6.7 Mutations

1. AA substitution of one nucleotide in the codon for amino acid 120 Correct
   A substitution keeps the number of nucleotides the same, so the reading frame is unchanged and only codon 120 is altered. If the new codon specifies valine, the enzyme has valine at position 120 and is otherwise identical, which is what was found.
2. BAn insertion of one nucleotide in the codon for amino acid 120
   A student who thinks an insertion changes only the codon where it occurs picks this. Inserting one nucleotide shifts the reading frame, so every codon from 120 onward is read differently, not just codon 120.
3. CA mutation in codon 120 that occurred because the cell needed valine
   A student who thinks mutations arise to meet a need picks this. A change in codon 120 does account for the valine, but it arose by chance, through a replication or repair error or DNA damage; the cell's need for valine did not cause it.
4. DA replacement of the amino acid glutamic acid by valine within the gene
   A student who thinks a gene contains the amino acids of its protein picks this. A gene is a sequence of nucleotides; a change of amino acid results from a change in the nucleotides of the codon.

CED 6.7.A.1.i · Read this in Fix

1. AThere is no variation, as every cell is a copy of the single original cell
   A student who thinks asexual reproduction always produces identical copies picks this. Mutations during replication make some descendants differ from the original cell.
2. BVariation is present, as random mutations arise when DNA is replicated Correct
   Every round of DNA replication carries a chance of uncorrected errors, so cells come to differ in some of their DNA sequences even without sexual reproduction or DNA transfer. Mutation is one source of genetic variation.
3. CVariation is present, as cells change their DNA to suit their environment
   A student who thinks mutations arise because organisms need them picks this. Variation is present, but it arises from random mutations, not from cells changing their DNA to fit their surroundings.
4. DThere is no variation, as any new mutation spreads to every cell at once
   A student who thinks all individuals in a population change together picks this. A mutation arises in one cell and is passed only to that cell's descendants, so cells with and without it coexist.

CED 6.7.B.1.ii · Read this in Fix

1. AThe cell has lost a whole copy of chromosome 4, so every gene on that chromosome is now missing
   A student who thinks any loss of chromosome material is the loss of a whole chromosome picks this. Only a segment was lost; the rest of chromosome 4 remains, so only the genes in the segment are affected.
2. BEach gene in the segment now has a point mutation, which can change the protein it codes for
   A student who thinks chromosome changes work by mutating the genes picks this. A deletion removes the segment's genes from one chromosome; it does not substitute nucleotides in them.
3. CEach gene in the lost segment is left with one copy, which can alter the amount of its product Correct
   This is an alteration in chromosome structure (a deletion). The cell still has both copies of chromosome 4, but one lacks the segment, so each gene in it is present in only one copy. The altered amounts of these gene products can disrupt development and cause a genetic disorder.
4. DEvery gene after the break point on chromosome 4 is now read in a shifted reading frame
   A student who thinks any change to DNA shifts the reading frame picks this. Each gene is read in codons from its own start codon; losing a segment does not shift the frame of genes elsewhere on the chromosome.

CED 6.7.B.2.iii · Read this in Fix

1. AThey were inherited from a common ancestor of the three groups and conserved Correct
   Processes that increase genetic variation, such as meiosis with crossing over and fertilization, are evolutionarily conserved: they arose in a common ancestor of these groups and have been maintained in its descendants, which is why distantly related organisms share them.
2. BEach group developed them separately, as each needed genetic variation
   A student who thinks traits arise because organisms need them picks this. New traits arise from random genetic changes, not from need; the shared processes are explained by inheritance from a common ancestor.
3. CPlants and fungi descended from animals and inherited the processes from them
   A student who thinks living groups descended from other living groups picks this. Plants, fungi and animals share common ancestors; none of the living groups is the ancestor of the others.
4. DThey are the most advanced way to reproduce, so every complex lineage reached them
   A student who thinks evolution is progress toward more advanced forms picks this. Evolution has no goal; the processes are shared because they were inherited and maintained.

