Calculating the energy transferred (1,500) or the energy lost from the producer to the primary consumer (135,000) instead of reading the question carefully.
Question 17
Back to top ↑A student carried out an experiment to investigate the effect of enzyme concentration on the rate of an enzyme-controlled reaction.
The student used a starch agar plate with five identically sized small wells cut into the agar. The wells were filled with identical volumes of different concentrations of amylase solution.
The starch agar plate was incubated overnight, and the plate was then flooded with iodine solution. Most of the agar stained blue, but there was a clear area around each well where starch had been digested by the amylase.
The results are shown in the table below.
| Percentage concentration of amylase | Diameter of clear area/mm |
|---|---|
| 1.0 | 27 |
| 0.1 | 24 |
| 0.01 | 15 |
| 0.001 | 12 |
| 0.0001 | 9 |
Which of the following factors could have affected the diameter of the clear area around the wells containing amylase?
- pH of the starch agar
- concentration of the amylase solution
- temperature at which the plates were incubated
Key Idea (💡): The diameter of the clear area depends on how fast the enzyme diffuses and digests the starch. Any factor that affects enzyme activity (pH, temperature, concentration) will affect the rate of digestion and thus the diameter.
Reveal the answer & worked solution — commit to an option first
Correct Answer: H. 1, 2, and 3
Fastest Approach (🚀):
pH affects enzymes (1 is true). Amylase concentration is literally the independent variable shown in the table to affect diameter (2 is true). Temperature affects kinetic energy and enzyme activity (3 is true). Therefore, 1, 2, and 3.
Step-by-Step Breakdown:
1. pH of the starch agar (Statement 1)
Enzymes have an optimum pH. Changes in pH can alter the charges on the amino acids in the active site, affecting substrate binding, or even denature the enzyme. Thus, the pH of the agar definitely affects the rate of digestion and the clear area diameter.
2. Concentration of the amylase solution (Statement 2)
The provided table explicitly proves that changing the concentration of the amylase solution changes the diameter of the clear area (e.g., 1.0% gives 27 mm, 0.0001% gives 9 mm). A higher concentration means more active sites are available to digest starch.
3. Temperature (Statement 3)
Temperature affects the kinetic energy of both the enzyme and substrate molecules. Higher temperatures (up to the optimum) increase the rate of diffusion through the agar and the frequency of successful collisions, leading to a larger clear area.
Therefore, all three factors affect the diameter of the clear area.
Common Mistake (⚠️):
Thinking that because concentration was the only variable intentionally changed (independent variable), it's the only one that could affect the result.
Question 18
Back to top ↑Which of the following statements about bacterial cell division is/are correct?
- Daughter cells show a large degree of genetic difference to the parent cell
- After each division, two daughter cells are produced by mitosis
- Chromosome replication occurs in the cytoplasm of the bacteria
Key Idea (💡): Bacteria are prokaryotes. They lack a nucleus, so their circular DNA is free in the cytoplasm. They divide asexually by binary fission, producing genetically identical daughter cells.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. 3 only
Fastest Approach (🚀):
1 is false (they are clones). 2 is false (they use binary fission, not mitosis). 3 is true (no nucleus, so DNA replicates in cytoplasm). Thus, 3 only.
Step-by-Step Breakdown:
1. Evaluate Statement 1
Bacterial cell division is an asexual process (binary fission). The DNA is copied and the cell splits, producing daughter cells that are genetically identical clones of the parent cell (barring rare mutations). They do not show a 'large degree' of genetic difference. (Statement 1 is false).
2. Evaluate Statement 2
Mitosis is a process specific to eukaryotes, involving the breakdown of a nuclear envelope and the use of spindle fibers to separate multiple linear chromosomes. Bacteria do not undergo mitosis; they undergo binary fission. (Statement 2 is false).
3. Evaluate Statement 3
Because bacteria are prokaryotes, they do not have a nucleus. Their genetic material (a singular circular chromosome) resides freely in the nucleoid region of the cytoplasm. Therefore, chromosome replication must occur in the cytoplasm. (Statement 3 is true).
Only statement 3 is correct.
Common Mistake (⚠️):
Believing bacteria undergo mitosis, simply because they divide into two cells.
Question 19
Back to top ↑The hormone ecdysone is synthesised in the prothoracic glands found in the upper thorax of some invertebrates and is released into the haemolymph. It is then transported to cells near the surface of the body and causes the loss of the exoskeleton so that a new exoskeleton can form.
