Unit 4 test

Unit 4 · Unit 4 end-of-unit test (also the test-out)

Suggested time: about 80 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question, then press Submit the test. Feedback and the scoring guides appear after you submit. For the three 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. After you submit, mark your own free-response work against each scoring guide. Suggested time: 80 minutes. Every count, concentration, time and reading in this test is imagined for the question unless the question says otherwise. Where a question gives the length of the whole cell cycle, use that length. To count chromosomes, count centromeres: a joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Question 1

A bacterium makes a blue pigment only when its cells are crowded. Each cell releases a signal molecule and carries a receptor for it. A researcher puts a dense culture in a mesh bag whose holes let molecules through but keep the cells in, and hangs it in a flowing stream: the culture stays colorless. The same culture sealed in a jar of the same water turns blue within an hour. The two cultures have nutrients and the same crowding.

Why is the culture in the bag colorless?

A. The stream carries the signal molecule away, so its concentration around the cells stays too low to bind their receptors B. The cells in the bag sense their neighbors by touch, and the mesh keeps them from pressing together C. The flowing water washes the pigment out of the cells as fast as they make it D. The cells in the bag carry fewer receptors than the cells in the jar

Question 2

Four cell types of a canary: the hormone each releases and the receptor each carries. The measurements are imagined.

Researchers measure four cell types of a canary: how much of one hormone each releases into the blood, and how much of the receptor for that hormone each carries. The table gives the results.

Which of the following gives the signaling cell type and the target cell types for this hormone?

A. Signaling: the gland cells. Targets: the liver cells, the skin cells and the muscle cells B. Signaling: the liver cells and the muscle cells. Targets: the gland cells C. Signaling: the gland cells. Targets: the gland cells, because they carry the hormone D. Signaling: the gland cells. Targets: the liver cells and the muscle cells

Question 3

In a gibbon, a nerve ending releases molecule H into the gap beside the tiny muscle cell at the base of a hair in its skin, and the muscle cell contracts within a millisecond, so the hair stands up. Cells of a gland in the gibbon’s abdomen release the same molecule H into the blood, and liver cells across the body respond about a minute later.

Which of the following classifies molecule H at the hair muscle and in the blood, in that order?

A. A neurotransmitter in both cases B. A hormone at the hair muscle, then a neurotransmitter in the blood C. A neurotransmitter at the hair muscle, then a hormone in the blood D. A hormone in both cases

Question 4

Four cell types of a koala: the distance of each from the gland and the time after the release at which each first responds. The measurements are imagined.

At time zero, cells of a gland in a koala release a hormone into the blood. The table gives, for four cell types that respond, the distance of each from the gland and the time at which it first responds. A fifth cell type, a skin cell of the hind foot, carries the receptor and lies 90 cm from the gland.

Which of the following predicts when the hind-foot skin cell first responds?

A. About 50 seconds after the release B. About 80 seconds after the release C. Several hours after the release D. It does not respond, because 90 cm is beyond the hormone’s reach

Question 5

In a heart muscle cell of a wombat, 15 receptors have a hormone bound. Over the next minute the cell’s cAMP rises from 340 molecules to 4,090 molecules.

What is the amplification at this step, as cAMP molecules made per bound receptor?

A. 23 cAMP per receptor B. 250 cAMP per receptor C. 273 cAMP per receptor D. 3,750 cAMP per receptor

Question 6

In cultured cells of a yak, relay protein J gains a phosphate group within a minute of a hormone binding the cells’ receptors. Researchers repeat the experiment twice with a radioactive tag. In the first experiment the tag is on the phosphate groups of the cells’ ATP; afterwards J carries radioactive phosphate. In the second the tag is on the hormone; afterwards J carries phosphate with no tag.

Which of the following do the two experiments show about the phosphate on J?

A. The hormone handed it to J through the receptor B. The receptor made it from the hormone and passed it on C. J took it from the fluid outside the cell D. A kinase transferred it from ATP onto J

Question 7

Four tubes: what each holds and the cAMP made in ten minutes. The counts are imagined.