CED 6.7.C.1.iii · Read this in Fix

1. AIt has Met and Lys, a new amino acid in place of Trp, then Gln and Gly
   A student who thinks every point mutation replaces exactly one amino acid picks this. The new codon, UGA, does not specify an amino acid; it is a stop codon, so translation ends after Lys.
2. BIt has only Met and Lys, as the third codon is now a stop codon Correct
   The mutant's third codon is UGA, which the table shows is a stop codon. Translation ends there, so the polypeptide has only the first two amino acids, Met and Lys. This is a nonsense mutation: a point mutation that causes a premature stop.
3. CIt has Met and Lys, then a new series of amino acids from codon 3 onward
   A student who thinks any change in a codon shifts the reading frame picks this. The mutation replaces one nucleotide, so the frame is unchanged; and translation stops at the new stop codon anyway.
4. DNo polypeptide is made, as a stop codon in an mRNA blocks all translation
   A student who thinks a mutation stops a gene from being expressed picks this. Translation starts at AUG as usual and continues until it reaches the stop codon, so a short polypeptide (Met-Lys) is made.

CED 6.7.A.1.iii · Read this in Fix

1. AIts mutation changed enzyme E's amino acid sequence, as every mutation does
   A student who thinks every mutation changes a protein's amino acid sequence picks this. Mutant 1's enzyme is as active per milligram as the wild type's; the data show a change in the amount of enzyme in the cells, which mutations can also cause.
2. BIts mutation is neutral, as each of its enzyme molecules still works as well as before
   A student who thinks a mutation matters only if it changes the protein's structure picks this. Each molecule works normally, but mutant 1 makes a quarter of the normal amount, so its cells have much less enzyme activity.
3. CIts mutation arose because the fungus needed less enzyme E in its cells
   A student who thinks mutations arise because an organism needs them picks this. The data show what the mutation does, not why it arose; mutations arise at random, and nothing in the data concerns a need for less enzyme.
4. DIts mutation lowers the amount of enzyme E, so the cells have less enzyme E activity Correct
   Mutant 1 has only 25% of the wild type's amount of enzyme E, but each milligram of its enzyme is as active as the wild type's (100%). Its mutation changes the amount of protein, not how well each molecule works, so its cells have about a quarter of the normal total activity.

CED 6.7.A.1 · Read this in Fix

1. AA silent mutation, as it leaves the fish's phenotype unchanged
   A student who thinks 'silent' means 'no effect on the phenotype' picks this. A silent mutation leaves the amino acid sequence unchanged; this mutation replaces an amino acid, so it is not silent, even though it is neutral.
2. BNot a mutation, as a DNA change counts only if it alters the phenotype
   A student who thinks a change in DNA is a mutation only if it has a visible effect picks this. Any change in the nucleotide sequence is a mutation, whatever its effect.
3. CA neutral mutation, as it has no detectable effect on enzyme or phenotype Correct
   The change in the DNA is a mutation, and it replaces an amino acid, but neither the enzyme's activity nor the fish's phenotype is detectably affected. A mutation with no detectable effect is neutral.
4. DA beneficial mutation, as it arose to meet a need of the fish
   A student who thinks mutations arise to meet an organism's needs picks this. Mutations arise at random, and the tests found no advantage: the change has no detectable effect on the enzyme or the phenotype.

CED 6.7.A.1 · Read this in Fix

1. AMet-Phe-Gly-Leu
   A student who thinks deleting one nucleotide removes one amino acid picks this: the wild-type sequence without Ala. One nucleotide is a third of a codon, so the deletion shifts the frame and changes every later codon.
2. BMet-Asp-Phe-Gly
   A student who thinks a deletion changes only the codon where it occurs picks this. The deletion moves every later codon boundary, so the codons after codon 2 are not UUC and GGA but UCG and GAC.
3. CMet-Ala-Phe-Gly
   A student who thinks a one-nucleotide change is too small to affect the protein picks this, the wild-type sequence. The deletion shifts the reading frame, so the amino acids from codon 2 onward differ.
4. DMet-Asp-Ser-Asp Correct
   Reading the mutant mRNA in codons of three from AUG gives AUG GAU UCG GAC, which the table shows specify Met, Asp, Ser and Asp. Deleting one nucleotide shifted the reading frame, so every codon after the deletion is read differently: a frameshift mutation.