Which of the following statements explains how ecdysone can act on cells near the surface of the body?
- Ecdysone is synthesised by specialised neurosecretory cells
- Ecdysone is soluble in haemolymph because it is a polar molecule
- Ecdysone is complementary to cell surface receptors on cells throughout the body of some invertebrates
Key Idea (💡): Ecdysone is synthesised in glands, not neurosecretory cells, so Statement 1 is false. It still acts via intracellular receptors, not cell-surface receptors, so Statement 3 is false. But unlike typical vertebrate steroid hormones, ecdysone (an ecdysteroid) is heavily hydroxylated, making it distinctly more polar and haemolymph-soluble than a typical steroid -- so Statement 2 is true.
Reveal the answer & worked solution — commit to an option first
Correct Answer: F. 2 only
Fastest Approach (🚀):
Statement 1 is false: the prompt says ecdysone is made in glands, not neurosecretory cells. Statement 3 is false: ecdysone still acts via intracellular receptors, not cell-surface receptors. Statement 2 is true: unlike typical steroid hormones, ecdysone is heavily hydroxylated, making it distinctly more polar and haemolymph-soluble.
Matches Option F.
Step-by-Step Breakdown:
1. Evaluate Statement 1
The prompt explicitly states ecdysone is synthesised in the prothoracic glands. Glands are part of the endocrine system and are composed of epithelial glandular cells, not neurosecretory (nerve) cells. While its release is triggered by a neurosecretory hormone (PTTH), ecdysone itself is not synthesised by neurosecretory cells. Statement 1 is false.
2. Evaluate Statement 2
Ecdysone is a steroid hormone, but unlike typical vertebrate steroid hormones such as testosterone or cortisol, ecdysone and other ecdysteroids are heavily hydroxylated -- carrying several hydroxyl groups across the steroid ring system. This makes them distinctly more polar and water-soluble than a typical steroid, which is why ecdysone can travel through the aqueous haemolymph without the same reliance on carrier proteins that the more lipophilic vertebrate steroids need. Statement 2 is true.
3. Evaluate Statement 3
Despite being more polar than a typical steroid, ecdysone is still fundamentally lipid-soluble enough to diffuse across the phospholipid bilayer of the cell membrane. It acts by binding to intracellular (nuclear) receptors, not cell-surface receptors. Statement 3 is false.
4. Conclusion
Only Statement 2 correctly explains how ecdysone can act on cells near the surface of the body.
Matches Option F.
Common Mistake (⚠️):
Assuming ecdysone, as a "steroid hormone," must be non-polar like typical vertebrate steroids -- ecdysteroids are a heavily-hydroxylated exception, making them notably more polar and water-soluble.
Question 20
Back to top ↑Four types of neurone are shown, labelled A to D. Diagram A has a cell body at one end with several short dendrites and a single long axon extending away. Diagram B has a highly branched, multipolar structure with the cell body surrounded by dendrites. Diagram C has a cell body sitting in the middle of one continuous axon. Diagram D has a single long axon, split into two branches, with the cell body extending off to the side on a short stalk.
Which of the diagrams A to D shows a sensory neurone?
Key Idea (💡): Different functional classes of neurones have distinct shapes. Motor neurones have the cell body at one end with many dendrites (multipolar). Sensory neurones typically have their cell body branching off the middle of the axon (pseudo-unipolar), residing in the dorsal root ganglion.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. Diagram D
Fastest Approach (🚀):
Look for the cell body hanging off the side of a long axon. This is Diagram D.
Step-by-Step Breakdown:
1. Analyze Diagram A
Diagram A shows a cell body at one end with multiple short dendrites and a single long axon extending away. This is the classic structure of a motor neurone.
2. Analyze Diagram B
Diagram B shows a highly branched, multipolar structure with the cell body surrounded by dendrites. This is typically an interneurone (relay neurone) found in the central nervous system.
3. Analyze Diagram C
Diagram C shows a cell body situated directly in the middle of a continuous axon pathway. This is a bipolar neurone, which acts as a sensory neurone in highly specific areas like the retina or olfactory system, but is not the general representation of a somatic sensory neurone.
4. Analyze Diagram D
Diagram D shows a single long axon (divided into a peripheral and central branch) with the cell body extending off the side on a short stalk. This is a pseudo-unipolar neurone. This is the classic morphology of a typical sensory neurone (like those whose cell bodies cluster in the dorsal root ganglion).
