1.5 Lipids

1. AIts hydrocarbon chains exert a repulsive force on the water molecules, pushing them away.
   A student who thinks hydrophobic molecules and water push each other away picks this. No repulsive force is involved: water molecules attract one another more strongly, through hydrogen bonds, than they attract the nonpolar chains, so the chains are excluded.
2. BIts carbon–carbon bonds are too strong, so surrounding water molecules cannot break them apart.
   A student who thinks dissolving breaks a substance's molecules apart picks this. When a molecular substance dissolves, whole molecules separate from one another and mix with water; no bonds within the molecules are broken. The fat does not dissolve because its nonpolar chains cannot interact favorably with water.
3. CEach of its molecules is itself oily, and oily substances do not mix with water.
   A student who gives single molecules the properties of the bulk substance picks this. A single molecule is not 'oily'; the fat's behavior in water comes from a property of its molecules, their nonpolar chains, which cannot form hydrogen bonds with water.
4. DIts long hydrocarbon chains are nonpolar, so they cannot form hydrogen bonds with water. Correct
   Fats are made mostly of long chains of carbon and hydrogen atoms that share electrons almost equally, so the chains carry almost no partial charges. Water molecules form hydrogen bonds with one another and with polar or charged molecules but not with nonpolar chains, so the fat molecules are excluded and gather into droplets.

CED 1.5.A.1 · Read this in Fix

1. AThey are polymers, so each is built by joining many similar monomers.
   A student who thinks all large biological molecules are chains of repeating monomers picks this. A fat is assembled from glycerol and three fatty acids, and steroids contain no fatty acids; lipids are not long chains of repeating monomers like proteins or polysaccharides.
2. BThey are rich in energy, so cells use them mainly as stores of fuel.
   A student who thinks of all lipids as fats used for energy picks this. Fats store energy, but phospholipids form membranes, cholesterol stabilizes animal cell membranes and steroid hormones act as signals.
3. CThey are mostly nonpolar, so they do not dissolve readily in water. Correct
   Lipids are a diverse group defined by a property rather than by a shared monomer: they are made mostly of carbon and hydrogen joined by nonpolar bonds, so they are hydrophobic and do not dissolve readily in water. A phospholipid has a polar head, but its two long tails are nonpolar.
4. DThey are harmful, so cells are healthier with as few of them as possible.
   A student who carries over health messages about fat and cholesterol picks this, treating harmfulness as what the molecules have in common. Lipids are essential: phospholipids and cholesterol are parts of cell membranes, and fats and steroid hormones have necessary functions; they are grouped as lipids because they are mostly nonpolar and hydrophobic.

CED 1.5.A.2 · Read this in Fix

1. AIts phosphate groups bond covalently to one another, joining the molecules in a sheet.
   A student who thinks a membrane is a sheet held together by bonds between its molecules picks this. Phospholipids in a bilayer are not covalently bonded to one another; the arrangement results from how their heads and tails interact with water.
2. BIt is built to make cell membranes, so it lines up in a sheet whenever a cell needs one.
   A student who explains a molecule's behavior by the cell's needs picks this. A molecule has no purpose: phospholipids form bilayers in water even when no cell is present, because of their charged heads and nonpolar tails.
3. CIts phosphate group forms a charged head that faces water, while its two tails avoid water. Correct
   The models show that a phospholipid has two fatty acid chains and, in place of the fat's third chain, a phosphate group. The phosphate group is charged and interacts with water, while the nonpolar tails do not, so in water the molecules line up tail to tail with their heads facing the water on both sides: a bilayer. A fat has no charged head, so its molecules simply cluster into droplets.
4. DIts two tails are attracted to water and so face it, and its phosphate head avoids water.
   A student who has the polarity of the head and tails reversed picks this. The fatty acid tails are nonpolar and are excluded by water; it is the charged phosphate head that interacts with water.