Researchers tear the membranes from cultured cells that respond to a hormone through a G protein-coupled receptor. They set up four tubes, with or without the membranes, the hormone and ATP, and measure the cAMP made in ten minutes. The table gives the results.

Which of the following do the four tubes show?

A. The hormone itself is converted into cAMP once it reaches the membrane B. ATP makes cAMP on its own whenever the hormone is present C. An enzyme in the membrane makes cAMP from ATP while the hormone is bound D. The membranes release stored cAMP when the hormone binds

Question 8

Which kinases carry a phosphate after the hormone arrives, when one kinase is blocked. The results are imagined.

In leaf cells of a begonia, a hormone binds a surface receptor and three kinases, W, N and T, relay the message in a phosphorylation cascade. Researchers block one kinase at a time with a drug, add the hormone, and test which kinases carry a phosphate. The table gives the results.

Which of the following is the order of the three kinases in the cascade?

A. Kinase T, then kinase N, then kinase W B. Kinase W, then kinase N, then kinase T C. Kinase N, then kinase W, then kinase T D. Kinase W, then kinase T, then kinase N

Question 9

A warthog’s fat cells respond to hormone T by releasing fatty acids; its salivary gland cells respond to the same hormone by releasing saliva. In both cell types the message passes from a receptor through a G protein, a cAMP-making enzyme and a kinase to a last protein. Researchers put the gene for the fat cells’ last protein, the one their kinase phosphorylates, into gland cells. Given hormone T, these gland cells now release fatty acids as well as saliva.

Which of the following does the result show about where the two cell types’ pathways differ?

A. At the response protein the kinase phosphorylates B. At the receptor that binds hormone T C. At the enzyme that makes cAMP D. At the G protein the receptor switches on

Question 10

Four trials on the cultured cells: the receptor carried, the ligand added and the result. The results are imagined.

In a cultured cell line, receptor 1 binds ligand 1 and starts pathway 1; receptor 2 binds ligand 2 and starts pathway 2. Researchers build a joined receptor from receptor 1’s ligand-binding domain and receptor 2’s intracellular domain, and put it into cells that carry neither original receptor. The table gives what the cells do when the researchers add each ligand.

Which of the following do the results show about the two domains?

A. Ligand answered: set by the intracellular domain. Pathway started: set by the ligand-binding domain B. Ligand answered: set by the ligand-binding domain. Pathway started: set by the ligand itself C. Ligand answered and pathway started: both set by the intracellular domain D. Ligand answered: set by the ligand-binding domain. Pathway started: set by the intracellular domain

Question 11

The mRNA for enzyme M, the enzyme M protein and the waxy coat at four times after the hormone arrives. The measurements are imagined.

In a dry spell a plant hormone reaches the outer cells of a currant bush’s leaves. Over a day the cells make enzyme M, and the leaf’s waxy coat thickens. Researchers measure, in the outer leaf cells, the mRNA for enzyme M and the amount of enzyme M protein at four times after the hormone arrives. The table gives the results.

Which measurement is the earliest evidence that the hormone changed which genes the leaf cells express?

A. The enzyme M protein at 3 hours B. The thickened waxy coat at 8 hours C. The mRNA for enzyme M at 30 minutes D. The mRNA for enzyme M at 0 minutes

Question 12

The pathway in the cauliflower root cells as a model. The shaded box is set in the plasma membrane.

The model shows a pathway in the root cells of a cauliflower. A hormone binds a receptor; the receptor switches on kinase P; kinase P switches off protein H by phosphorylating it. While protein H is active it holds the response off, so the cells lengthen only after protein H is switched off.

Which of the following changes would make the root cells lengthen all the time, with no hormone present?

A. A mutation that leaves the cells with no kinase P B. A mutation that leaves the cells with no protein H C. A mutation that leaves the cells with no receptor D. A blocker that fills the receptor’s binding site

Question 13

In a muskrat, hormone N released after a meal binds receptors on the muscle cells of the gut wall, and the muscle contracts more often for about an hour, until the blood has carried the hormone away. A mutant muskrat carries a receptor whose binding site grips hormone N so tightly that the hormone is never released. The mutant’s hormone N and its gut muscle are otherwise normal.