CED 6.7.A.1.ii · Read this in Fix

1. AStrain A's is likelier to function, as it lost one nucleotide, too small a change to matter
   A student who thinks a one-nucleotide change is too small to matter picks this. Losing one nucleotide shifts the reading frame, so strain A's polypeptide differs from codon 10 onward; strain B's is the one likely to retain function.
2. BStrain B's is likelier to function, as it lacks one amino acid but the rest is read in frame Correct
   Deleting three nucleotides removes one whole codon, so strain B's polypeptide lacks one amino acid but every later codon is read in the original frame. Deleting one nucleotide shifts the frame from codon 10 onward, changing almost all of strain A's polypeptide, which is very unlikely to function.
3. CBoth are equally likely to function, as each deletion changes only the amino acid at codon 10
   A student who thinks a deletion changes only the codon where it occurs picks this. Strain A's deletion shifts the frame and changes every codon after it; only strain B's affects codon 10 alone.
4. DNeither polypeptide is made, as any deletion in a gene stops the gene from being expressed
   A student who thinks a mutation stops a gene from being expressed picks this. Both genes are still transcribed and translated from the start codon; the deletions change the polypeptides made.

CED 6.7.A.1.ii · Read this in Fix

1. AThe mutant's protein P has the same amino acid sequence as the normal protein P Correct
   A silent mutation is one that changes the nucleotide sequence without changing the amino acid sequence, because the new codon specifies the same amino acid. An unchanged amino acid sequence in protein P is the direct evidence for the claim.
2. BThe mutant plant has the same phenotype as a plant without the substitution
   A student who thinks 'silent' means 'no effect on the phenotype' picks this. A mutation that changes an amino acid can also leave the phenotype unchanged; that would make it neutral, not silent.
3. CThe mutant plant's cells contain none of the protein encoded by gene P
   A student who thinks a silent mutation switches a gene off picks this. A silent mutation leaves the protein's amino acid sequence unchanged; finding no protein P would argue against the claim.
4. DThe mutant's gene P contains the same amino acids as normal gene P
   A student who thinks a gene contains the amino acids of its protein picks this. Genes are made of nucleotides; the evidence for a silent mutation is in the amino acid sequence of the protein.

CED 6.7.A.1.iv · Read this in Fix

1. AThe changed base pair arose because the cell needed a different nucleotide at that site
   A student who thinks mutations arise because a cell needs them picks this. The model shows a copying error that was not corrected; nothing about it is directed by a need, and replication errors occur at random.
2. BThe model shows a mutation only if radiation or a chemical caused that mispairing
   A student who thinks mutations are caused only by outside agents picks this. Errors in DNA replication and repair cause mutations without any outside agent; the model shows such an error.
3. CThe model shows that a mispairing left unrepaired leads to a permanent change Correct
   In the first round, T is placed opposite G and the mismatch is not repaired. In the second round, the strand with T serves as a template and is paired with A, giving a molecule with an A–T pair where the original had G–C. An uncorrected replication error has become a permanent change in the base-pair sequence that will be copied in later rounds.
4. DThe model shows that any replication error makes a cell less able to survive
   A student who thinks all mutations are harmful picks this. The model shows how a sequence change arises; it says nothing about its effect, which may be detrimental, neutral or beneficial.

CED 6.7.B.1 · Read this in Fix

1. APlates of yeast handled in the same way but not exposed to chemical X Correct
   Mutations also arise without chemical X, from replication and repair errors. Plates treated identically except for chemical X show how many mutant colonies arise anyway, so the student can tell whether exposure increases the number.
2. BKeeping all of the exposed plates at the same temperature and pH
   A student who thinks keeping conditions constant provides a control picks this. Constant conditions make the comparison fair, but there is still nothing to compare the exposed plates with.
3. CMore plates of yeast exposed to the same concentration of chemical X
   A student who thinks repeating the treatment is enough picks this. More exposed plates give a better estimate of the result with chemical X but no measure of how many mutants arise without it.
4. DOther plates of yeast exposed to chemical X and UV light together
   A student who thinks changing more factors gives a clearer test picks this. UV light also causes mutations, so any extra mutant colonies could come from either agent.

CED 6.7.B.1 · Read this in Fix

1. AA 10 s exposure had no effect on the frequency of mutants, as its bars overlap those for 0 s
   A student who thinks overlapping error bars prove that two means are the same picks this. Overlap means a difference has not been shown, not that there is none; a small effect of 10 s is possible.
2. BEvery exposure increased the frequency of mutants, as each mean is above the 0 s mean
   A student who treats any difference between means as real picks this. The 10 s mean (6) is above the 0 s mean (4), but the error bars overlap, so the difference could be due to chance.
3. CUV caused mutations in genes that help yeast survive UV, as the cells needed them
   A student who thinks mutations arise because organisms need them picks this. UV damages DNA at random; the data show how many mutants arose, not that the mutations were in genes that help survival.
4. DExposures of 20 s and 30 s raised mutant frequency, as their bars are clear of the 0 s bar Correct
   The ±2 SE bars for 20 s (12–18) and 30 s (22–32) do not overlap the bar for 0 s (2.5–5.5), so these increases are likely real. The bar for 10 s (4–8) overlaps the bar for 0 s, so these data do not show an effect of 10 s, although they do not rule one out; the claim is therefore limited to 20 s and 30 s.

