Chapter 14

1. A model organism is easy to grow and manipulate in the laboratory and has representative characteristic(s) common to a larger group of organisms (e.g., a mouse for mammals).

2. Srb, Horowitz, Beadle, and Tatum treated wild-type Neurospora with X rays to cause mutations. Strains that could not grow without the addition of an amino acid (arginine) were isolated. The strains were separately incubated without arginine but with intermediates along the proposed biochemical pathway that makes arginine. Growth in one of the intermediates meant that the mutation for that strain must occur in a gene encoding an enzyme that acts before that substance in the pathway. In this way, the collection of mutant strains led to identification of the intermediates in the pathway and their order, and ultimately to the enzyme that acts to convert each intermediate into the next.

   A-15

3. Enzymes: 4 → 2 → 3 → 1 → 5

   Compounds: C → F → E → D → G → T

1. The central dogma states that DNA is transcribed to RNA, which gets translated into protein, a process that is unidirectional.

2. In retroviruses, the genome is RNA. To replicate, the RNA is converted to a DNA copy, which is then transcribed into RNA. This violated the original central dogma in that the DNA-to-RNA conversion was thought to be unidirectional.

1. If the code were just single letters (A or T or G or C), each “letter” would translate to one amino acid (41). But there are 20 amino acids, so a single-letter code could not unambiguously identify all amino acids. A doublet code would create 16 possible codons (42), not enough to uniquely identify all 20 amino acids. A triplet code using four letters has 64 unique possibilities (43), more than enough for 20 amino acids and stop codons. The code is redundant (there is more than one codon per amino acid) but not ambiguous (a codon does not stand for more than one amino acid).

2. RNA polymerase binds to DNA at the promoter. The DNA is unwound to expose the bases. The enzyme has binding sites for substrates, the ribonucleoside triphosphates. The enzyme then adds nucleotides to a growing chain by complementary base pairing to template DNA.

3. DNA must be replicated exactly. Any error in DNA in a gene that encodes a protein will result in an error in the RNA that is transcribed from that DNA region. This could result in a different codon and therefore a different amino acid at that location in the protein. The protein’s function may change. RNA is made in many copies. So an error in an RNA could result in an error in the protein translated from it, but since there are many more normal copies of that RNA, there would be plenty of the normal protein for normal function.

1. At the 5′ end, a “cap” of modified GTP is added. This facilitates binding of mRNA to the ribosome and protects the mRNA from hydrolysis by ribonucleases. At the 3′ end a poly A tail is added, with 100–300 A nucleotides. This assists in export of mRNA from the nucleus.

2. snRNP particles bind to consensus sequences at the 5′ and 3′ splice sites on pre-mRNA in the nucleus. The two ends approach each other, mediated by the snRNPs. mRNA is cut first at the 5′ end and then at the 3′ end, releasing the intron. The two exons are then joined.

3. 192 amino acids would be encoded by 576 nucleotides. Add start and stop codons for a total of 582 base pairs in DNA. The DNA has extra base pairs for the promoter and terminator of transcription, introns, and a sequence for mRNA binding to the ribosome.

1. rRNA acts as a scaffold for proteins to make the ribosome structure, with binding sites for tRNA. An rRNA has a nucleotide sequence region complementary to a region on mRNA so the two RNAs can bind and begin translation. An rRNA acts as the catalyst for peptide bond formation.

2. 1. 3′-TACGGGCCCAATTCTTAAAATTTTACT-5′

   2. The bottom strand is transcribed: It has sequences that are transcribed into start (AUG) and stop (UGA) codons in RNA

   3. mRNA: AUGCCCGGGUUAAGAUAUUUUAAAUGA
      Polypeptide: Met-Pro-Gly-Leu-Arg-Tyr-Phe-Lys

3. A polysome is formed when more than one ribosome is bound to mRNA at the same time. This can occur because ribosomes move along mRNA is a 5′-to-3′ direction, translating as they go, much like a cafeteria line. Polysomes allow more proteins to be made at a given time from an mRNA.