CED 1.5.A.1 · Read this in Fix

1. AIts carbon atoms are joined by double bonds wherever possible.
   A student who has the meaning of 'saturated' reversed picks this. A saturated fatty acid has no double bonds between its carbon atoms; a fatty acid with one or more double bonds is unsaturated.
2. BIts carbon atoms are joined to one another only by single bonds. Correct
   In a saturated fatty acid, every bond between carbon atoms in the chain is a single bond, so each carbon carries as many hydrogen atoms as it can: the chain is saturated with hydrogen.
3. CIt has taken up as much water as its carbon chain can hold.
   A student who reads 'saturated' in the everyday or solution sense picks this. The term refers to hydrogen, not water: fatty acid chains are nonpolar and do not take up water.
4. DEach of its molecules is a solid, which is why it forms solid fats.
   A student who gives single molecules the properties of the bulk substance picks this. 'Solid' describes many molecules packed together, not one molecule; saturated chains are straight and pack closely, which is why fats rich in them tend to be solid at room temperature.

CED 1.5.A.1.i · Read this in Fix

1. AThe kinks in its tails keep its molecules from packing closely, so they attract each other less. Correct
   The model shows that each of fat B's tails has a kink at a double bond, so the tails cannot line up side by side as fat A's straight tails do. Molecules that are farther apart attract one another less, so they separate at a lower temperature, and fat B is liquid at a temperature at which fat A is solid.
2. BThe double bonds in its tails break at 20 °C, splitting each tail into two shorter pieces.
   A student who thinks melting breaks the bonds inside molecules picks this. When a fat melts, its molecules move apart from one another; the covalent bonds within each tail, including the double bonds, stay intact.
3. CEach of its molecules is softer and more fluid than fat A's, so the molecules are themselves liquid.
   A student who gives single molecules the properties of the bulk substance picks this. One molecule is not liquid or solid; fat B is liquid because its kinked molecules are loosely packed and attract each other less.
4. DIts tails bend so that it stays liquid, as the organism needs a fat that can flow freely.
   A student who explains a structure by the organism's needs picks this. The kinks are caused by double bonds in the chains; the fat is liquid because of how its molecules pack, not because the tails bend for a purpose.

CED 1.5.A.1.ii · Read this in Fix

1. AIts fatty acids would have fewer double bonds, and the oil would be more liquid at room temperature.
   A student who thinks double bonds make chains stiffer and hold molecules together more strongly picks this. It is the reverse: double bonds put kinks in the chains, so chains with fewer double bonds pack more closely and the oil is more solid.
2. BThe plant would be unable to make any fatty acids at all, so its seeds would store no oil.
   A student who thinks a mutation destroys a whole product picks this. The enzyme converts single bonds into double bonds in fatty acids that have already been made; without it the plant still makes fatty acids, but they are more saturated.
3. CThe oil would be unchanged, as the plant would alter its DNA to restore the enzyme it needs.
   A student who thinks organisms change their genes when they need to picks this. Mutations are not reversed in response to an organism's needs; the seeds would contain more saturated fatty acids.
4. DIts fatty acids would contain fewer double bonds, and the oil would be more solid at room temperature. Correct
   Without the enzyme, single bonds in the fatty acid chains are not converted into double bonds, so the fatty acids are more saturated. Chains with fewer double bonds have fewer kinks, pack more closely and attract each other more, so the oil melts at a higher temperature and is more solid at room temperature.

CED 1.5.A.1.iii · Read this in Fix

1. AMelting point rises as the number of double bonds rises, as double bonds hold the chains rigid.
   A student who thinks double bonds make chains stiffer and hold molecules together more strongly picks this. The graph shows the opposite trend: fatty acids with more double bonds have lower melting points, because the kinks keep the molecules from packing closely.
2. BMelting point falls as the number of double bonds rises, steeply at first, then less steeply. Correct
   Melting point falls from about 69 °C with no double bonds to about 13 °C with one, −5 °C with two and −11 °C with three. The first double bond lowers it by about 56 °C, the second by about 18 °C and the third by about 6 °C, so each added double bond lowers the melting point by less than the one before.
3. CMelting point falls by about the same number of degrees with each double bond that is added.
   A student who assumes the relationship is a straight line picks this. The decreases are about 56 °C, 18 °C and 6 °C, not equal; the graph curves and levels off.
4. DMelting point falls with more double bonds, as the double bonds break when the fatty acid melts.
   A student who thinks melting breaks the bonds within molecules picks this. When a fatty acid melts, its molecules move apart from one another, but their covalent bonds, including the double bonds, stay intact; the data show only how melting point changes.

