Chapter 8 Practice Problems Flashcards
existences of an intermediate messenger between DNA and protein
protein synthesis occurs in the cytoplasm, while DNA resides in the nucleus
the genetic code is nonoverlapping
single base substitutions affect only one amino acid in the protein chain
the codon is more than one nucleotide
two mutations affecting the same Amino acid can recombine to give wild type
the genetic code is based on triplets of bases
one or two base deletions (or insertions) in a gene disrupt its function three base deletions (or insertions) are often compatible with functions
stop codons exist and terminate translation
artificial messages containing certain codons produce shorter proteins than messages not containing those codons
the amino acid sequence of a protein depends on the base sequence of an mRNA
artificial messages with different base sequences gave rise to different proteins in an in vitro translation system
a group of three mRNA bases signifying one amino acid
codon
the linear sequence of amino acids in the polypeptide corresponds to the linear sequence of nucleotide pairs in the gene
colinearity
the grouping of mRNA bases in threes to be read as codons
reading frame
addition or deletion of a number of base pairs other than three into the coding sequence
frameshift mutation
most amino acids are not specified by a single codon
degeneracy of the genetic code
UAA, UGA, or UAG
nonsense codon
AUG in a specific context
initiations codon
the strand of DNA having the base sequence complementary to that of the primary transcript
template strand
the strand of DNA that has the same base sequence as the primary transcript
intron
removing base sequences corresponding to introns from the primary transcript
RNA splicing
using information in the nucleotide sequence of a strand of DNA to specify the nucleotide sequence of a strand of RNA
transcription
using the information encoded in the nucleotide sequence of an mRNA molecule to specify the amino acid sequence of a polypeptide molecule
translation
produces different mature mRNAs from the same primary transcript
alternative splicing
a transfer RNA molecule to which the appropriate amino acid has been attached
charged tRNA
copying RNA into DNA
reverse transcription
How would the artificial mRNA 5’…GUGUGUGU…3’ be read according to each of the following models for the genetic code?
a. two-base, not overlapping
b. two base, overlapping
c. three base, not overlapping
d. three base, overlapping
e. four base, not overlapping
a. … GU GU GU GU GU… or …UG UG UG UG UG….
b. …GU UG GU UG GU UG GU UG GU…
c. …GUG UGU GUG UGU GUG…
d. …GUG UGU GUG UGU GUG UGU GUG UGU…(overlapping gives more coding information)
e. …GUGU GUGU… or …UGUG UGUG…
An example of a portion of the T4 rIIB gene in which Crick and Brenner had recombined one + and one - mutation is shown here
wild type 5’ AAA AGT CCA TCA CTT AAT GCC 3’
mutant 5’ AAA GTC CAT CAC TTA ATG GCC 3’
a. where are the + and - mutations in the mutant DNA?
b. the double mutant produces wild-type plaques. What alterations in amino acids occurred in this double mutant/
c. how can you explain the fact that amino acid are different in the double mutant than in the wild type sequence, yet the phage has a wild type phenotype?
a.the 4th A from the 5′ end of the wild-type sequence is deleted (−) in the mutant, and a G is inserted (+) just upstream of the GCC at the 3′ end of the wild-type sequence
b. The amino acids in the wild-type and mutant proteins are shown below:
wild type: Lys Ser Pro Ser Leu Asn Ala
mutant: lys val his leu met ala
The four red amino acids between the – and + mutations other than Leu are different from wild-type.
c. Those four amino acids must not be in a crucial functional domain of the protein (for example, they do not carry out catalysis), nor do they alter the overall structure of the protein significantly enough to affect its function.
Consider Crick and Brenner’s experiments which showed that the genetic code is based on nucleotide triplets
a. Crick and Brenner obtained FC7, an intragenic suppressor of FC0 that was a mutation in a second site in the rIIB gene near the FC0 mutation. Describe a different kind of mutation in the rIIB gene these researchers might have recovered by treating the FC0 mutant with proflavin and looking for restored rIIB+ function
b. how could Crick and Brenner tell the difference between the occurrence described in part a and an intragenic suppressor like FC7/
c. when FC& was separated from FC0 by recombination, the result was two rIIB- mutant phages: one was FC7 and the other was FC0. How could they discriminate between the rIIB- recombinants that were FC7 and those that were FC0?
d. explain how Crick and Brenner could obtain different deletion or addition mutations so as to make the various combinations such as ++,–,+++, and —
a. A reversion of FC0 back to the wild-type sequence (deletion of the inserted base) could also have restored rIIB+ function.
b. A true revertant cannot produce rIIB− progeny through recombination with wild-type rIIB+ phage because the true revertant and the wild-type have the same base pair sequence.
c. Only FC7 mutants can recombine with FC0 to produce rIIB+ phage.
d. To obtain many deletion mutations, several different intragenic suppressors of FC0 could have been obtained, including FC7; all of them are single-base deletions like FC7. Each would have been separated from FC0 by recombination with a true wild-type (rIIB+) strain Multiple addition mutations (like FC0) could have been obtained by isolating suppressors of deletion mutants like FC7, recombining them away from FC7,
The HbB^S (sickle cell) allele of the human B-globin gene changes the sixth amino acid in the B-globin chain from glutamic acid to valine. In HbB^C, the sixth amino acid in B-globin is changed from glutamic acid to lysine. What would be the other of these two mutations within the map of the B-globin gene?