Working No test statistic is calculated; the decision rests on the ±2 SE error bars. 0 s: mean 4, bar 2.5–5.5. 10 s: mean 6, bar 4–8. 20 s: mean 15, bar 12–18. 30 s: mean 27, bar 22–32. 10 s vs 0 s: the bars overlap (4–5.5 is common to both), so a difference has not been shown, though one is not ruled out. 20 s vs 0 s and 30 s vs 0 s: the bars do not overlap (12 > 5.5; 22 > 5.5), so these increases are likely real. Conclusion: 20 s and 30 s of UV increased the frequency of mutants; an effect of 10 s is not shown by these data.

CED 6.7.B.1 · Read this in Fix

1. AResistant cells become more common, as resistance benefits cells in every environment
   A student who thinks a mutation is beneficial in itself picks this. Whether a mutation is beneficial depends on the environment; here, with no antibiotic, it slows division.
2. BResistant cells lose the mutation, as the resistance it gives is not needed here
   A student who thinks a trait that is not needed is lost by the individuals that have it picks this. Each cell keeps its DNA; the proportion of resistant cells falls because they reproduce more slowly.
3. CResistant cells become less common, as the mutation is detrimental where Y is absent Correct
   Without antibiotic Y, resistance gives no advantage, and resistant cells divide more slowly, so they leave fewer descendants than nonresistant cells. In this environment, the mutation is detrimental, and the proportion of resistant cells falls over generations.
4. DThe proportion of resistant cells stays the same, as selection acts only when Y is present
   A student who thinks selection happens only when something kills organisms picks this. Slower division is enough: nonresistant cells leave more descendants, so the population changes.

CED 6.7.B.1.i · Read this in Fix

1. ABoth errors can be passed to the adult's children, as each changes the adult's chromosomes
   A student who thinks changes in body cells are passed to offspring picks this. Only cells that give rise to gametes pass changes to children; skin cells do not.
2. BThe error in mitosis gives every cell of the adult's body an abnormal chromosome number
   A student who thinks an error in one cell changes the whole organism picks this. The error is passed only to the descendants of that one skin cell.
3. CNeither error can affect a child, as fertilization restores the normal chromosome number
   A student who thinks fertilization always restores the normal number picks this. A sperm with an extra or missing chromosome gives a zygote with an abnormal number; fertilization adds the gametes' chromosomes, it does not correct them.
4. DOnly the error in meiosis can give a child an abnormal chromosome number in every cell Correct
   Nondisjunction in meiosis can produce a sperm with an extra or missing chromosome; a zygote formed from it has the abnormal number, and every cell of the child inherits it by mitosis. The skin-cell error affects only that cell's descendants in the adult's skin and is not passed to children.

CED 6.7.B.2 · Read this in Fix

1. AThree copies of chromosome 1 and two copies of chromosome 2 Correct
   Sperm X carries two copies of chromosome 1 (the pair failed to separate in meiosis I) and one copy of chromosome 2. A normal egg carries one of each, so the zygote has three copies of chromosome 1 and two of chromosome 2: an abnormal number of one chromosome (aneuploidy).
2. BThree copies of chromosome 1 and three copies of chromosome 2
   A student who thinks nondisjunction of one pair adds an extra copy of every chromosome picks this. Only the chromosome 1 pair failed to separate; sperm X has the normal single copy of chromosome 2, so the zygote has two copies of it.
3. CTwo copies of chromosome 1 and two copies of chromosome 2
   A student who thinks fertilization always restores the normal number picks this. The zygote receives two copies of chromosome 1 from sperm X and one from the egg, three in all.
4. DTwo copies of each chromosome, with altered genes on chromosome 1
   A student who thinks changes in chromosome number work by mutating genes picks this. Nondisjunction changes how many copies of chromosome 1 the zygote has; it does not change the nucleotide sequence of its genes.