CED 1.5.A.1.iv · Read this in Fix

1. AFatty acids with more double bonds will melt at lower temperatures than those with fewer.
   A student who thinks the null hypothesis is the result the investigator expects picks this. This is the alternative hypothesis, a predicted effect; the null hypothesis states that there is no effect.
2. BFatty acids with more double bonds will melt at higher temperatures than those with fewer.
   A student who thinks the null hypothesis is the opposite of the expected result picks this. A rise in melting point is still an effect of the number of double bonds; the null hypothesis states that there is no effect.
3. CThe number of double bonds in a fatty acid has no effect on its melting point. Correct
   A null hypothesis states that the independent variable has no effect on the dependent variable, so that any differences in the results would be due to chance. Here the independent variable is the number of double bonds and the dependent variable is the melting point.
4. DNone of the fatty acids will melt at any of the temperatures used in the investigation.
   A student who thinks a null hypothesis predicts that nothing happens at all picks this. 'Null' refers to no effect of the independent variable: the fatty acids can melt, but the null hypothesis predicts that their melting points do not differ with the number of double bonds.

CED 1.5.A.1.iv · Read this in Fix

1. A37% Correct
   Percent decrease = (60 − 38)/60 × 100 = 37%. At 10 °C the membranes contain a smaller proportion of saturated fatty acids and so a larger proportion of unsaturated ones; their kinked chains pack less tightly, which keeps the membrane fluid rather than stiff at the lower temperature.
2. B22%
   A student who reports the difference in percentage points as the percent change picks this. 60% − 38% = 22 percentage points; relative to the original value this is 22/60 × 100 = 37%.
3. C58%
   A student who divides the change by the new value picks this: 22/38 × 100 = 58%. Percent change is calculated relative to the original value, 60%.
4. D63%
   A student who gives the new value as a percentage of the original picks this: 38/60 × 100 = 63%. The new proportion is 63% of the original, which is a decrease of 37%.

Working From the graph, 60% of the fatty acids are saturated at 30 °C and 38% at 10 °C. Percent change = (original − new)/original × 100 = (60 − 38)/60 × 100 = 36.7%, which is 37% to two significant figures, a decrease. Distractors: the difference in percentage points, 60 − 38 = 22, reported as 22%; dividing the change by the new value, 22/38 × 100 = 58%; the new value as a percentage of the original, 38/60 × 100 = 63%.

CED 1.5.A.1.iv · Read this in Fix

1. AFat mass fell in both groups, so both groups used their stored fat as their main source of energy.
   A student who treats any difference between two means as real picks this. The fed group's mean fell by 3 g, but its error bars overlap widely, so this difference could be due to chance variation between samples.
2. BFat mass fell in the hibernating group, but a change in the fed group is not supported by the data. Correct
   In the hibernating group, mean fat mass fell from 42 g to 21 g, and the ±2 SE bars (39–45 g and 18–24 g) do not overlap, so the decrease is likely real: stored fat supplied energy while the animals hibernated without food. In the fed group the means differ by only 3 g and the bars (37–43 g and 33–41 g) overlap, so a change is not shown.
3. CFat mass in the fed group stayed exactly the same, as the error bars for that group overlap.
   A student who thinks overlapping error bars prove that two means are equal picks this. Overlap means that a difference has not been shown, not that there is none; the sample means here are 40 g and 37 g.
4. DFat mass fell in the hibernating group, and the mass of the fat was converted into energy.
   A student who thinks matter is turned into energy when fat is used picks this. Breaking down fat releases chemical energy for the cell's use, but the fat's atoms are not converted into energy; they leave the body, mainly in carbon dioxide and water.

