Practice questions · Topic 5.5

Unit 5 · Practice for the Topic 5.5 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Environmental effects on phenotype, summed up

Video coming soon

The same genes, different looks; conditions change which genes the cells use, not the genes; phenotypic plasticity; the six named cases; how clones in two conditions separate genes from environment.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one investigation one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question shows means with error bars, read the legend first: every error bar here represents ±2SE.

Question 1

A nine-banded armadillo always gives birth to four young with the same DNA, because all four grow from one fertilized egg. A wildlife center rears one set of four. It feeds two of them twice as much food as the other two, in one enclosure at the same temperature. The two fed more grow much heavier.

What do the four young show?

A. The two young fed less inherited alleles for a small body B. The smaller ration changed the DNA of the two young that ate it C. The same genes gave two body masses, one on each ration D. Body mass in armadillos is set by the food alone

Question 2

A sea snail moved from water without crabs into water where crabs that eat snails live grows its new shell thicker. Its DNA is the same in both waters.

What does the presence of the crabs change inside the snail's cells?

A. The sequence of the DNA in the snail's shell-building genes B. Which alleles for shell thickness the snail's cells carry C. The number of chromosomes in each cell of the snail D. How strongly the cells turn on their shell-building genes

Question 3

Which of the following changes to a ram can be passed on to a lamb the ram fathers?

A. The ram’s fleece grows thick over a cold winter B. A change in the DNA of one of the ram’s sperm cells gives one gene a new version C. A change in the DNA of one of the ram’s skin cells gives one gene a new version D. The ram’s leg muscles grow large from climbing steep pasture

Question 4

Marbled crayfish are all one clonal line: every marbled crayfish has the same DNA.

Which of the following observations shows phenotypic plasticity in marbled crayfish?

A. Marbled crayfish kept in one tank at one temperature all grow to about the same size B. Marbled crayfish and a second crayfish species kept in one tank grow to different sizes C. A marbled crayfish grows larger as it gets older D. Marbled crayfish kept in warm water grow larger than those kept in cool water

Question 5

A nursery planted cuttings of one hydrangea in two beds. The soil in the first bed is acid, and its plants flower blue; the soil in the second bed is alkaline, and its plants flower pink. A gardener takes cuttings from the pink-flowered plants in the alkaline bed and plants them in her own garden, where the soil is acid.

Predict the flower color of the gardener's plants, and give the reason.

A. Pink, because a cutting keeps the alleles of the plant it was cut from B. Blue, because the soil decides the color and the gardener's soil is acid C. Blue, because the acid soil changes the plants' color alleles D. Pink, because a plant keeps the color of the bed it came from, whatever soil it is moved to

Question 6

The grower's design: two groups of twelve spider-plant plantlets from one plant, with the hours of light each group gets; water, soil and temperature are the same for every plant.

A grower tests whether the hours of light a day change the leaf length of spider plants. The parent plant has always had 8 hours of light a day. Her design is in the table below.

Which group is the control, and why?

A. Group A, because it grows under the parent plant's usual light B. Group B, because it gets the most light C. Both groups, because both are plantlets of one plant D. Neither group, because a control must get no light at all

Question 7

A student grows 16 cuttings of one blackcurrant bush in shade and 16 in sun and measures every stem. The sun group has a mean stem length of 30.0 cm with a standard deviation s of 4.0 cm.

Which of the following is the ±2SE error bar of the sun group’s mean?

A. 29.0 cm to 31.0 cm B. 28.0 cm to 32.0 cm C. 26.0 cm to 34.0 cm D. 22.0 cm to 38.0 cm

Question 8

A grower grows cuttings of one plant in two conditions with everything else the same, and measures a trait in each group. The two means carry ±2SE error bars, and the bars stay clear of each other.

What can be concluded?

A. The two groups must have carried different alleles B. The difference between the two means is exactly its true size C. The difference is unlikely to be chance, so the condition is credited D. Nothing about the condition

Question 9

A student says that because the environment can change a phenotype, an organism's genotype must change over its lifetime too.

What is wrong with the student's statement?

A. The set of genes in use changes, not the genes themselves; the genotype stays fixed B. Nothing; when the cells change which genes they use, the genotype must change with them C. The environment changes the genotype only in the body cells, and the gametes keep the original D. The genotype does change over a lifetime, and the changed genotype is passed to the offspring

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 5 points

Does compost in the soil change how many flower heads a chrysanthemum makes? A student tests this with 20 plants she rooted from cuttings of one chrysanthemum. She pots 10 in the ordinary soil the parent plant grows in and 10 in the same soil with compost mixed in, keeps every pot in one greenhouse with the same light, water and temperature, and after twelve weeks counts the flower heads on each plant and finds each group’s mean. The means are drawn below; each error bar represents ±2SE.

Mean number of flower heads per plant for chrysanthemum cuttings from one plant grown in ordinary soil and in soil with compost, 10 plants per group, with the ends of each error bar printed beside it; error bars represent ±2SE.

(a) Identify the independent variable in the student’s test. (1 point)

Hint: Which one condition did the student deliberately change between the two groups?

A full-credit answer: The independent variable is the soil, ordinary or with compost mixed in, because it is the one condition the student deliberately changed.

Check the box for each point your answer earns

the soil (ordinary or with compost) as the independent variable.

Common slip: Naming the number of flower heads. That is what was counted, the dependent variable.

(b) Identify the dependent variable and one control variable. (1 point)

Hint: What did the student count, and which conditions did she hold the same in both groups?