CED 6.7.B.2.i · Read this in Fix

1. APlants with an extra chromosome carry mutated genes that slow their growth
   A student who thinks a change in chromosome number works by mutating genes picks this. The data show effects of an extra chromosome; an extra copy of normal genes changes the amounts of gene products without any gene being mutated.
2. BAn extra chromosome is associated with lower survival and slower development Correct
   Seedlings with an extra copy of chromosome 3 or chromosome 5 survived to flowering less often (50% and 35%, against 94%) and took longer to flower (55 and 62 days, against 41). A change in chromosome number is associated with developmental limitations.
3. CPlants with an extra chromosome do better, having more genetic material
   A student who thinks more chromosomes mean more genetic information and a more capable organism picks this. The extra chromosome carries extra copies of existing genes, and the plants with it survived less often and flowered later.
4. DThe extra chromosome arose because the seedlings needed more genes
   A student who thinks genetic changes arise because organisms need them picks this. The data show effects of the extra chromosome, not its cause; nondisjunction occurs by chance.

CED 6.7.B.2.ii · Read this in Fix

1. A0.035 per generation
   A student who divides the value at the end of the interval by the interval picks this: 0.70/20 = 0.035. The rate uses the change in the proportion, 0.70 − 0.10 = 0.60.
2. B0.030 per generation Correct
   Rate = (0.70 − 0.10)/(30 − 10) = 0.60/20 = 0.030 per generation. The antibiotic kills or stops the growth of nonresistant cells, so cells with the resistance mutation leave more descendants and the mutation becomes more common: the environment selects for it.
3. C0.020 per generation
   A student who divides the change by the later time instead of the elapsed time picks this: 0.60/30 = 0.020. The interval is 30 − 10 = 20 generations.
4. D0.300 per generation
   A student who counts the plotted intervals as the time picks this: 0.60/2 = 0.300. The two intervals between the points span 20 generations, read from the axis.

Working From the graph, the proportion is 0.10 at generation 10 and 0.70 at generation 30. Rate = change in proportion/change in time = (0.70 − 0.10)/(30 − 10) = 0.60/20 = 0.030 per generation. Distractors: dividing the later value by the interval, 0.70/20 = 0.035; dividing the change by the later time, 0.60/30 = 0.020; using the 2 plotted intervals as the time, 0.60/2 = 0.300.

CED 6.7.C.1 · Read this in Fix

1. AIts genetic variation is unchanged, as only a new mutation can add a new gene
   A student who thinks only mutation can add a new gene to a population picks this. Horizontal transfer, such as conjugation, brings in genes that arose in other cells or species.
2. BEvery cell becomes resistant at once, as transferred genes spread to all cells together
   A student who thinks all individuals in a population change together picks this. Only the few cells that received the plasmid, and their descendants, carry the gene.
3. CRecipient cells fuse with donor cells, making offspring with genes of both species
   A student who thinks conjugation is sexual reproduction picks this. In conjugation, DNA passes from donor to recipient; the cells do not fuse, and no offspring are produced by the transfer.
4. DIts genetic variation increases, as it gains a gene from another species Correct
   Conjugation transfers DNA from one cell to another. The recipient cells now carry a resistance gene that no cell in the population had before, so the population's genetic variation has increased by horizontal transfer from another species, not by a new mutation in this population.

CED 6.7.C.1.i · Read this in Fix

1. AMix the two strains in a tube without host cells; later test the viruses there for Ab and aB
   A student who thinks viruses can copy and recombine their genes outside cells picks this. Viruses replicate only inside host cells, so no recombinants would form in the tube, and the plan tests nothing about infection.
2. BInfect all of the cells with both strains together; then test the progeny for Ab and aB
   A student who thinks a comparison group is unnecessary picks this. Without cells infected by one strain alone, the plan cannot show that recombinants arise only when both strains share a cell.
3. CInfect cells with both strains and other cells with one strain each; test progeny for Ab and aB Correct
   Viruses with the new combinations Ab and aB can arise only by recombination. Comparing cells infected with both strains with cells infected with one strain each shows whether these combinations appear only when both strains are in the same cell, which is what the hypothesis predicts.
4. DInfect cells with both strains, holding temperature constant as the control; test for Ab
   A student who thinks keeping conditions constant provides a control picks this. A constant temperature makes the conditions fair but gives no comparison: there are still no single-strain infections.

CED 6.7.C.1.ii · Read this in Fix