Unit 6: Gene Expression and Regulation. Practice questions with scoring guidelines.

• It does not differ significantly: the ±2SE intervals, 90 to 230 for virus 2 and 60 to 180 for the control, overlap.
• No significant difference, because the error ranges 160 ± 70 and 120 ± 60 overlap.

• Virus 1 has a DNA genome and virus 2 an RNA genome: virus 1 particles took up thymidine (5,300 compared with 120 in the control) but not uridine (150, like the control's 140), and virus 2 particles took up uridine (4,800) but not thymidine (160, overlapping the control); thymine is found only in DNA and uracil only in RNA.
• Virus 1 uses DNA, because only thymine-containing label rose far above the control in its particles; virus 2 uses RNA, because only uracil-containing label rose far above the control in its particles.

• It measures the background radioactivity from labeled cell material that is purified along with virus particles (or from free label), so only radioactivity above the control level can be credited to nucleic acid in the virus particles.
• As a negative control: without virus, any radioactivity measured comes from the cells or the method, so comparing with it shows whether the label was really built into the viruses' nucleic acid.

• Both viruses' particles would be far more radioactive than the control, because adenine is a base in both DNA and RNA nucleotides, so it would be built into the DNA genome of virus 1 and the RNA genome of virus 2.
• Particles of both viruses would be labeled above the control: unlike thymine and uracil, adenine is found in both DNA and RNA, so this label cannot tell the two genomes apart.

• To show that any loss of infection is caused by destroying the RNA specifically, not by adding an enzyme or by the incubation itself
• To rule out a trace of contaminating DNA or protein in the extract as the carrier of the heritable information: if destroying DNA or protein has no effect, the infection depends on the RNA
• They are controls: if only the RNA carries the information, infection continues after DNase or protease treatment, so a drop after RNase treatment can be attributed to the loss of RNA

• RNase group: no (or almost no) plants with spots; DNase and protease groups: about the same number as the untreated group
• Only the RNase treatment reduces infection, to about zero like the buffer-only group, while the DNase and protease groups are infected about as often as the untreated group

• Virus V coat protein
• The coat protein of virus V, not the coat protein of virus W

• The coat proteins of the new particles are made in the infected cells from the genetic information in the RNA, and the RNA came from virus V, so it codes for V coat protein
• The W coat protein that entered the plant carries no heritable information and is not passed on; only the V RNA is copied and passed to the new particles, so its genes determine their coat
• The RNA is the hereditary material, so the new particles have the traits (coat protein) encoded by the V RNA rather than those of the protein coat it arrived in

• In mutant worms gene D mRNA falls from 72 on day 2 to 12 on day 4 and 3 on day 8, whereas in wild-type worms it stays high (95 to 100)
• It drops to almost nothing in the mutant after day 2 but stays at about 100 in wild type

• Deleting gene D would remove factor D altogether, so none of its target genes, such as gene E, would be switched on even while V is present; changing only the binding site lets factor D still be made in response to V, so only its effect on its own gene is tested
• To test only factor D's control of its own gene: with the binding site changed, factor D is still made (day 1 values match wild type) and can still act on other genes, so any later difference is caused by losing factor D binding at gene D

• Factor D binds the regulatory DNA of its own gene and keeps gene D switched on: in the mutant, where factor D cannot bind there, gene D mRNA falls to 12 by day 4 and 3 by day 8 after V is gone, while in wild type it stays at about 100
• Gene D activates itself (positive feedback) once V has started it: both strains make the same gene D mRNA on day 1 (70 and 68, overlapping intervals), but only the wild type, with an intact binding site, keeps it high after day 2

• High, similar to the day 2 mutant level (about 66) or above, because V keeps gene D transcribed, so factor D is still made and keeps switching on gene E
• Gene E mRNA would stay high rather than falling to 4, because V alone can drive transcription of gene D, supplying the factor D that gene E needs; the changed binding site matters only once V is gone

• The substitution changes one codon in the mRNA transcribed from the gene, so during translation a tRNA carrying a different amino acid pairs with that codon (a missense change)
• The altered DNA triplet is transcribed into an altered codon that specifies a different amino acid, so one amino acid in the polypeptide changes while the rest of the sequence stays the same

• The new amino acid changes the shape (or chemical properties) of the part of enzyme V where W binds, so W binds poorly and does not block the enzyme, while the rest of the enzyme still works and converts clotting proteins
• The altered enzyme V is not blocked by W because the changed amino acid alters the site W binds to, so in the liver cells the enzyme goes on converting clotting proteins to their working form

• The resistant rats will clot more slowly (bleed more): with little vitamin K their form of enzyme V works more slowly than the normal form, so their liver cells release fewer working clotting proteins into the blood
• Clotting will be worse in the resistant rats, because the resistant enzyme converts fewer clotting proteins when vitamin K is scarce, so less working clotting protein reaches their blood

• It will decrease: without W the substitution gives no advantage, and on a low-vitamin-K diet rats carrying it clot poorly and are more likely to die from bleeding before reproducing, so they pass the substitution to fewer offspring
• It will fall, because in this environment the substitution is harmful: rats carrying it survive and reproduce less than rats without it, so a smaller proportion of each generation inherits it

Beyond this unit. Award if given; the point can be earned without it.

• U7 It will decrease, because natural selection now acts against the allele: carriers have lower fitness than noncarriers when W is absent and vitamin K is scarce, so the allele frequency falls