Therefore, Diagram D is the sensory neurone.
Common Mistake (⚠️):
Confusing the motor neurone (A) with the sensory neurone (D).
Question 21
Back to top ↑Cyanobacteria are photosynthetic prokaryotes.
Scientists exposed cyanobacteria to light of different colours and intensities and made the following observations:
- Most cyanobacteria are blue in colour
- At a low light intensities, glucose production in cyanobacteria is low.
- When light intensity reaches a certain level the rate of glucose production in cyanobacteria stops increasing.
Which of the following statements, A to D, correctly explains these observations?
Key Idea (💡): If cyanobacteria appear blue, they are reflecting blue light and absorbing other wavelengths (like red). At low intensities, increasing the light increases the rate of glucose production, meaning light intensity is the limiting factor.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. The pigments in a cyanobacteria absorb red light and light intensity is a limiting factor for the rate of photosynthesis
Fastest Approach (🚀):
Blue colour = reflects blue, absorbs red. Eliminates A and C. Low light limits glucose production, so it is a limiting factor. Eliminates B. Leaves D.
Step-by-Step Breakdown:
1. Determine absorbed wavelengths
The colour an object appears is the colour of light it reflects. Since cyanobacteria are blue in colour, their pigments are reflecting blue light. This means they must be absorbing other wavelengths of light, such as red, to use for photosynthesis. Therefore, they do not absorb blue light.
2. Determine the limiting factor
A limiting factor is a variable that restricts the rate of a reaction. The observations state that at low light intensities, glucose production is low, and increasing it (up to a certain point) increases production. This indicates that at those lower levels, light intensity is a limiting factor. Once it reaches a high level and plateaus, some other factor (like CO$_2$ or temperature) has become limiting.
Statement D correctly identifies that they absorb red light and that light intensity is a limiting factor.
Common Mistake (⚠️):
Thinking that because an organism is blue, it absorbs blue light. It actually reflects it.
Question 22
Back to top ↑The thyroid gland absorbs any iodine that enters the body, so the radioactive isotope kills the cancerous cells in the thyroid gland. The I¹³¹ is then excreted from the body.
Different body fluids excrete different proportions of I¹³¹: urine has the highest rate of excretion, followed by faeces, then a smaller amount via sweat, with none excreted via expired air.
Which of the following A to D, correctly explains the different proportions of I¹³¹ in each body fluid?
Key Idea (💡): Urine and sweat are aqueous (water-based) solutions. If a substance is excreted in high amounts via urine and sweat, but zero amounts in expired air, it must be highly soluble in water and not volatile.
Reveal the answer & worked solution — commit to an option first
Correct Answer: A. I¹³¹ is very soluble in water
Fastest Approach (🚀):
The graph shows I131 leaves mostly via urine. Urine is mostly water. Thus, I131 must be highly water-soluble. Option A.
Step-by-Step Breakdown:
1. Analyze the graph
The graph shows that I$^{131}$ is excreted primarily in urine. It is also excreted in faeces and sweat, but not at all in expired air.
2. Evaluate the options
- A. I$^{131}$ is very soluble in water: Urine and sweat are primarily water. For a substance to be filtered by the kidneys and excreted in high concentrations in urine, it must dissolve well in blood plasma and ultimately in the water of the urine. This perfectly explains the graph.
- B. Cross capillary walls: While true (it must cross capillaries in the glomerulus), this doesn't explain why it favors urine over other routes.
C. Kidneys are more efficient: This is a description of the graph, not an explanation of the underlying chemical reason.
D. Thyroid blood supply: This explains how the iodine reaches the thyroid, not how it is excreted.
Therefore, its high solubility in water is the fundamental reason it is excreted via urine and sweat.
Common Mistake (⚠️):
Choosing C because it seems to describe the graph perfectly, without realizing the question asked for an explanation of the proportions, not a description.
Question 23
Back to top ↑Which of the following statements, A-D, does not provide evidence for natural selection?
Key Idea (💡): Evidence for natural selection usually involves demonstrating how a population has adapted to a specific selective pressure (like predators, antibiotics, or pesticides). Organisms remaining identical over long periods does not actively provide evidence for natural selection driving evolutionary change.
