Unit 1 · Unit 1 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.
Along a polypeptide's backbone, every amino acid contributes a C=O group and an N–H group. Oxygen pulls shared electrons harder than carbon does, and nitrogen pulls shared electrons harder than hydrogen does. In many proteins, a C=O group on one stretch of the backbone lies close to an N–H group on another stretch.
Which of the following best explains why the two groups hydrogen-bond to each other?
Mosquito larvae hang from the underside of a still pond surface. A gardener spreads a thin film of oil over the surface, and the larvae sink.
Which of the following best explains why the larvae sink after the gardener adds the oil?
A student heats 50 g of each of three liquids from 20 °C in identical beakers on identical hot plates. Hexane is a hydrocarbon. The table gives each liquid's temperature after 2 minutes and the energy needed to turn 1 g of it into vapor.
Which of the following best explains why the student used the same mass, 50 g, of each liquid?
A student heats 50 g of each of three liquids from 20 °C in identical beakers on identical hot plates. Hexane is a hydrocarbon. The table gives each liquid's temperature after 2 minutes and the energy needed to turn 1 g of it into vapor.
Based on the data, which liquid has the highest specific heat capacity?
A student heats 50 g of each of three liquids from 20 °C in identical beakers on identical hot plates. Hexane is a hydrocarbon. The table gives each liquid's temperature after 2 minutes and the energy needed to turn 1 g of it into vapor.
Which of the following best explains why the water's temperature rose more slowly than the other two liquids' temperatures?
A student heats 50 g of each of three liquids from 20 °C in identical beakers on identical hot plates. Hexane is a hydrocarbon. The table gives each liquid's temperature after 2 minutes and the energy needed to turn 1 g of it into vapor.
A drop of each liquid evaporates from a person's skin. Which of the following liquids cools the skin most for each gram that evaporates?
A cell adds one more amino acid to the end of a growing polypeptide.
Which of the following describes the bond that joins the new amino acid to the chain?
A sealed glass terrarium holds moss, soil, water, air and a few woodlice. No one adds or removes anything for a year. The moss grows and roughly doubles in mass.
Which of the following best explains where the atoms in the moss's new mass came from?
A growing mouse is fed for a month on food in which every nitrogen atom is a heavy form of nitrogen that can be traced. Its carbon, hydrogen, oxygen, sulfur and phosphorus come from ordinary sources.
Which of the following classes of molecules must contain the heavy nitrogen?
A cell is placed in a solution that supplies every element it needs except phosphorus. It goes on taking in sugar.
Which of the following molecules can the cell still build?
A chain of 60 glucose units was cut by hydrolysis into 5 pieces of 12 units each.
How many water molecules were consumed?
A bacterium joins amino acids one after another to build a protein. Hours later the same bacterium breaks the protein back down into free amino acids.
Which of the following best describes the role of water in the two reactions?
A student measures the mass of a 2.00 g sample of starch and of a 2.00 g sample of cellulose, breaks each completely into glucose by hydrolysis, and measures the mass of the glucose collected from each. The results are in the table.
Why does the glucose collected from each sample have a greater mass than the sample had?
A student measures the mass of a 2.00 g sample of starch and of a 2.00 g sample of cellulose, breaks each completely into glucose by hydrolysis, and measures the mass of the glucose collected from each. The results are in the table.
Starch is a plant's food store; cellulose is the material of a plant's cell walls. Which conclusion do the results support?
Researchers studied the membranes of six fish species, each living in water at a different temperature. For each species they measured the percentage of the membrane's fatty-acid tails that carry at least one double bond. The graph shows the results.
Which of the following best describes the relationship shown in the graph?
Two glucose stores, molecule J and molecule K, each hold 1,000 glucose units. Molecule J is one unbranched chain. Molecule K is a compact cluster with many short side chains.
From which molecule can a cell remove more glucose units at once, and why?
Wood, which is mostly cellulose, holds up a heavy roof for centuries, while a block of dried starch crumbles under a light load.
Which of the following best explains why cellulose is so much stronger than starch?
Four liquids are stirred into separate beakers of water: ethanol, C₂H₅OH, which carries one O–H group; glycerol, which carries three O–H groups; a cooking oil, whose molecules are mostly long hydrocarbon chains; and a solution of glucose, whose molecules carry five O–H groups.
Which of the following liquids separates into its own layer instead of mixing into the water?
A molecule found in an animal's body is one small three-carbon molecule joined to three long hydrocarbon tails. Each tail is joined to the three-carbon molecule through the carboxyl group at the tail's end.
Which of the following best predicts how this molecule behaves in water, and why?
