Unit guidesAP BiologyFree response

AP Biology free response: six question types, each marked point by point

Free response is 50% of your score: 6 questions in 90 minutes, 34 points. Each type asks for something different, and each one rewards a habit you can practice.

94 original practice questions, all six types. Write an answer and every rubric point comes back marked, with the reason. Grading is free until January, and marking your own against the rubric is always free.

Write one and get it graded

The six types

Interpreting and evaluating experimental results

Long · 9 points · about 25 min

An experiment and its data. You explain the biology behind it, pick apart the design, work with the numbers, and predict what happens next.

Before part (a), name the independent variable, the dependent variable and the control to yourself. Half the points lean on knowing which is which.

An answer, marked point by point

Deer antlers are bony structures that regrow every year, and during the growing season the cells at the tip of a growing antler divide rapidly. Researchers hypothesized that a signaling protein secreted in the antler tip, which they call AGF, stimulates these cells to divide. AGF binds to a receptor tyrosine kinase in the plasma membrane, and the researchers suspected that a protein kinase inside the cell, called K, relays the signal. They grew antler-tip cells in culture dishes, removed all growth factors from the medium for 24 hours, and then gave each dish one of the five treatments shown in the table for 18 hours. They then measured the percentage of cells that had entered S phase, detected by a labeled nucleotide that becomes part of a cell's DNA only if the cell replicates its DNA, and the amount of mRNA for a cyclin (relative to medium only, set at 1.0). Each value is the mean of five dishes.

Effects of AGF and a kinase K inhibitor on antler-tip cells (n = 5 dishes per value)

Treatment	Cells that entered S phase (%, mean ± 2SE)	Relative cyclin mRNA level (mean ± 2SE)
1. Medium only	4 ± 1	1.0 ± 0.1
2. AGF	38 ± 3	6.2 ± 0.5
3. AGF + antibody that blocks the AGF-binding site of the receptor	5 ± 1	1.1 ± 0.2
4. AGF + drug that inhibits kinase K	9 ± 2	1.4 ± 0.2
5. Drug that inhibits kinase K, no AGF	3 ± 1	0.9 ± 0.1

(a)Describe how the binding of AGF to its receptor starts a signal inside the cell. Explain how an increase in cyclin levels can allow a cell to pass the G1 checkpoint and enter S phase.2 pts
(b)Identify the treatment that serves as the control for determining the effect of AGF. Explain why the researchers included treatment 5.2 pts
(c)Calculate the percent decrease in the percentage of cells that entered S phase for treatment 4 relative to treatment 2. Make a claim about whether kinase K acts before or after the increase in cyclin mRNA in the AGF pathway. Support your claim with data from the table.3 pts
(d)Predict how the percentage of cells that enter S phase in medium only would change if the cells carried a mutation that makes kinase K active all the time, even without AGF. Justify your prediction.2 pts

Part (a) — a student wrote

When AGF binds to the receptor, the receptor changes shape and phosphorylates itself, which activates relay proteins like kinase K in a phosphorylation cascade. When there are more cyclins, the cell can pass the G1 checkpoint and start copying its DNA in S phase.

1/1

Describes that AGF binding changes the receptor (its shape, or two receptors pairing up) so that the receptor's kinase phosphorylates the receptor itself, or phosphorylates or activates relay proteins inside the cell, starting the signal.

This earns the point because it says the receptor changes shape and phosphorylates itself, which starts the cascade inside the cell.

0/1

Explains that cyclin binds to and activates a cyclin-dependent kinase (CDK), and links the active cyclin-CDK complex to moving the cell past the G1 checkpoint into S phase (for example, by phosphorylating target proteins).

No point: it says more cyclin lets the cell pass the checkpoint but never says that cyclin binds and activates a CDK that phosphorylates target proteins.

Part (b) — a student wrote

Treatment 1 is the control because it has no AGF in it. Treatment 5 was included to make the experiment more valid and to have something to compare treatment 4 to.

1/1

Identifies treatment 1 (medium only) as the control for the effect of AGF.

This earns the point because treatment 1, the medium-only dishes, is correctly named as the control for the effect of AGF.

0/1

Explains that treatment 5 tests whether the drug by itself changes S-phase entry or cyclin mRNA (for example, by harming the cells), so that the reduction in treatment 4 can be attributed to blocking kinase K in the AGF pathway.