Reveal the answer & worked solution — commit to an option first
Correct Answer: B. Fossils of animals identical to species living today can be found in shallow rocks
Fastest Approach (🚀):
A, C, and D all describe populations changing in response to an environmental threat (snails, DDT, antibiotics). B describes stasis. B is the answer.
Step-by-Step Breakdown:
1. Evaluate A (Clover and Snails)
Snails eating clover act as a selective pressure. Clover plants that randomly mutate to produce higher cyanide survive and reproduce. This is classic natural selection in action.
2. Evaluate C (Insecticide Resistance)
DDT acts as a strong selective pressure. Insects with resistance alleles survive and pass them on, rapidly shifting the population. This is direct, observable evidence of natural selection.
3. Evaluate D (Antibiotic Resistance)
The emergence of MRSA (methicillin-resistant Staphylococcus aureus) after the introduction of antibiotics is another textbook example of natural selection driven by human-introduced selective pressures.
4. Evaluate B (Identical Fossils)
Finding fossils that are identical to living species shows evolutionary stasis (a lack of morphological change over time). While this can occur under stabilizing selection, it does not actively demonstrate the process of a population adapting to a new environment, which is the primary evidence used to support natural selection driving evolution.
Therefore, B does not provide evidence.
Common Mistake (⚠️):
Overthinking stabilizing selection and assuming any fossil is evidence for natural selection.
Question 24
Back to top ↑The diagram below shows a pathogen: a particle with an RNA genome enclosed in a protein capsid, surrounded by a lipid envelope studded with glycoproteins, and containing the enzymes reverse transcriptase and protease.
Which of the options A to D, is the disease caused by this pathogen?
Key Idea (💡): The diagram clearly shows a virus (capsid, lipid envelope, RNA). More specifically, it contains the enzyme 'reverse transcriptase'. This is the defining feature of a retrovirus, the most famous of which is HIV.
Reveal the answer & worked solution — commit to an option first
Correct Answer: A. HIV/AIDs
Fastest Approach (🚀):
Reverse transcriptase is unique to retroviruses. HIV is a retrovirus. Thus, HIV/AIDs.
Step-by-Step Breakdown:
1. Identify the type of pathogen
The diagram shows a structure with a protein capsid and genetic material (RNA) surrounded by a lipid envelope. This is the structure of a virus.
2. Identify the specific virus
The key label is the enzyme reverse transcriptase. Viruses that contain RNA and reverse transcriptase are called retroviruses. They use this enzyme to convert their RNA into DNA once inside the host cell. The Human Immunodeficiency Virus (HIV), which causes AIDS, is the classic example of a retrovirus.
3. Rule out others
Potato blight: Caused by a protoctist (Phytophthora infestans).
Ringworm: Caused by a fungus.
Tuberculosis: Caused by a bacterium (Mycobacterium tuberculosis*).
Therefore, the disease is HIV/AIDs.
Common Mistake (⚠️):
Guessing randomly without analyzing the labels on the diagram, specifically 'reverse transcriptase'.
Question 25
Back to top ↑The graph below shows readings from a calorimeter as pigment leaks out of beetroot membranes at different temperatures: absorbance rises with temperature in an S-shaped (sigmoidal) curve, then plateaus at higher temperatures from a point labelled P onwards.
Which statement, A to D, explains why the absorbance stops increasing at point P?
Key Idea (💡): High temperatures damage the cell membrane (denaturing proteins, increasing fluidity). This causes the initial steep rise in absorbance as pigment leaks out. However, the leakage will eventually stop when the concentration of pigment inside the vacuole equals the concentration in the surrounding water. At this point, there is no longer a concentration gradient to drive net diffusion.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. Pigment is in equal concentration inside and outside the cells.
Fastest Approach (🚀):
Options A, B, and C explain why the pigment leaks (the steep part of the curve). Option D explains why it stops leaking (equilibrium). Thus, D.
Step-by-Step Breakdown:
1. Understand the steep part of the curve
As temperature increases, the kinetic energy of the phospholipids increases, creating gaps. At very high temperatures, transmembrane proteins denature. This completely destroys the selective permeability of the membrane, allowing the red betalain pigment to leak out rapidly. Statements A, B, and C all describe mechanisms for this increase in permeability and leakage.
2. Understand the plateau (Point P)
The question asks why the absorbance stops increasing. Even with a completely destroyed, fully permeable membrane, diffusion requires a concentration gradient. Pigment will move out of the cell until the concentration outside the cell equals the concentration inside the cell.