A full-credit answer: The dependent variable is the number of flower heads per plant, the quantity counted; one control variable is the light, kept the same for every plant.
The water, the temperature and the twelve weeks of growth are control variables too, and any one of them counts.

Check the box for each point your answer earns

flower heads per plant as the dependent variable and any one condition held the same for every plant (light, water, temperature, the growing time) as a control variable.

Common slip: Giving the compost as the dependent variable. The compost is what the student changed; the flower heads are what she counted.

(c) Identify the control group and describe what it shows. (1 point)

Hint: Which group lacks the change under test, and what does its result let the student compare against?

A full-credit answer: The control group is the 10 plants in ordinary soil.
It shows how many flower heads the cuttings make without compost, so the compost group’s result can be compared against it and any difference credited to the compost.

Check the box for each point your answer earns

the ordinary-soil group as the control, giving the count without compost to compare against.

Common slip: Calling the compost group the control. The control lacks the change under test.

(d) Justify the student’s use of cuttings from one chrysanthemum rather than plants grown from a packet of seeds. (1 point)

Hint: What would plants from many seeds differ in that cuttings of one plant do not?

A full-credit answer: Cuttings from one plant all have the same DNA, so both groups share one genotype and a difference in flower heads between them cannot come from their genes.
Plants from a packet of seeds would carry different genotypes, and a difference could be genes or compost.

Check the box for each point your answer earns

cuttings of one plant share one genotype, so a difference between the groups cannot be genetic and is credited to the soil.

Common slip: Saying cuttings are used because they flower sooner. The point is that their genes are the same.

(e) Evaluate the claim that the compost raised the flower-head count, using the error bars drawn above. (1 point)

Hint: Read the bottom of the higher bar’s error bar and the top of the lower bar’s error bar, and ask whether the two bars overlap.

A full-credit answer: The mean was 12.0 flower heads per plant in ordinary soil and 17.5 with compost.
Read at the bottom of the compost bar’s error bar: 15.9. Read at the top of the ordinary-soil bar’s: 13.4.
13.4 sits below 15.9, so the ±2SE bars do not overlap, and the difference is very unlikely to be chance.
The plants share one genotype, and only the soil differed.
So the claim is supported: the compost raised the flower-head count.

Check the box for each point your answer earns

the judgement (the claim is supported) AND the ground (the ±2SE bars do not overlap, so the difference is very unlikely to be chance). Accept with or without: one genotype and one condition changed, so the compost is credited.

Common slip: Arguing only that 17.5 is larger than 12.0. Two means can differ by chance; the non-overlapping bars are what make chance very unlikely.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

A grower has 30 rose plants grown from cuttings of one rose. He wants to know whether the brightness of the light changes the size of the flowers. He grows 15 plants in bright light and 15 in dim light, with the same water, soil and temperature for every plant, and at flowering he measures the diameter of the first flower on each plant and finds each group's mean. The means are drawn below; each error bar represents ±2SE.

Mean flower diameter in centimeters of rose plants from one clonal line grown in bright light and in dim light, 15 plants per group; error bars represent ±2SE.

(a) Identify the independent variable and the dependent variable. (1 point)

A full-credit answer: The independent variable is the brightness of the light, bright or dim, the one condition changed.
The dependent variable is the flower diameter, the quantity measured.

Check the box for each point your answer earns

light brightness as the independent variable and flower diameter as the dependent variable.

Common slip: Swapping the two. The grower changed the light and measured the flowers.

(b) Describe the feature of the grower's design that lets a difference between the groups be credited to the light rather than to the plants' genes. (1 point)

A full-credit answer: Every plant is a cutting of one rose, so all 30 plants share one genotype.
A difference between the bright-light group and the dim-light group therefore cannot come from their genes, and with water, soil and temperature the same the light is the only condition that differs.

Check the box for each point your answer earns

the plants are cuttings of one rose, so both groups share one genotype and a difference cannot be genetic; with everything else the same the light is the only difference.

Common slip: Naming the equal group sizes as the feature. Fifteen in each group helps the comparison, but it is the shared genotype that rules out the genes.

(c) Predict what the two means and their ±2SE error bars would look like if dim light gave larger flowers. (1 point)

A full-credit answer: The dim-light mean would sit above the bright-light mean.
The two ±2SE error bars would not overlap: the bottom of the dim-light bar would sit above the top of the bright-light bar.
Then the difference would be unlikely to be chance, and with one genotype and one condition changed the light would be credited.

Check the box for each point your answer earns

the dim-light mean higher and the two ±2SE bars not overlapping.

Common slip: Describing the means farther apart with the bars still overlapping. Overlapping bars leave the difference open however far apart the means look; the bars must clear each other.

(d) Evaluate the claim that dim light gives larger flowers, using the error bars drawn above. (1 point)

A full-credit answer: The mean flower diameter was 7.2 cm in bright light and 7.6 cm in dim light.
The error bars represent ±2SE.
The bright-light bar runs from 6.6 to 7.8 cm and the dim-light bar from 7.0 to 8.2 cm.
The two bars overlap, so the difference between the mean values could be chance.
So the claim is not supported by these data: they do not show a difference between the two light levels.

Check the box for each point your answer earns

the judgement (the claim is not supported) AND the ground (the ±2SE bars overlap, so the difference between the means could be chance).

Common slip: Arguing that 7.6 is larger than 7.2, so dim light gives bigger flowers. Two means can differ by chance; the overlapping bars mean chance has not been ruled out.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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