No point: 'more valid' and 'something to compare' never say that treatment 5 rules out the drug lowering division on its own.

Part (c) — a student wrote

(38 - 9) / 38 = 0.763, so there is a 76% decrease. K acts before the cyclin mRNA increases. When the drug blocked K, cyclin mRNA was only 1.4 instead of 6.2 with AGF alone, so the signal never got to the cyclin gene.

1/1

Calculates a 76% (76.3%) decrease in S-phase entry for treatment 4 relative to treatment 2: (38 - 9) / 38 x 100.

This earns the point because the setup is right and the answer, a 76% decrease, is correct.

1/1

Claims that kinase K acts before (upstream of) the increase in cyclin mRNA, so K is needed for AGF to raise cyclin mRNA.

This earns the point because it claims that K acts before the increase in cyclin mRNA.

1/1

Supports the claim with cyclin mRNA values showing that blocking K prevents most of the rise that AGF causes: 1.4 with AGF and the drug versus 6.2 with AGF alone.

This earns the point because it cites the cyclin mRNA levels of 1.4 with the drug and 6.2 with AGF alone, although 'never got to the cyclin gene' overstates them, since 1.4 is still a little above the 1.0 of medium only.

Part (d) — a student wrote

The cells with the mutation would enter S phase more often than normal cells, because the mutation makes the cells divide more, like cancer cells.

1/1

Predicts that the percentage of cells entering S phase in medium only would increase (above the 4% of normal cells) for cells with a permanently active kinase K.

This earns the point because it predicts that the mutant cells enter S phase more often than normal cells, which is the medium-only comparison the prompt sets up.

0/1

Justifies by reasoning that a permanently active K passes the signal on to the rest of the pathway (for example, turning on cyclin transcription) even when no AGF has bound the receptor.

No point: 'the mutation makes the cells divide more' restates the prediction and never explains that an always-active K relays the signal without AGF binding the receptor.

This answer earns 6 of 9.

Interpreting and evaluating experimental results with graphing

Long · 9 points · about 25 min

The same kind of experiment, plus a graph you build from the data. On the exam you draw it on a grid; here you describe it — the points are for the same things.

Categories on the x-axis mean a bar graph; a continuous variable means a line or scatter plot. Label both axes with units, plot every value, and draw error bars when the data give you variability.

An answer, marked point by point

Herbicide H kills plants by blocking an enzyme they need to make certain amino acids. A farm began spraying its fields with herbicide H once each spring and has sprayed every year since. An annual weed grows among the crops; each weed plant lives for one season and reproduces only by seed. Every three years, beginning just before the first spraying, researchers collected seeds from the weed population in the fields and stored them in cold, dry conditions. After 12 years of spraying, they tested all five seed collections together. Because seed age can affect seedling health, they first grew one generation of plants from each collection in a greenhouse, keeping each collection's plants separate so that they mated only with one another, and used the fresh seeds from that generation in the test. They germinated these seeds at the same time in the same greenhouse, with identical soil, water, light, and temperature. When the seedlings were three weeks old, the researchers sprayed them with the dose of herbicide H used on the farm and counted the seedlings alive 14 days later. Each collection was tested in 5 trays of 40 seedlings. The results are in the table.

Years of spraying before seed collection (years)	Mean seedling survival after herbicide H (percent)	±2 SE (percent)
0	2.0	1.5
3	7.5	3.5
6	25.0	6.0
9	58.0	7.0
12	84.0	5.0

±2 SE is two standard errors of the mean, calculated from the 5 trays for each collection.

(a)Explain how spraying herbicide H every year could cause the percentage of herbicide-resistant weeds in the population to increase over time.1 pt
(b)Explain why the researchers grew and tested seedlings from all five collections at the same time in the same greenhouse, rather than testing each collection's seedlings on the farm in the year its seeds were collected.1 pt
(c)Describe the graph you would construct to show how seedling survival changed with years of spraying. State the type of graph and why it suits these data, what each axis shows (with units), the values you would plot (give the numbers), and how you would show the variability given in the table.4 pts
(d)Calculate the frequency of the R allele, as a decimal, in the 12-year seed collection, assuming that resistance is caused by a dominant allele R, that seedlings carrying one or two copies of R survive the herbicide while seedlings with no R allele die, and that the collection is in Hardy-Weinberg equilibrium for this gene. Then predict how the frequency of R would change over many generations if the farm stopped spraying, given that without the herbicide, resistant plants produce about 20 percent fewer seeds than susceptible plants, and justify your prediction.3 pts

Part (a) — a student wrote

The herbicide kills the weeds that aren't resistant. The resistant ones survive and make seeds, and their offspring inherit the resistance, so every year a bigger percentage of the weeds is resistant.