Glutamic acid (in the wild-type protein) can be encoded by either GAA or GAG. In sickle cell anemia, this amino acid is changed to valine by a single base change. Valine is encoded by GUN with N representing any of the four bases. Therefore, the second base of the triplet was altered from A to U in the HbβS allele. In HbβC the glutamic acid codon (GAA or GAG) is changed to a lysine (AAA or AAG). The change here is in the first base of the codon. The mutation causing the HbβC allele therefore precedes the HbβS mutation in the sequence of the β-globin gene when reading the RNA-like strand in the 5′-to-3′ direction. (Stated another way, the HbβC mutation is upstream of the HbβS mutation by one nucleotide pair.)
The amino acid sequence of part o f a protein has been determined: N…Gly Ala Pro Arg Lys…C
a mutation has been induced in the gene encoding this protein using the mutagen proflavin. The resulting mutant protein can be purified and its amino acid sequence determined. The amino acid sequence of the wild type protein from the N terminus of the protein to the glycine in the preceding sequence. Starting with this glycine, the sequence of amino acids is changed to the following: N…Gly His Gln Gly Lys…C. Using the amino acid sequences, one can determine the sequence of 14 nucleotides from the wild type gene encoding this protein. what is this sequence?
wild type mRNA: GGN GCN CCN AGA/G AAA/G
mutant mRNA; GGN CAU/C CAA/C GGN AAA/G
A high concentration of Mg2+ ions in the test tube, much higher than that found in cells, allows ribosomes to initiate translation at any position on an RNA molecule. Predict the outcomes of in vitro translation with each of the following synthetic mRNAs at both high and low Mg2+ concentrations:
a. poly-UG (UGUGUG…)
b. poly-CAUG (CAUGCAUGCAUG..)
c. poly-GUAU (GUAUGUAUGUAUGUAU..)
a. polypeptides will be produced only at high Mg2+ concentrations. The codons in the RNA will be read as …UGU GUG UGU GUG… and so polypeptides with alternating Cys and Val amino acids will be made.
b. At low Mg2+ concentrations, translation will start with AUG, and the RNA will be read as AUG CAU GCA UGC AUG …; polypeptides with repeats of Met-His-Ala-Cys will be synthesized. At high Mg2+ concentrations, because translation can initiate anywhere on the RNA, the same polypeptides will be produced except they will not all begin with Met. (RNAs not read starting with AUG are read as UGC AUG CAU GCA… or GCA UGC AUG CUA…)
c. At low Mg2+ concentrations, translation will start with AUG: AUG UAU GUA UGU… and thus polypeptides with the repeat Met-Tyr-Val-Cys will be made. At high Mg2+ concentrations, translation can start anywhere on the RNA. The result is only polypeptides containing Met, Tyr, Val, and Cys in that order, but they will not all begin with Met. (RNAs that are not read starting with AUG are read as UGU AUG UAU GUA… or GUA UGU AUG UAU…)
Identify all the amino acid specifying codons in the genetic code where a point mutation (a single base change) could generate a nonsense codon.
UGA = stop; so UCA, UUA UAG = Stop; so UAU, UAC UAA = stop; so, UCA, UUA, UAC, UAU
RNAs consisting only of Us and Gs could be synthesized in vitro, but they would have random sequences. Suppose a pool of random sequence RNA has synthesized in a reaction mixture containing three times as much UTP as GTP, and that the resulting RNAs were translated in vitro
a. how many different codons exist in the RNAs
b. how many different amino acids would you find in the polypeptides synthesized?
c. why are your answers to a and b not the same?
d. how often would you expect to find each of the codons in a?
e. in what proportions would you expect to find each of the amino acids in the polypeptides?
f. if you did this experiment-that is, synthesized random sequence RNAs containing 3:1 ration of U:G and quantified the amount of each amino acid in the polypeptides produced - prior to knowledge of the genetic code table, what would the results have told you?
a. Eight codons exist in RNAs made up of Us and Gs in random order: UUU, UUG, UGU, GUU UGG, GUG GGU, GGG.