Working No test statistic is calculated; the decision rests on the ±2 SE error bars. Hibernating: start 42 g (bar 39–45 g), end 21 g (bar 18–24 g); the bars do not overlap (24 < 39), so the decrease is likely a real difference. Active and fed: start 40 g (bar 37–43 g), end 37 g (bar 33–41 g); the bars overlap (37–41 g), so the 3 g difference between the means may be due to chance and a change is not supported. Overlap does not show that the means are exactly equal, and the mass of fat that is broken down is not converted into energy: its atoms leave the body mainly in carbon dioxide and water.

CED 1.5.A.2.i · Read this in Fix

1. AIts fat is turned into heat at the same rate as heat escapes from the body.
   A student who thinks fat is converted into heat picks this. Breaking down stored fat releases energy, some of it as heat, but the fat's matter is not converted into heat; the blubber keeps the seal warm mainly as an insulating layer that slows heat loss.
2. BIt keeps the cold of the water from moving in and cooling the seal's body.
   A student who thinks of cold as something that flows into objects picks this. Only heat moves, from warmer to cooler places; the blubber keeps the seal warm by slowing the flow of heat out of its body.
3. CIt stops heat from leaving the body at all, so the seal loses no heat to the sea.
   A student who thinks an insulator blocks heat completely picks this. Insulation slows heat transfer but does not stop it; the seal still loses heat to the water and replaces it with heat from its metabolism.
4. DIt reduces how quickly the seal's body loses heat to the sea around it. Correct
   Fat conducts heat poorly, so a thick layer of blubber reduces the rate at which heat passes from the warm body to the cold water. The seal still loses some heat, but slowly enough that heat released by its metabolism keeps its body temperature near 37 °C.

CED 1.5.A.2.i · Read this in Fix

1. AIts blood glucose would fall below normal between meals, as the hormone is absent. Correct
   The hormone helps maintain blood glucose between meals, when no glucose is entering the blood from food, an example of homeostasis. Without it, nothing would offset the use of glucose by the body's cells at these times, so its concentration would fall lower than normal.
2. BIts muscles would shrink, as steroid hormones act mainly to build muscle tissue.
   A student who knows 'steroids' only as muscle-building drugs picks this. Steroid hormones support many functions, including growth and development, energy metabolism and homeostasis; this hormone regulates blood glucose.
3. CIt would have less stored energy, as steroids are lipids that its cells use as fuel.
   A student who thinks all lipids are energy stores picks this. Steroid hormones act as chemical signals at low concentrations; they are not a significant fuel, and losing this one affects the regulation of blood glucose.
4. DOther organs would start to make the hormone, as the body needs to have it.
   A student who thinks the body produces whatever it needs picks this. Other organs do not take over a function because it is missing; the cells that made the hormone are destroyed, so its effect on blood glucose is lost.

CED 1.5.A.2.ii · Read this in Fix

1. AThey would work better, as cholesterol is harmful to the cells.
   A student who carries over health messages about cholesterol in the diet and blood picks this. Cholesterol in the plasma membrane is essential to animal cells; removing it harms them.
2. BThey would run short of fuel, as cholesterol is stored for energy.
   A student who thinks all lipids are energy stores picks this. Cholesterol is a structural component of animal cell membranes, not a fuel.
3. CTheir membranes would be less stable, as cholesterol supports their structure. Correct
   In animal cells, cholesterol helps keep the membrane structurally stable. With most of it removed, the membranes would be less stable, so they would be more easily damaged and more likely to leak.
4. DTheir membranes would be unchanged, as the phospholipids are bonded together.
   A student who thinks a membrane is a sheet held together by bonds between its molecules picks this. The phospholipids of a bilayer are not covalently bonded to one another, and cholesterol among them helps keep the membrane's structure stable.