1/1

Explains natural selection: resistant weeds survive the spraying and reproduce, passing heritable resistance to their offspring, so resistance becomes more common each generation.

This earns the point because it links the survival of resistant weeds to their reproduction and to offspring inheriting resistance, which raises the percentage each year.

Part (b) — a student wrote

They grew them all in the same greenhouse so it would be a fair test with all the conditions controlled.

0/1

Explains that identical conditions remove environmental differences between years, so differences in survival can be attributed to the seed collections themselves, that is, to inherited (genetic) differences, rather than to the growing environment.

No point: 'a fair test with all the conditions controlled' never says that identical conditions let survival differences be traced to inherited differences among the seed collections.

Part (c) — a student wrote

I would make a line graph because years is continuous data. The x-axis would be years of spraying (years) and the y-axis would be the percent of seedlings that survived (%). I would plot the points (0, 2.0), (3, 7.5), (6, 25.0), (9, 58.0), and (12, 84.0) and connect them, so the line goes up the whole time.

1/1

Chooses a line graph or scatterplot and justifies it by years of spraying being a continuous (numerical) independent variable.

This earns the point because a line graph suits a continuous independent variable, and the response gives that reason.

1/1

Puts years of spraying on the x-axis (in years) and mean seedling survival on the y-axis (in percent).

This earns the point because both axes are named with units, years of spraying on x and percent survival on y.

1/1

Plots all five means at their years: (0, 2.0), (3, 7.5), (6, 25.0), (9, 58.0), and (12, 84.0).

This earns the point because all five year-mean pairs are listed correctly.

0/1

Shows variability with error bars on each point extending 2 SE above and below the mean, using the table's ±2 SE values.

No point: the description never mentions error bars, so the ±2 SE variability in the table is missing from the graph.

Part (d) — a student wrote

Since the survivors are resistant, p^2 = 0.84, so p = the square root of 0.84 = 0.92. If the farm stopped spraying, R would get less common over many generations, because resistant plants make 20 percent fewer seeds without the herbicide, so they have fewer offspring and pass fewer R alleles to the next generation.

0/1

Calculates the frequency of R as 0.6: q^2 = 1 - 0.84 = 0.16, q = 0.4, p = 1 - 0.4 = 0.6.

No point: survivors include Rr plants as well as RR plants, so 0.84 equals p^2 + 2pq, and the correct route is q^2 = 1 - 0.84 = 0.16, q = 0.4, p = 0.6.

1/1

Predicts that the frequency of R decreases over generations once spraying stops.

This earns the point because it predicts that R becomes less common over many generations once spraying stops.

1/1

Justifies with selection against R: without the herbicide, resistant plants make fewer seeds, so plants carrying R leave fewer offspring and pass fewer R alleles to each generation.

This earns the point because it ties the resistant plants' lower seed production to fewer offspring and fewer R alleles in the next generation.

This answer earns 6 of 9.

Scientific investigation

Short · 4 points · about 10 min

A described investigation. You explain the biology it tests, identify part of its design, predict a result, and justify the prediction.

A null hypothesis says the variable has NO effect. A control is the group that differs only in the variable being tested. Keep the two straight and you bank two points.

An answer, marked point by point

Protein BP7 is a small protein made by a soil bacterium. In the folded protein, a leucine, an amino acid with a nonpolar R group, sits in the interior of the protein. Researchers made an altered version of BP7 in which this interior leucine is replaced by lysine, an amino acid with a positively charged R group; every other amino acid is unchanged. At 25°C, both versions fold into the same overall shape. To compare how stable the two versions are, the researchers will slowly heat a solution of each version from 25°C to 95°C and record the unfolding temperature, the temperature at which half of the protein molecules have unfolded.

(a)Explain why amino acids with nonpolar R groups tend to be found in the interior of a folded protein that is surrounded by water.1 pt
(b)Identify one variable that the researchers should keep the same for the solutions of both protein versions.1 pt
(c)Predict how the unfolding temperature of the altered protein will compare with that of the normal protein.1 pt
(d)Justify your prediction.1 pt

Part (a) — a student wrote

Nonpolar R groups are hydrophobic, so they get pushed away from the water around the protein and cluster together in the middle. The polar and charged R groups end up on the outside where they can interact with water.

