Mutations

In strain B the fourth codon changes from GAA to GAG. Both code for glutamic acid, so the amino acid sequence is unchanged: a silent mutation.

• Strain A: Strain A's deleted base shifts the reading frame, so the chain changes after alanine and stops early.
• Correct: Strain B: Correct: GAA and GAG both code for glutamic acid.
• Strain C: Strain C's third codon becomes UGA, a stop codon, so the chain ends after two amino acids.
• Strain D: Strain D's fourth codon becomes GUA, valine instead of glutamic acid.
• Strain E: Strain E gains a codon, AAG, so its polypeptide has an extra lysine.

Deleting U7 shifts the reading frame: AUG GCA GGG AAC UAA… gives Met-Ala-Gly-Asn, then UAA, a stop codon that was not in frame before.

• Correct: Met-Ala-Gly-Asn: Correct: a frameshift changes every codon after it and here reaches a stop early.
• Met-Ala-Trp-Glu-Leu-Asn-Gly: This is the normal polypeptide; one deleted base changes the codons after it.
• Met-Ala-Glu-Leu-Asn-Gly: Removing one base does not remove one whole codon; it regroups the rest.
• Met-Ala: The frame shifts at codon 3, but the stop codon comes two codons later, not at once.

Codons are read in threes. Inserting AAG adds one lysine and leaves every later codon in its original frame. Deleting one base regroups all later bases into new codons, which here also creates an early stop.

• Correct: Three added bases add a codon and keep the frame; one lost base shifts each later codon.: Correct: a multiple of three keeps the frame.
• Added bases are cut out of the mRNA by spliceosomes, but lost bases are not put back by the cell.: Spliceosomes remove introns, recognized by sequences at their ends; they do not remove inserted bases.
• Insertions change just the DNA, while deletions also change the mRNA and the protein made from it.: Both insertions and deletions in DNA are copied into the mRNA.
• A change near the start of an mRNA has a larger effect than one farther along, whatever its size.: The two changes are close together (positions 7 and 12); position does not explain the difference.

Chemical Q raises revertants from 22 to 380 with or without the extract. Chemical P has no effect alone (24) but raises them to 410 with liver extract, so a product the liver enzymes make from P is the mutagen.

• Correct: Chemical Q causes mutations by itself; chemical P causes them after liver enzymes change it.: Correct: Q acts directly; P must be converted first.
• Chemical P causes mutations by itself; chemical Q causes them after liver enzymes change it.: This reverses the two chemicals.
• Neither chemical causes mutations, because some revertant colonies appear on each plate.: The background of about 22 is the spontaneous rate; the treated plates are 17-19 times higher.
• The liver extract causes the mutations, and the chemicals simply help the cells to grow.: The extract alone gives 25, the same as solvent alone.

Spontaneous mutations happen at a low rate in every population: errors that escape proofreading and mismatch repair, and damage from normal chemistry. Among millions of cells, a few happen to restore the gene.

• Correct: Mutations also arise with no mutagen, from rare copying errors and everyday DNA damage.: Correct: the background is the spontaneous mutation rate.
• The solvent itself is a weak mutagen, which caused a few of the spread cells to mutate.: Nothing suggests the solvent is mutagenic; a spontaneous background is expected.
• Some cells sensed the lack of histidine and changed their gene so they could grow.: Mutations are not directed by need; cells cannot choose to change a gene.
• Some cells were contaminants from the air, carrying a working histidine gene.: The plates are kept sterile and lack histidine for exactly this reason; the counts are steady across plates.

If contact with phage caused resistance, every plate would show a similar small share, as in set 2. Separate cultures vary hugely because mutations happen at random times during growth: an early one gives a jackpot of resistant offspring, a late one a few, and often none happens at all. Set 2's samples share one history, so they are similar.

• Correct: Hypothesis 2: an early mutation in one culture leaves many resistant descendants, spreading set 1's counts.: Correct: random mutations before exposure predict jackpots between independent cultures.
• Hypothesis 1: each plate's phage turned a similar share of cells resistant, which explains the even counts of set 2.: Set 2's steady counts are expected under both hypotheses; set 1's spread is what tells them apart.
• Hypothesis 1: the cultures in set 1 met different amounts of phage, so their counts differed widely from one another.: Each plate was covered with phage in the same way; only the cultures' separate histories differ.
• Hypothesis 2: the cells in set 2 grew in one culture, so they could share their resistance with each other.: Resistance came from mutations inside cells; the spread in set 1, not sharing in set 2, is the evidence.