Once equilibrium is reached, there is no net movement of pigment. Therefore, the absorbance of the surrounding water stops increasing. This is described by Statement D.
Common Mistake (⚠️):
Choosing A, B, or C because they correctly describe the effect of temperature on membranes, failing to realize they explain the rise, not the plateau.
Question 26
Back to top ↑The diagram below shows part of the plasma membrane: label 1 points to an integral protein, label 2 points to a carbohydrate chain, and label 3 points to the fatty acid tails of the phospholipids.
Which of the label lines points to a structure that could contain a sulphur atom?
Key Idea (💡): Sulphur is a key element found in the amino acids cysteine and methionine, which are used to build proteins. Carbohydrates and the fatty acid tails of lipids are made only of Carbon, Hydrogen, and Oxygen.
Reveal the answer & worked solution — commit to an option first
Correct Answer: D. only 1
Fastest Approach (🚀):
Identify the structures: 1=Protein, 2=Carbohydrate, 3=Lipid tail. Recall elemental composition: Proteins have S, carbs and lipids do not. Thus, only 1.
Step-by-Step Breakdown:
1. Identify the structures in the diagram
Arrow 1: Points to a large, globular structure embedded in the membrane. This is an integral protein.
Arrow 2: Points to a branching chain extending from the surface. This is a carbohydrate chain (part of a glycoprotein or glycolipid).
- Arrow 3: Points to the hydrophobic tails of the phospholipid bilayer. These are fatty acid chains.
2. Determine elemental composition
Carbohydrates (2): Composed of Carbon, Hydrogen, and Oxygen (CHO).
Fatty acid tails (3): Composed of long hydrocarbon chains (Carbon and Hydrogen, with a little Oxygen at the ester bond).
- Proteins (1): Composed of Carbon, Hydrogen, Oxygen, Nitrogen, and Sulphur (CHONS). The sulphur is found in the R-groups of specific amino acids like cysteine, which forms disulfide bridges vital for tertiary structure.
Therefore, only the protein (structure 1) could contain a sulphur atom.
Common Mistake (⚠️):
Thinking that all macromolecules in the membrane contain complex elements, or misidentifying the structures in the diagram.
Question 27
Back to top ↑The following are a series of organic molecules and the chemical processes that occur to convert them into different molecules.
Which of the rows A-D, is correct?
A. nucleic acid --hydrolysis--> nucleotide --hydrolysis--> polynucleotide
B. alpha-glucose --condensation--> amylopectin --hydrolysis--> alpha-glucose
C. amino acid --condensation--> dipeptide --hydrolysis--> polypeptide
D. beta-glucose --condensation--> cellulose --condensation--> maltose
Key Idea (💡): Condensation reactions link monomers together (releasing water). Hydrolysis reactions break polymers apart into monomers (using water). Alpha-glucose is the monomer for starch (amylopectin).
Reveal the answer & worked solution — commit to an option first
Correct Answer: B. Row B
Fastest Approach (🚀):
A is wrong (nucleotides don't hydrolyze into polynucleotides). C is wrong (dipeptides don't hydrolyze into polypeptides). D is wrong (cellulose is broken down by hydrolysis, and maltose is made of alpha-glucose). B is perfectly correct.
Step-by-Step Breakdown:
1. Evaluate Row A
Nucleic acids (polymers) hydrolyse into nucleotides (monomers). However, a nucleotide cannot undergo hydrolysis to form a polynucleotide. Forming a polynucleotide requires a condensation reaction. (Row A is false).
2. Evaluate Row C
Amino acids undergo condensation to form a dipeptide. However, a dipeptide does not undergo hydrolysis to form a polypeptide. Hydrolysis breaks things down. A dipeptide would hydrolyse back into amino acids. (Row C is false).
3. Evaluate Row D
Beta-glucose condenses to form cellulose. However, cellulose does not undergo condensation to form maltose. It would need to undergo hydrolysis, and even then, it would yield cellobiose or beta-glucose, not maltose (which is a disaccharide of alpha-glucose). (Row D is false).
4. Evaluate Row B
Alpha-glucose monomers join together via condensation reactions to form the polymer amylopectin (a component of starch). When amylopectin is digested, it undergoes hydrolysis reactions to break back down into its alpha-glucose monomers. This row is entirely factually correct.
Common Mistake (⚠️):
Confusing the direction of condensation (building) vs hydrolysis (breaking).