CED 1.5.A.2.iii · Read this in Fix

1. ACompare it with preparation 3, whose vesicle membranes also contain cholesterol
   A student who thinks a control should receive the same treatment as the experimental sample picks this. Preparation 3 also contains cholesterol and differs from preparation 1 in temperature, so it cannot show the effect of cholesterol.
2. BCompare it with preparation 4, which has no vesicles, so nothing can affect the dye
   A student who thinks the control is always the sample with nothing in it picks this. Preparation 4 differs from preparation 1 in having no membranes at all, so it cannot isolate the effect of cholesterol.
3. CUse preparation 1 alone, as its leakage by itself shows what cholesterol does
   A student who thinks a treated sample alone shows a treatment's effect picks this. Without the leakage from membranes that lack cholesterol under the same conditions, the leakage from preparation 1 cannot be attributed to cholesterol.
4. DCompare it with preparation 2, whose vesicles differ only in lacking cholesterol Correct
   Preparations 1 and 2 are made and tested in the same way at the same temperature, 37 °C; the only difference is that preparation 2's membranes lack cholesterol. More leakage from preparation 2 than from preparation 1 would support the claim that cholesterol makes membranes less leaky.

Working No calculation. Preparation 1 (phospholipids + cholesterol, 37 °C) must be compared with a preparation that differs from it only in the factor being tested, cholesterol: preparation 2 (phospholipids only, 37 °C). Preparation 3 also contains cholesterol and differs in temperature; preparation 4 has no vesicles; a single preparation gives no comparison.

CED 1.5.A.2.iii · Read this in Fix

1. AArrangement W Correct
   The heads are charged and interact with water; the tails are nonpolar and are excluded by water. In arrangement W, two layers of phospholipids lie tail to tail, so the tails are shielded from water on both sides while the heads face the water: a lipid bilayer, as in cell membranes.
2. BArrangement X
   A student who has the polarity of the head and tails reversed picks this. In arrangement X the tails face the water on both sides and the heads are buried inside, the opposite of what charged heads and nonpolar tails do.
3. CArrangement Y
   A student who thinks a membrane is a single layer of phospholipids picks this. In a single layer surrounded by water, the tails on one side are exposed to water; two layers, tail to tail, shield all the tails.
4. DArrangement Z
   A student who thinks the polar head makes the whole phospholipid dissolve picks this. The two long nonpolar tails keep phospholipids from dissolving as separate molecules; they gather into a bilayer with the tails away from water.

CED 1.5.A.2.iv · Read this in Fix

1. AIts lipids form one layer, a single molecule thick, around the cell.
   A student who thinks a membrane is a single layer of lipid picks this. A single layer would cover an area about equal to the cells' surface area; in every sample the lipids covered about twice that area.
2. BIts lipids form a layer two molecules thick around each cell. Correct
   In every sample the lipids, spread one molecule thick, cover about twice the surface area of the cells (1.86/0.92 = 2.0, 2.17/1.10 = 2.0 and 1.71/0.85 = 2.0). Because the cells have no internal membranes, all of this lipid came from the plasma membrane, so the membrane contains enough lipid to cover each cell twice: a layer two molecules thick, the lipid bilayer.
3. CIts amount of lipid per unit area differs from one sample to the next.
   A student who treats any difference between values as a real difference picks this. The ratios of lipid area to cell surface area, 2.02, 1.97 and 2.01, differ by less than 3%, small enough to arise from measurement error; every sample gives a ratio of about 2, so the data support the same layer two molecules thick in each.
4. DIts lipid molecules are arranged with their tails facing the water.
   A student who has the polarity of the head and tails reversed picks this. The hydrophobic tails face the interior of the bilayer, away from the water; the hydrophilic heads face the water on both sides.

Working Ratio of lipid area to cell surface area: sample 1, 1.86/0.92 = 2.02; sample 2, 2.17/1.10 = 1.97; sample 3, 1.71/0.85 = 2.01. In each sample the lipid covers about twice the cells' surface area; since the cells have no internal membranes, the plasma membrane's lipid forms a layer two molecules thick.

CED 1.5.A.2.iv · Read this in Fix