Which of the following predicts the mutant muskrat’s phenotype after a meal?

A. The gut muscle contracts more often for a shorter time than normal B. The gut muscle never contracts more often, because the hormone cannot leave the receptor C. The gut muscle contracts more often only after a larger meal than normal D. The gut muscle contracts more often for many hours, long after the hormone has left the blood

Question 14

In root cells of a cowpea plant, a hormone’s message passes receptor → kinase 1 → kinase 2 → kinase 3 → the genes for root hairs are expressed. A mutant line carries an always-active kinase 2, locked in its active shape. The mutant’s roots grow hairs all the time. Researchers test four drugs on the mutant.

Which drug stops the mutant’s root cells from expressing the root-hair genes?

A. A drug that holds kinase 3 in its inactive shape B. A drug that holds kinase 1 in its inactive shape C. A drug that fills the receptor’s binding site D. A drug that destroys the hormone in the fluid around the roots

Question 15

A degu, a small rodent, holds the potassium in its blood near 4.5 millimoles per liter. When the blood potassium rises after a meal, cells of a gland near the degu’s kidneys release hormone W into the blood. Hormone W makes kidney cells pass more potassium into the urine. As the blood potassium falls back toward 4.5 millimoles per liter, the gland cells release less hormone W.

Which kind of loop is this?

A. Positive feedback B. A response with no feedback C. Negative feedback D. Amplification

Question 16

Cells of a hemp plant make a purple pigment in three enzyme steps: enzyme 1 turns the starting molecule into a yellow intermediate, enzyme 2 turns the yellow intermediate into a red intermediate, and enzyme 3 turns the red intermediate into the purple pigment. When the pigment is plentiful it binds enzyme 3 at a site away from the active site and slows it. Enzymes 1 and 2 keep working at their usual rate.

When the pigment is plentiful, which molecule builds up in the cells?

A. The starting molecule B. The yellow intermediate C. The red intermediate D. The purple pigment

Question 17

In a bacterium of warm lakes, heat makes some proteins misfold. A sensor protein detects misfolded proteins and switches on the genes for refolding proteins, which fold the damaged proteins back into shape. As the misfolded proteins disappear, the sensor switches the genes off again. A mutant strain has no sensor protein. Both strains are moved from 30 °C water to 45 °C water.

Which of the following predicts what happens in the mutant strain’s cells at 45 °C?

A. The refolding proteins climb higher than in the normal strain, because nothing switches them off B. The cells make refolding proteins at once, because heat switches the genes on directly C. The misfolded proteins refold on their own once the sensor is gone D. Misfolded proteins build up, because the refolding proteins stay at their resting level

Question 18

A body cell of Japanese knotweed holds 88 chromosomes. The cell copies its DNA and enters mitosis. To count chromosomes, count centromeres: a joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

How many chromosomes does the cell hold in anaphase, once the sister chromatids have separated?

A. 44 B. 88 C. 176 D. 352

Question 19

A cell of one type holds 34 pg of DNA in G2. It goes through mitosis and cytokinesis.

How much DNA does each daughter cell hold just after cytokinesis?

A. 8.5 pg B. 17 pg C. 34 pg D. 68 pg

Question 20

A cell from the root of a lotus plant, part way through a division.

The drawing shows one cell from the root of a lotus plant, fixed and stained part way through a division.

Which stage of mitosis does the drawing show?

A. Telophase B. Anaphase C. Metaphase D. Prophase

Question 21

One 17-hour cycle of a cultured sloth cell: the nucleus, the cell’s mass and the DNA per cell over four stretches of hours. The readings are imagined.

Researchers follow one cultured sloth cell through its 17-hour cycle, recording its nucleus, its mass and its DNA per cell. The table gives the record.