1/1

Explains that nonpolar (hydrophobic) R groups are excluded from the surrounding water, so they cluster together in the protein's interior (hydrophobic interactions); saying that polar and charged R groups take the water-facing surface and leave the nonpolar ones buried together is equivalent.

This earns the point because it says hydrophobic R groups are pushed away from water and cluster in the interior while polar and charged R groups face the water.

Part (b) — a student wrote

The temperature of both solutions should be kept constant during the experiment so that it is a fair test.

0/1

Identifies one condition of the solutions or the heating procedure, other than the independent or dependent variable, that could change the unfolding temperature and should be kept the same for both versions (for example, pH, salt concentration, protein concentration, or rate of heating).

No point: the researchers deliberately raise the temperature from 25°C to 95°C, so it cannot be kept constant; pH, salt concentration, or protein concentration would have earned it.

Part (c) — a student wrote

The altered protein will unfold at a lower temperature than the normal protein.

1/1

Predicts that the altered protein (with lysine) will have a lower unfolding temperature than the normal protein.

This earns the point because it predicts a lower unfolding temperature for the altered protein.

Part (d) — a student wrote

Lysine has a charged R group, so it is hydrophilic and does not fit with the nonpolar R groups in the center. This breaks up the hydrophobic interactions holding the middle of the protein together, so it takes less heat to unfold it.

1/1

Justifies by explaining that lysine's charged, hydrophilic R group disrupts the hydrophobic interactions of the nonpolar interior, so the folded structure is held together less strongly and less heat is needed to unfold it.

This earns the point because it links lysine's charged, hydrophilic R group to breaking up the hydrophobic interactions, so less heat is needed to unfold the protein.

This answer earns 3 of 4.

Conceptual analysis

Short · 4 points · about 10 min

A biological situation, usually with something changed or broken. Describe it, explain the mechanism, predict what the change does, and justify.

Follow the chain one link at a time: the change, what it does to the molecule or cell, and what that does to the organism. Every link you skip is a point you might not get.

An answer, marked point by point

In a chloroplast, the light-dependent reactions take place in the thylakoid membranes, and the Calvin cycle takes place in the stroma. Researchers isolated intact chloroplasts from pea leaves and kept them in bright light with carbon dioxide available. They then added compound U, which forms channels that let hydrogen ions (H+) move freely across the thylakoid membrane without passing through ATP synthase. Compound U has no direct effect on photosystem II, photosystem I, or the other proteins of the electron transport chain.

(a)Describe how the light-dependent reactions establish a higher concentration of H+ in the thylakoid space than in the stroma.1 pt
(b)Explain why adding compound U decreases the rate of ATP synthesis in the chloroplasts.1 pt
(c)Predict the effect of adding compound U on the rate of carbon fixation by the Calvin cycle.1 pt
(d)Justify your prediction.1 pt

Part (a) — a student wrote

Light excites electrons in photosystem II, and as the electrons move down the electron transport chain their energy is used to pump H+ into the thylakoid space. This makes more H+ inside the thylakoid than in the stroma.

1/1

Describes a way the light-dependent reactions add H+ to the thylakoid space: energy from electron flow along the electron transport chain pumps H+ from the stroma into the thylakoid space, or photosystem II splits water and releases H+ into the thylakoid space.

This earns the point because it says the electron transport chain uses the electrons' energy to pump H+ into the thylakoid space.

Part (b) — a student wrote

Compound U lets H+ go across the thylakoid membrane, so the chloroplast can't make as much ATP.

0/1

Explains that compound U lets H+ leak across the membrane, collapsing the H+ gradient, so less H+ flows through ATP synthase, which depends on that flow to make ATP.

No point: it restates what compound U does and never explains that losing the H+ gradient stops the H+ flow through ATP synthase that powers ATP synthesis.

Part (c) — a student wrote

Carbon fixation will go down.

1/1

Predicts that the rate of carbon fixation by the Calvin cycle decreases.

This earns the point because it gives a clear direction, a decrease in carbon fixation.

Part (d) — a student wrote

Because the Calvin cycle needs energy from the light reactions to work.