A dishwashing detergent molecule has a small head carrying a full charge and one long hydrocarbon tail. In water, detergent molecules gather into tiny balls with every tail pointing inward and every head on the outside.
Why do the molecules arrange themselves this way?
The drawing shows a student's model of a phospholipid bilayer lying between the watery fluid outside a cell and the watery cytosol inside it. Heads are drawn as circles and tails as wavy lines.
Which of the following best identifies the error in the model?
The model shows a short strand of DNA, three nucleotides long. Phosphate groups are drawn as circles, sugars as pentagons and bases as rectangles carrying their letters. The two ends of the strand are marked X and Y.
The sugar at end Y carries a free –OH group; end X carries a free phosphate group. One more nucleotide joins the strand. Where does it join?
The model shows a short strand of DNA, three nucleotides long. Phosphate groups are drawn as circles, sugars as pentagons and bases as rectangles carrying their letters. The two ends of the strand are marked X and Y.
Which bond joins the sugar of one nucleotide to the phosphate group of the next along this strand?
The model shows a short strand of DNA, three nucleotides long. Phosphate groups are drawn as circles, sugars as pentagons and bases as rectangles carrying their letters. The two ends of the strand are marked X and Y.
A second DNA strand pairs with this strand along its whole length. Written from its own 5′ end, what does the second strand read?
The drawing shows a student's model of a piece of double-stranded DNA, four base pairs long, with the ends of both strands labeled and the paired bases joined by dashed lines.
What, if anything, is wrong with the model?
The DNA of a fungus is analyzed. Its two strands are paired along their whole length, and thymine makes up 27% of its bases.
What percentage of the fungus's bases is cytosine?
Researchers grow one batch of animal cells so that its membranes end up with far less cholesterol than normal. Under gentle shaking, these cells tear far more easily than normal cells do.
Which of the following best explains the observation?
A change in a protein's sequence replaces leucine, whose R group is a hydrocarbon, with lysine, whose R group ends in –NH₃⁺, at one position deep inside the folded protein.
Which of the following best predicts the effect of the change on the protein?
Glutamate's R group is –CH₂–CH₂–COO⁻. Serine's R group is –CH₂–OH.
Is each amino acid's R group charged, polar or nonpolar?
An antibody is a protein built from four polypeptide chains, two long and two short. Each chain folds on its own, and the four folded chains then fit together into one molecule. A change in the sequence of the short chains stops them fitting to the long chains, though each chain still folds.
Which of the following best describes the effect of the change on the antibody's structure?
(a)(i) Identify the independent variable in the researchers' experiment. (1 point)
A full-credit answer: The independent variable is the concentration of the antifreeze protein in the tube.
Check the box for each point your answer earns
Scoring note: 'the protein' alone does not earn the point; the temperature at which ice began to grow is the dependent variable and does not earn the point.
Common slip: Naming the temperature at which ice began to grow. That is what the researchers measured, the dependent variable; the independent variable is what they changed.
(a)(ii) Justify the researchers' inclusion of the tubes with no protein. (1 point)
A full-credit answer: The tubes with no protein show the temperature at which the salt solution alone begins to freeze.
So any lowering of that temperature in the other tubes can be attributed to the protein.
Check the box for each point your answer earns
Accept one of the following: 'they show what happens with no protein, for comparison'; 'they give the baseline freezing temperature'.
Scoring note: 'they are 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 control shows in this experiment: the freezing temperature of the solution with no protein.
(b)(i) Describe the relationship between the concentration of protein and the temperature at which ice began to grow. (1 point)
A full-credit answer: As the concentration of protein increases, the temperature at which ice begins to grow decreases.
With no protein, ice grew at −0.7 °C; with 8 mg per mL, it grew only at −2.0 °C.
Check the box for each point your answer earns
Accept: 'the more protein, the lower the temperature at which ice grew', with or without values from the table.
Scoring note: a direction is required; 'the protein changes the temperature' does not earn the point.
Common slip: Saying the protein lowers the temperature of the solution. The protein lowers the temperature at which ice begins to grow, not the temperature of the tube.
(b)(ii) Describe the pattern in the data at concentrations of 4 mg per mL and above. (1 point)
A full-credit answer: At 4 mg per mL and above, the temperature at which ice begins to grow changes very little.
Doubling the concentration from 4 to 8 mg per mL lowers it only from −1.9 °C to −2.0 °C.
Check the box for each point your answer earns
Accept one of the following: 'the effect levels off'; 'the effect reaches a plateau'; 'each added milligram lowers the temperature less than the one before'.