How long does G2 last in these cells?

A. 1 hour B. 3 hours C. 6 hours D. 7 hours

Question 22

A student counts 420 cells on a slide from the root tip of a pansy seedling and finds 21 of them in one of the four stages of mitosis. Take the whole cell cycle in this root as 28 hours.

How long does mitosis take in these cells?

A. 0.05 h B. 1.4 h C. 5.0 h D. 5.6 h

Question 23

Of the cells on a slide from a fuchsia’s leaf, 4% are in mitosis. Of the cells on a slide from the fuchsia’s root tip, 7% are in mitosis.

What is the percent change in the percent of cells in mitosis, from the leaf to the root tip?

A. +3% B. +43% C. +75% D. +175%

Question 24

A skin cell of a platypus divides, and each of its two daughter cells receives a complete genome identical to the parent cell’s.

Which of the following is the step that guarantees each daughter cell receives one copy of every chromosome?

A. The cell counts its chromosomes at metaphase and stops the division if the number is odd B. The nuclear envelope re-forms around whichever chromosomes have reached each pole of the cell C. Cytokinesis divides the cytoplasm into two halves of equal volume and equal contents D. Anaphase sends one sister chromatid of every pair to each pole of the cell

Question 25

In one tissue, prophase takes 60 minutes and anaphase takes 15 minutes. A slide of the tissue is fixed at one instant, and 40 of its cells are caught in prophase.

About how many of its cells are caught in anaphase?

A. 4 B. 10 C. 40 D. 160

Question 26

Four groups of cultured dormouse cells: how long each waits at the end of G2. The times are imagined.

Researchers damage the DNA of cultured dormouse cells after S phase and time how long each cell waits at the end of G2 before entering mitosis. One group of cells carries extra copies of a DNA-repair enzyme. In every group the apoptosis response has been switched off, so a held cell never removes itself. The table gives the results.

Which of the following do the results show about the G2 checkpoint?

A. The hold ends when the DNA has been repaired, however long the repair takes B. The hold is a timer that lasts six hours in every cell with damaged DNA C. The checkpoint itself repairs the damage, and more repair enzyme speeds the checkpoint up D. The hold ends when the cell has grown large enough to divide

Question 27

A normal cell in G1 has its DNA damaged beyond repair. Its checkpoints and its apoptosis response all work.

Which of the following is the sequence of events in this cell?

A. The G1 checkpoint holds the cell; the cell repairs the damage; the cell passes into S phase B. The cell dismantles itself at once, before any checkpoint holds it C. The G1 checkpoint holds the cell; repair fails; the cell dismantles itself in an orderly way D. The G1 checkpoint holds the cell; repair fails; the cell passes into S phase with the damage

Question 28

Normal ibex cells and a tumor cell line from the same tissue: four measurements. The values are imagined.

Researchers compare a tumor cell line from an ibex with normal cells of the same tissue. The table gives four measurements.

Which of the following controls on the cycle is still working in the tumor line?

A. The requirement for a growth signal before the cycle starts B. The G1 checkpoint’s hold on damaged DNA C. The apoptosis response of a cell whose damage cannot be repaired D. The M checkpoint’s hold until every chromosome is attached from both poles

Question 29

Four tubes of G2 extract from marten cells: what was added to each and the result. The results are imagined.

Researchers make extracts of cultured marten cells held in G2, each extract holding nuclei. To three of the extracts they add proteins, and they record whether the nuclear envelopes break down, the first event of mitosis. The table gives the results.

Which of the following do the results show about how the cyclin–CDK complex moves a cell into mitosis?

A. The active complex phosphorylates the proteins that start mitosis, and the phosphates must stay on B. The active complex breaks the nuclear envelope down directly, with none of the proteins that start mitosis taking part C. The CDK starts mitosis on its own, and the cyclin only speeds it up D. The active complex removes phosphates from the proteins that start mitosis

Question 30

Researchers give cultured piranha cells a drug that stops the centrosome from being copied in interphase. Treated cells enter mitosis with one centrosome instead of two, and the spindle fibers grow from one pole only. The cells’ checkpoints all work.