0/1

Justifies by the Calvin cycle's need for ATP from the light-dependent reactions: with less ATP, the cycle's reactions (turning fixed carbon into sugar and regenerating the CO2 acceptor) slow down.

No point: 'needs energy' is too vague, because the justification has to name ATP from the light reactions as what the Calvin cycle runs short of.

This answer earns 2 of 4.

Analyze a model or visual representation

Short · 4 points · about 10 min

A diagram or model of a process. Say what it shows, explain a relationship in it, use it to predict, and say what it leaves out.

Read every arrow as a sentence ("A activates B"). When a part asks for a limitation, name something real the model cannot show — not "it is only a model".

An answer, marked point by point

A soil amoeba is a single-celled eukaryote that lives in soil water with a pH of about 7 and feeds on bacteria. After the amoeba engulfs a bacterium, the bacterium is digested inside a food vacuole by hydrolytic enzymes delivered from lysosomes. These enzymes are most active between pH 4.5 and pH 5.0 and lose most of their activity above pH 6.5. Figure 1 is a model of how one engulfed bacterium is digested.

Figure 1. Model of the digestion of an engulfed bacterium in a soil amoeba. Each arrow points from a step to its result. Step 1: Plasma membrane folds around a bacterium -> a food vacuole containing the bacterium forms inside the cell Step 2: A lysosome fuses with the food vacuole -> hydrolytic enzymes enter the food vacuole Step 3: Proton pumps in the food vacuole membrane use ATP -> H+ ions move from the cytosol (pH 7.2) into the food vacuole (pH 4.7) Step 4: Hydrolytic enzymes in the food vacuole -> bacterial proteins, nucleic acids and polysaccharides are broken down into monomers Step 5: Transport proteins in the food vacuole membrane -> amino acids, nucleotides and sugars move into the cytosol

(a)Based on the pH values in Figure 1, describe how H+ ions move relative to their concentration gradient in Step 3.1 pt
(b)Explain why the amino acids, nucleotides and sugars in Step 5 require transport proteins to cross the food vacuole membrane.1 pt
(c)Use the model to predict how a drug that blocks the proton pumps would affect the time needed to digest an engulfed bacterium.1 pt
(d)Justify your prediction in part (c).1 pt

Part (a) — a student wrote

The H+ ions go into the food vacuole where the pH is lower, so they are moving from a high concentration to a low concentration, down their gradient. The pump uses ATP to move them.

0/1

Describes that H+ ions move against their concentration gradient, from the cytosol (lower H+ concentration, pH 7.2) into the food vacuole (higher H+ concentration, pH 4.7).

No point: a lower pH means a higher H+ concentration, so the pumps move H+ against its gradient, the opposite of what this response says.

Part (b) — a student wrote

Amino acids, nucleotides and sugars are too big and polar to slip between the phospholipids, and the middle of the membrane is hydrophobic, so they need a transport protein to get through.

1/1

Explains that these molecules are polar or charged and so cannot cross the hydrophobic (nonpolar) interior of the phospholipid bilayer without a transport protein.

This earns the point because it connects the monomers' polarity to the hydrophobic middle of the membrane, which is why a transport protein is needed.

Part (c) — a student wrote

It would take longer for the bacterium to be digested.

1/1

Predicts that digestion of the bacterium takes longer (is slower) when the proton pumps are blocked.

This earns the point because it predicts the correct direction: digestion takes longer when the pumps are blocked.

Part (d) — a student wrote

Because the proton pump is blocked, the enzymes would not be able to break down the bacterium as fast.

0/1

Justifies the prediction by connecting the blocked pumps to a higher (less acidic) pH in the food vacuole, at which the hydrolytic enzymes have much lower activity.

No point: it restates that the pump is blocked but never says the vacuole pH stays high, above the range where the enzymes work.

This answer earns 2 of 4.

Analyze data

Short · 4 points · about 10 min

A table or graph. Read a value or a trend, calculate something, use the data to back or sink a claim, and predict.

Quote the numbers. "The rate rose from 2.1 to 6.4 mg/min" earns what "the rate went up" does not.

An answer, marked point by point

A student synthesized five double-stranded DNA fragments. Each fragment is 200 base pairs long, and the fragments differ only in the percentage of their base pairs that are guanine-cytosine (G-C) pairs rather than adenine-thymine (A-T) pairs. The student slowly heated each fragment in the same salt solution and recorded its melting temperature (Tm), the temperature at which half of the DNA molecules have separated into single strands. Each fragment was tested three times, and Table 1 shows the mean melting temperature of each fragment.