Scoring note: restating the two values with no statement of the pattern does not earn the point.
Common slip: Writing that the protein stops working above 4 mg per mL. It still works, lowering the temperature to −2.0 °C; adding more protein simply adds little.
(c)(i) Explain how the R groups on the protein's flat face bind the protein to the surface of an ice crystal. (1 point)
A full-credit answer: Each –OH group is polar: its oxygen is δ− and its hydrogen δ+.
Ice is made of water molecules, which carry partial charges too.
So the –OH groups form hydrogen bonds with the water molecules at the ice surface.
The row of –OH groups binds the flat face to the ice at many points.
Check the box for each point your answer earns
Accept: 'the –OH R groups hydrogen-bond to the ice', with the partial charges implied.
Scoring note: 'the R groups stick to the ice' with no bond named, or 'covalent bonds to the ice', does not earn the point.
Common slip: Saying the face bonds covalently to the ice. No electrons are shared; the –OH groups are attracted to the ice's water molecules by hydrogen bonds.
(c)(ii) Explain why the nonpolar R groups are found in the interior of the folded protein rather than on the surface facing the blood. (1 point)
A full-credit answer: The blood is mostly water.
Water's partial charges attract polar and charged R groups, so those stay at the surface.
Nonpolar R groups carry no partial charges, so water is not attracted to them.
The water molecules stay attracted to one another and exclude the nonpolar R groups, so those cluster in the interior.
Check the box for each point your answer earns
Accept one of the following: 'hydrophobic R groups cluster away from the water'; 'polar R groups hydrogen-bond with the water and stay at the surface, leaving the nonpolar R groups inside'.
Scoring note: a fold explained with no reference to water does not earn the point; 'nonpolar R groups attract one another strongly' does not earn the point.
Common slip: Saying the nonpolar R groups pull one another inward. Water's exclusion of them, and its attraction to the polar R groups, drives the fold; the R-group interactions then hold it.
(c)(iii) Explain why the pattern you described in (b)(ii) occurs. (1 point)
A full-credit answer: The protein works by covering the surface of each ice crystal.
At 4 mg per mL there is enough protein to cover the surface of every crystal that forms.
Extra protein finds no free ice surface to bind.
So adding more protein lowers the temperature no further.
Check the box for each point your answer earns
Accept one of the following: 'the ice surface is already fully occupied'; 'all the binding places on the ice are used up'.
Scoring note: 'the protein is used up' does not earn the point.
Common slip: Saying the protein is used up at high concentration. Nothing is used up; there is more protein than ice surface for it to bind.
(d)(i) The researchers heat a sample of the 8 mg per mL solution to 70 °C for ten minutes; the sample turns cloudy. They cool it back to its starting temperature and repeat the measurement. Predict the temperature at which ice crystals begin to grow in the heated sample. (1 point)
A full-credit answer: Ice crystals begin to grow at about −0.7 °C, the same temperature as in the tubes with no protein.
Check the box for each point your answer earns
Accept: 'higher than −2.0 °C, close to the no-protein value'.
Scoring note: a temperature or a direction of change is required.
Common slip: Predicting −2.0 °C because the protein is still there. The protein's chains are still there, but its fold, and with it the working face, is gone.
(d)(ii) Justify your prediction. (1 point)
A full-credit answer: Heating broke the weak interactions that held each chain's fold in place.
The unfolded chains tangled together, which made the sample cloudy.
These tangled chains do not refold on cooling.
So the flat face of –OH groups no longer exists, and the protein cannot bind ice.
The solution freezes as if no protein were present.
Check the box for each point your answer earns
Accept: 'the protein has permanently lost its fold and does not refold on cooling, so the ice-binding face is lost' (a student who uses the exam's own term for a protein that has lost its fold earns the point).
Scoring note: 'the peptide bonds broke' or 'the protein was used up' does not earn the point; the cloudiness, the tangling and the unchanged sequence need not be mentioned.
Common slip: Claiming the protein works again once it is cold. Cooling restores the temperature, not the fold; the tangled chains stay tangled.
(a) Identify the fat whose fatty-acid tails are the most saturated. (1 point)
A full-credit answer: Cocoa butter has the most saturated tails.
Its tails average 0.4 double bonds each, the fewest in the table.
Check the box for each point your answer earns
Accept: 'the fat with 0.4 double bonds per tail'.
Common slip: Picking the fat that sets at the highest temperature without reading the double-bond column. The setting temperature follows from the double bonds; the double-bond column is the evidence the task asks for.