Which of the following predicts what happens to the treated cells in mitosis?

A. They complete mitosis as normal, because one pole can pull the sister chromatids apart on its own B. They are held before their sister chromatids separate, because each chromosome is attached to fibers from one pole only C. They are held at the end of G2, because a missing centrosome counts as damaged DNA D. They divide into two daughter cells, one of which receives every chromosome

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 · Interpreting and Evaluating Experimental Results · 9 points

Suppose the seeds of one kind of plant, from dry hillsides that burn every few decades, germinate in the season after a fire. Researchers hypothesize that a molecule in smoke, molecule D, binds a receptor in the seed’s cells and starts a pathway that ends in the cells expressing the genes whose proteins start the embryo growing. They soak seeds in water alone, in water that smoke has been bubbled through, in water with purified molecule D, or in water with hormone E, a plant hormone known to make these seeds germinate. Seven days later they record the percent of seeds germinated (Figure 1). In a second experiment they repeat the treatments on seeds of a mutant line whose gene for the receptor is mutated (Figure 2). Every count is imagined.

Figure 1, normal seeds, and Figure 2, receptor-mutant seeds: the percent of seeds germinated after 7 days under the four treatments. Error bars represent ±2SE.

(a) Once the smoke has washed away, molecule D leaves its receptor. Describe one way a cell ends its response after the ligand has left. (1 point)

A full-credit answer: Phosphatases remove the phosphates the kinases added to the relay proteins.
Without their phosphates the relay proteins return to their resting shape and switch off, so the pathway stops.

Check the box for each point your answer earns

one of the following: the receptor returns to its resting shape once the ligand has left (and no new ligand binds); an enzyme destroys the second messenger, so its concentration falls; phosphatases remove the phosphates the kinases added, so the relay proteins switch off.

Scoring note: ‘the signal stops’ with no off-switch named does not earn the point.

Common slip: Saying the response stops because the ligand is gone. The ligand leaving is the start; the point needs what the cell does to switch the pathway off.

(b)(i) Identify the dependent variable in the researchers’ first experiment. (1 point)

A full-credit answer: The percent of seeds germinated after 7 days.

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the percent (or number) of seeds germinated after 7 days.

Common slip: Naming the treatment. The treatment is the independent variable, the thing the researchers changed.

(b)(ii) Based on Figure 1, describe the germination of the seeds treated with molecule D compared with the seeds in water alone. (1 point)

A full-credit answer: About 58% of the seeds treated with molecule D germinated, against about 4% in water alone.
The error bars do not overlap, so the difference is not due to chance.

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far more seeds germinated with molecule D than in water alone (accept the values, about 58% against 4%).

Accept also: the error bars do not overlap.

(b)(iii) Justify the researchers’ including a treatment of water alone in the experiments. (1 point)

A full-credit answer: Water alone shows how many seeds germinate with no smoke, molecule D or hormone E present.
So a rise in germination in another treatment can be attributed to the molecule added, not to the soaking itself.

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water alone shows the germination with no added molecule, so the effect of each added molecule can be separated from the effect of soaking (a comparison for the other treatments).

Scoring note: ‘it is the control’ with no purpose stated does not earn the point.

Common slip: Writing ‘it is the control’ and stopping. The point needs what the water shows: germination with nothing added.

(c)(i) Justify the researchers’ treating one sample of the mutant seeds with hormone E. (1 point)

A full-credit answer: Hormone E makes these seeds germinate through its own receptor, so it shows whether the mutant seeds can still germinate at all.
If they germinate with hormone E but not with molecule D, the fault is in the receptor for molecule D, not in the seeds’ ability to grow.

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hormone E shows that the mutant seeds can still germinate (the later steps and the embryo work), so a failure with molecule D can be attributed to the mutated receptor.

Common slip: Saying hormone E is ‘another control’ and stopping. Name what it shows: the mutant seeds can still germinate.