Table 1. Melting temperatures of five 200-base-pair DNA fragments

Fragment	G-C base pairs (% of all base pairs)	Mean melting temperature, Tm (degrees C)
1	30	81.4
2	40	85.6
3	50	89.9
4	60	94.1
5	70	98.2

(a)Describe the relationship between the percentage of G-C base pairs in a fragment and its melting temperature.1 pt
(b)Calculate the percentage of the bases in fragment 4 that are adenine (A).1 pt
(c)Evaluate the claim that the fragments with higher melting temperatures have stronger covalent bonds in their sugar-phosphate backbones.1 pt
(d)Explain, in terms of base pairing, why a higher temperature is needed to separate the strands of fragment 5 than to separate the strands of fragment 1.1 pt

Part (a) — a student wrote

As the percent of G-C base pairs goes up, the melting temperature goes up too, from 81.4 degrees at 30% to 98.2 degrees at 70%.

1/1

Describes a positive relationship: the melting temperature increases as the percentage of G-C base pairs increases.

This earns the point because it states the direction of the relationship and backs it with values from the table.

Part (b) — a student wrote

Fragment 4 is 60% G-C, so the other 40% is A-T. That means 40% of the bases are adenine.

0/1

Calculates that 20 percent of the bases in fragment 4 are adenine.

No point: 40 percent is the share of A-T base pairs, and since each A-T pair holds one A and one T, adenine is only 20 percent of the bases.

Part (c) — a student wrote

The claim is wrong because heating only breaks the hydrogen bonds between the bases, not the backbone, so backbone strength doesn't matter.

1/1

Rejects the claim with a correct reason: melting separates the strands by breaking hydrogen bonds between paired bases while the covalent backbone stays intact, or the fragments have identical backbones and differ only in their base pairs.

This earns the point because it rejects the claim and says heating breaks only the hydrogen bonds between the bases, leaving the backbone intact.

Part (d) — a student wrote

G-C pairs have 3 hydrogen bonds and A-T pairs only have 2. Fragment 5 is 70% G-C and fragment 1 is only 30% G-C, so fragment 5 has more hydrogen bonds between its strands and it takes more heat energy to break them all.

1/1

Explains that G-C pairs have three hydrogen bonds and A-T pairs two, so fragment 5 (70 percent G-C) has more hydrogen bonds than fragment 1 (30 percent G-C), holding its strands together more strongly so that more heat energy is needed to separate them.

This earns the point because it links the three-versus-two hydrogen bond difference to fragment 5's higher G-C content and to the extra heat energy needed.

This answer earns 3 of 4.

Read the verb first

Every part starts with a task verb, and the verb decides what earns the point.

Identify: Name it. A precise word or phrase. No explanation needed, and none is rewarded.
Describe: Say what it is or what happens. The relevant detail — the what, not the why.
Explain: Say how or why it happens. The cause or mechanism. If there is no "because" doing work, it is a description.
Predict: Commit to an outcome. A direction: increases, decreases, stays the same. "It would change" earns nothing.
Justify: Back a prediction or claim. The evidence or principle that makes it true, connected to the claim.
Calculate: Work it out. The setup and the answer, with units. A bare number is a gamble.
State the null hypothesis: Say what "no effect" would look like. No difference in the dependent variable across the independent variable, named.
Make or support a claim: Take a position the data can test. A claim you could be wrong about, backed by specific numbers from the question.
Evaluate: Judge a claim against the evidence. A verdict — supported or not — and the evidence that decides it.
Represent or construct: Build the graph or model. Every element a reader needs: type, labels, units, plotted values, error bars.

What earns a point

One idea, one point. Every rubric row rewards one specific idea. Extra sentences that hit no row earn nothing — and a wrong one can cost you.
Name the thing. "It affects the enzyme" is vague. "The inhibitor binds the active site" is a point.
Connect cause to effect. Explain and justify rows want the link, not a list of terms. Write the "because".
Answer the part you are on. A point is scored against the part that asks for it, so keep (a) in (a) and (b) in (b).
Numbers need setup and units. Show what you divided by what. "2.5" is not an answer; "2.5 mm per day" is.
Contradictions cancel. Write "increases" and "decreases" for the same thing and you have written neither.
Restating the question earns zero. Readers skip past it. Start with the answer.

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