(b) Describe the relationship between the average number of double bonds per tail and the temperature at which the fat sets. (1 point)
A full-credit answer: As the average number of double bonds per tail increases, the temperature at which the fat sets decreases.
Cocoa butter, with 0.4 double bonds per tail, sets at 34 °C; grapeseed oil, with 1.5, sets at −10 °C.
Check the box for each point your answer earns
Accept: 'fats with more double bonds set at lower temperatures' (an inverse relationship), with or without values from the table.
Scoring note: a direction is required; 'the double bonds affect the setting temperature' does not earn the point.
Common slip: Describing one fat instead of the relationship. The point needs how the setting temperature changes as the double bonds increase across the table.
(c) A classmate looks at the table and claims that any fat whose tails average fewer than one double bond will be solid at room temperature, 20 °C. Evaluate the classmate's claim, using the data in the table. (1 point)
A full-credit answer: The claim is not supported.
Duck fat's tails average 0.8 double bonds, fewer than one.
Yet duck fat sets at 14 °C, so at 20 °C it is liquid.
So a fat with fewer than one double bond per tail can be liquid at 20 °C.
Check the box for each point your answer earns
Scoring note: the judgement alone does not earn the point; the point requires a fat from the table named with its setting temperature.
Common slip: Agreeing because cocoa butter and palm oil are solid at 20 °C. One fat that breaks the claim is enough to refute it; duck fat does.
(d) Explain how the relationship you described in (b) results from the structure of the fatty-acid tails. (1 point)
A full-credit answer: Each double bond puts a kink in a fatty-acid tail.
Kinked tails cannot lie close against their neighbors along their whole length.
The weak attractions between tails add up only where the tails lie close, so kinked tails attract one another less.
The molecules must be cooled further, and so slowed further, before those weaker attractions can hold them in place.
So a fat with more double bonds sets at a lower temperature.
Check the box for each point your answer earns
Accept: 'weaker attractions between kinked tails, so a lower setting temperature'.
Scoring note: 'double bonds are weaker bonds' or 'the kinks make the molecules move faster' does not earn the point; the kink alone, with no link to the attractions between tails, does not earn the point.
Common slip: Saying the double bonds themselves are weak and let the molecules slide. The double bonds are inside each tail; what changes is how closely the tails can lie against one another.
(a) Identify the sample that is two strands paired along their whole length. (1 point)
A full-credit answer: Sample P is two strands paired along their whole length.
Check the box for each point your answer earns
Common slip: Choosing sample Q because it has more of one base. The answer rests on which sample's paired bases are present in equal amounts.
(b) Support your answer to (a), using the base percentages as evidence. (1 point)
A full-credit answer: In two paired strands every adenine is paired with a thymine and every guanine with a cytosine.
So the paired bases must be present in equal amounts.
Sample P has adenine 31% and thymine 31%, guanine 19% and cytosine 19%: the amounts match.
Sample Q has adenine 22% against uracil 30%: the amounts do not match.
So sample P's bases are paired and sample Q's are not.
Check the box for each point your answer earns
Accept: the contrast with sample Q (adenine 22% against uracil 30%; guanine 25% against cytosine 23%) as the evidence, with the same pairing reason.
Scoring note: percentages restated with no pairing reason do not earn the point.
Common slip: Quoting the percentages with no reason. The point needs why paired bases must be equal: every base on one strand is paired with one partner on the other.
(c) Determine which sample is RNA, and justify your decision using the table. (1 point)
A full-credit answer: Sample Q is RNA.
It contains uracil and no thymine.
RNA carries uracil in place of DNA's thymine.
Check the box for each point your answer earns
Scoring note: sample Q with no feature named does not earn the point; 'because it is a single strand' alone does not earn the point (a single strand of DNA is possible).
Common slip: Resting the decision on the single strand alone. A single strand is usual for RNA but does not settle it; the uracil does.
(d) Looking at the table, the student says: 'A nucleic acid whose bases are present in unequal amounts will stay single whatever strand it is mixed with.' Evaluate the student's claim. (1 point)
A full-credit answer: The student's claim is not supported.
Unequal amounts show only that the strand has no partner strand in the sample.
They do not show that its bases are unable to pair.
Each base can still pair with its partner base, adenine with uracil or thymine and guanine with cytosine, when a strand carrying those partner bases lies alongside it.
So a strand with unequal amounts of bases can still pair.
Check the box for each point your answer earns
Scoring note: the judgement alone does not earn the point; agreeing that unequal amounts make pairing impossible does not earn the point.
Common slip: Agreeing with the student because the amounts differ. The amounts describe the strand as it was found; pairing depends on a complementary strand being present.