(c)(ii) Based on Figure 2, describe the difference between the effects of molecule D and of hormone E on the mutant seeds. (1 point)

A full-credit answer: In the mutant seeds molecule D produced about the same germination as water alone, about 4%, while hormone E produced about 64%.

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molecule D did not raise germination in the mutant seeds (about the same as water alone), while hormone E did (about 64%).

(c)(iii) The bound receptor for molecule D makes the seed’s cells express the genes for the proteins that start the embryo growing. Based on the data for molecule D in Figures 1 and 2, predict the amount of mRNA from those genes in mutant cells given molecule D compared with normal cells given molecule D. (1 point)

A full-credit answer: The mutant cells make far less of that mRNA than normal cells given molecule D.

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less (or none) in the mutant cells.

Common slip: Predicting the same amount. The mutant seeds did not germinate with molecule D, so the genes were not switched on.

(d)(i) The researchers make a third line of seeds whose receptor for molecule D is normal on the outside of the membrane but lacks most of its intracellular domain. Predict the effect of molecule D on the expression of the embryo-growth genes in this line compared with normal seeds. (1 point)

A full-credit answer: The genes are expressed far less, or not at all, in the third line.

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less (or no) expression of the embryo-growth genes in the third line.

(d)(ii) Justify your prediction in part (d)(i). (1 point)

A full-credit answer: Molecule D still binds the receptor, because the binding site on the outside is normal.
The intracellular domain is the part that acts on the next molecule inside the cell.
With most of it missing, the receptor’s shape change reaches nothing inside, so the pathway never starts and the genes are not switched on.

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the intracellular domain is what passes the message inside (acts on the next molecule), so without it binding leads to no transduction and no change in gene expression.

Scoring note: ‘the receptor is broken’ with no domain named does not earn the point.

Free-response score: 0 of 9

Free response 2 · Analyze Data · 4 points

Researchers grow a line of cultured cells from an angelfish for two days with no growth factor, so almost every cell rests in G0. At 0 hours they add the growth factor. Every 4 hours they fix a sample and record the percent of the cells in S phase and the percent in mitosis (Figure 1). In two more cultures they add the growth factor at 0 hours and wash it out of the medium at 6 hours or at 14 hours; Table 1 gives the highest percent in S phase and the highest percent in mitosis that each culture reached in 24 hours. Every count is imagined.

Figure 1, the percent of cells in S phase and in mitosis over 24 hours with the growth factor kept in the medium (error bars represent ±2SE); Table 1, the highest percents reached in three cultures.

(a) Based on Figure 1, identify the time at which the largest percent of the cells was in S phase. (1 point)

A full-credit answer: 12 hours after the growth factor was added.

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12 hours.

(b) Based on Figure 1, describe the trend in the percent of cells in mitosis over the 24 hours. (1 point)

A full-credit answer: The percent in mitosis stays near 1% for the first 12 hours, rises after 12 hours to a peak of about 18% at 20 hours, and falls to about 4% by 24 hours.

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near zero (or flat) at first, a rise to a peak at about 20 hours, then a fall (a direction and a time are required).

(c) The researchers hypothesize that these cells need the bound growth factor only until they pass the G1 checkpoint, after which the later stages finish on their own. Support the hypothesis using the data in Table 1. (1 point)

A full-credit answer: The culture whose growth factor was washed out at 14 hours still reached 54% in S phase and 17% in mitosis, about the same as the culture that kept the growth factor.
So the cells that had passed the G1 checkpoint by 14 hours finished S phase and mitosis with no growth factor.
The culture washed out at 6 hours, before the cells reached S phase, never rose above 5% in S phase.

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the 14-hour wash-out culture reached about the same peaks as the culture that kept the growth factor (the cells finished S phase and mitosis without it), while the 6-hour wash-out culture never rose above 5% in S phase (a comparison between the wash-out cultures and the kept culture is required).

Common slip: Quoting only the 6-hour culture. That shows the growth factor is needed early; the 14-hour culture shows it is not needed later.