1. Signal sequences are translated to regions in a protein that bind to recognition molecules and/or receptors associated with a particular destination in the cell.

2. Post-translational modifications are made to specific amino acids at particular locations on a protein. The locations are determined by the three-dimensional shape of the protein, allowing an enzyme for the modification to bind to the amino acid at that location. So overall, the amino acid sequence of the protein determines which amino acids will be modified posttranslationally, and the amino acid sequence is determined by the genetic code and DNA sequence for the gene.

1. 34105: gene a  33442: gene b  36703: gene c.

2. The mutant strains may have had some residual enzyme activity, allowing for growth.

3. Arginine is part of proteins and is essential for the tertiary structure of proteins. Replacing it with other amino acids is not possible.

4. The double mutant cannot do the reactions ornithine → citrulline and citrulline → arginine. So adding ornithine or citrulline does not allow growth, since arginine is not produced.

1. Minus poly U mRNA: Charged tRNA does not bind to the codon at the ribosome, so no polypeptide can be made.

   Minus ribosomes: There are no locations for adjacent charged tRNAs to bind and also no peptidyl synthetase to catalyze peptide bond formation.

   Minus ATP: tRNA cannot get charged with amino acids, and mRNA cannot translocate along the ribosome. No polypeptides are made.

   Plus RNase: mRNA is destroyed. See Minus poly U mRNA, above.

   Plus DNase: All the components for polypeptide synthesis are present.

   Radioactive glycine instead of phenylalanine: Charged glycine tRNA binds to mRNA codon GGU, GGA, GGG, or GGC. These codons are not present in the poly U mRNA, so no radioactive polypeptides are made.

   Mixture of 19 radioactive amino acids minus phenylalanine: Charged tRNAs are made, but their codons are not present, so no polypeptides are made.

2. The data show that RNA (poly U) was essential for protein synthesis, since its absence resulted in no protein synthesis.

3. According to the genetic code (see Figure 14.5), UUU is the codon for phenylalanine. The other amino acids were not added to protein in the poly U experiment because only the UUU codon was present.

1. Leu-1 is a mutation in the protein catalyzing step C. The product of this step (α-ketoisocaproate) is the first in the pathway able to overcome blockage in this mutant; therefore, Leu-1 must be a mutation in the enzyme at step C. Leu-2 is a mutation in the protein catalyzing step B. The product of this step (3-isopropylmalate) is the first in the pathway able to overcome blockage in this mutant; therefore, Leu-2 must be a mutation in the enzyme at step B.

2. The mutant cells carrying the Leu-1 mutation are deficient in enzyme C but carry the wild type enzyme B. The mutant cells carrying the Leu-2 mutation are deficient in enzyme B but carry the wild type enzyme C. Therefore, when the two cells fuse, the diploid cell contains one copy of wild type enzyme B and one copy of wild type enzyme C, which restores the wild type phenotype.

3. The table summarizes the predicted enzyme activities:

   	Predicted enzyme activity
   Enzyme	Leu-1 (haploid)	Leu-2 (haploid)	Fused cells (diploid) Leu-1, Leu-2
   A	Wild type	Wild type	Wild type
   B	Wild type	None	Wild type
   C	None	Wild type	Wild type
   D	Wild type	Wild type	Wild type

4. One mutant strain is affected in the gene encoding the α subunit and the other mutant strain is affected in the gene encoding the β subunit. Thus, haploid cells cannot produce a functioning α₂β₂ enzyme because these cells have only one copy of each gene, one of which is normal and one of which is not. Diploid cells, however, have two copies of each gene with one normal gene for α supplied by one of the mating pair and one normal gene for β, supplied by the other of the mating pair. The presence of a copy of a normal gene for each subunit allows the diploid cells to produce normal polypeptide chains of α and β, which associate to form a functional α₂β₂ enzyme.