(d) Explain how the growth factor bound to its receptor leads a cell to pass the G1 checkpoint. (1 point)

A full-credit answer: The bound receptor starts a signal transduction pathway inside the cell.
The pathway’s signal is one of the conditions the G1 checkpoint waits for, so the hold that kept the cyclin–CDK complex switched off ends.
The active complex phosphorylates the proteins that start S phase, and the cell moves on.

Check the box for each point your answer earns

the bound receptor’s pathway (signal transduction) meets the G1 checkpoint’s growth-signal condition, so the cyclin–CDK complex is switched on (activated) and phosphorylates the proteins that start S phase.

Accept: the pathway’s signal ends the hold on the cyclin–CDK complex, and the active complex drives the cell into S phase.

Common slip: Saying the growth factor ‘makes the cell divide’ and stopping. The point needs the pathway and the complex it switches on.

Free-response score: 0 of 4

Free response 3 · Analyze Data · 4 points

Suppose liver cells of a jerboa, a small desert rodent, release hormone Y into the blood when the blood carries a high concentration of iron. Hormone Y reaches the cells lining the gut, which move iron from digested food into the blood. Researchers breed a line of jerboas whose liver cells make no hormone Y; the line’s gut cells and receptors are normal. They keep eight normal jerboas and eight hormone-lacking jerboas on the usual diet and eight of each on a high-iron diet for a month, then measure the blood iron and the hormone Y in each animal (Table 1). Every value is imagined.

Table 1: the mean blood iron and the mean hormone Y in the blood of the four groups of jerboas, each with ±2SE.

(a) Based on Table 1, identify the group of jerboas with the highest blood iron. (1 point)

A full-credit answer: The hormone-lacking jerboas on the high-iron diet.

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the hormone-lacking jerboas on the high-iron diet (290 μg/dL).

(b) Based on the data for the normal jerboas, describe the relationship between the blood iron and the hormone Y in the blood. (1 point)

A full-credit answer: When the blood iron rises from 105 to 125 μg/dL on the high-iron diet, hormone Y rises from 10 to 34 units.
A higher blood iron concentration goes with more hormone Y.

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hormone Y rises when the blood iron rises (a positive relationship), with or without the values.

(c) The researchers hypothesize that hormone Y is the response in a negative-feedback loop controlling the blood iron. Evaluate the hypothesis by comparing the four groups in Table 1. (1 point)

A full-credit answer: The data support the hypothesis.
In normal jerboas the high-iron diet raised the blood iron only from 105 to 125 μg/dL while hormone Y rose from 10 to 34 units: the response grew as the iron rose and the iron was held near its usual value.
In jerboas with no hormone Y, the same diet raised the blood iron to 290 μg/dL, and on the usual diet it sat at 160 μg/dL.
Removing the response let the iron climb.

Check the box for each point your answer earns

supported, because the hormone rises with the iron and the iron is held near its usual value in normal animals, while without the hormone the iron climbs far higher (both comparisons: within the normal line and between the lines).

Scoring note: ‘supported, because the hormone rises with iron’ alone does not earn the point; the hormone-lacking line is the test of the loop.

Common slip: Evaluating from the normal jerboas alone. The hormone-lacking line shows what happens when the response is removed.

(d) Explain how hormone Y, released by liver cells, changes what the gut-lining cells do. (1 point)

A full-credit answer: The blood carries hormone Y from the liver to the gut lining.
The gut-lining cells carry a receptor that binds hormone Y, so they are its target cells.
Binding starts a signal transduction pathway inside each cell, and the response is that the cell moves less iron into the blood.

Check the box for each point your answer earns

the blood carries hormone Y to the gut cells, which carry its receptor; binding starts a pathway (transduction) whose response is less iron moved into the blood (the receptor and the response are both required).

Accept: the response named as fewer iron transporters at the cell surface, or less iron taken up.

Common slip: Saying the hormone ‘tells the gut cells to stop’ with no receptor named. A hormone acts only on cells that carry its receptor.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 30 correct.