b. The eight codons in part (a) correspond to six different amino acids: Phe, Leu, Val, Cys, Trp, and Gly.
c. The answers to part (a) and (b) are different because the genetic code is degenerate in that the same amino acid is specified by several different codons: UUU = Phe, UUG = Leu, UGU = Cys, GUU or GUG = Val, UGG = Trp, GGU and GGG = Gly
d. The frequency of U is 3/4, and the frequency of G is 1/4. Therefore: f(UUU) = (3/4)3 = 27/64; f(UUG) = f(UGU) = f(GUU) = 3/4 × 3/4 × 1/4 = 9/64; f(UGG) = f(GUG) = f(GGU) = 3/4 × 1/4 × 1/4 = 3/64; f(GGG) = (1/4) 3 = 1/64.
e. Phe (UUU) = 27/64; Leu (UUG) = 9/64; Cys (UGU) = 9/64; Val (GUU or GUG) = 9/64 + 3/64 = 12/64; Trp (UGG) = 3/64; Gly (GGU or GGG) = 3/64 + 1/64 = 4/64
f. By doing this experiment, you would have been able to determine that UUU specifies Phe; that Val, Leu, and Cys are specified by codons composed of G+2U; and that Trp and Gly are specified by codons composed of 2G+U.
How many possible open reading frames exist that extend through the following sequence?
5’…CTTACAGTTTATTGATACGGAGAAGG…3’
3’…GAATGTCAAATAACTATGCCTCTTCC…5’
3 reading frames each sequence
Charles Yanofsky isolated many differed trap- mutants.
a. explain how he could identify Trp- auxotrophs of E. coli using replica planting
b. assuming that the role of TrpA enzyme in the tryptophan biosynthesis pathway was known, explain how Yanofsky could have identified trap-mutants among his Top- auxotrophs.
a. Yanofsky could have screened for Trp− auxotrophs by growing wild-type bacteria in the presence of a mutagen, and then plating the culture to obtain single colonies on a plate containing minimal medium supplemented with tryptophan. Each colony on the plate would then be replica-plated onto a minimal medium plate; colonies that fail to grow on the replica plate are Trp− auxotrophs.
b. trpA− colonies should be able to grow only on minimal medium containing the end product tryptophan or any intermediate downstream of TrpA action (that is, an intermediate whose conversion into tryptophan does not depend on TrpA). The trpA− cells should not grow on minimal medium supplemented with any upstream intermediate.
you identified a proflavin generated allele of a gene that produces a 110-amino acid polypeptide rather than the usual 157 amino acid protein. After subjecting this mutant allele to extensive proflavin mutagenesis, you are able to find a number of intragenic suppressors located in the part of the gene between the sequences encoding the N terminus of the protein and the original mutation, but no suppressors located in the region between the original mutation and the sequences encoding the usual C terminus of the preotein. why do you think this is the case?
the initial frameshift mutation changes the reading frame and there is a stop codon in this alternate reading frame. The second frameshift mutation of the opposite sign cannot restore function if it occurs after this stop codon. Because no suppressors can be identified between the original mutation and the premature stop codon, we can predict that the stop codon is likely to be very close to the site of the original frameshift mutation
a. can a tRNA exist that has the anticodon sequence 5’ IAA?if so which amino acid would it carry?
b. answer the same question for the anticodon 5’ xm^5s^2UAA
c. consider the DNA sequence of the gene encoding the tRNA. What is the sequence of the RNA like strand of the tRNA gene corresponding to the tRNA’s anticodon? what is the sequence of the template strand of the gene for these same three nucleotides?
a. no; a tRNA with the anticodon 5′ IAA cannot exist in an organism obeying the normal wobble rules shown in Figure 8.21b. Such an anticodon would recognize codons 5′ UUU (Phe), 5′ UUC (Phe), and 5′ UUA (Leu). A single tRNA species must carry only a single amino acid, and so the genetic code table tells you that a tRNA with the anticodon 5′ IAA cannot exist because it would not correspond to a single amino acid.
b. Yes; a tRNA with the anticodon 5′ xm5s2U would recognize the codons 5′ UUA and 5′ UUG, both of which specify Leu. Therefore, this tRNA could exist and it would carry the amino acid Leu.
c. For part (a), the RNA-like strand of the part of the tRNA gene encoding the anticodon is 5′ AAA; remember that A in the 5′ position of the anticodon is enzymatically altered to I after transcription. The template strand sequence is 5′ TTT.
For part (b), the RNA-like strand is 5′ TAA, and the template strand is 5′ TTA. Remember that U in the 5′ position of the anticodon is enzymatically altered after transcription, in different ways in different tRNAs.
remembering that the wobble base of the tRNA is the 5’ base of the anticodon:
a. in human tRNAs, what are the sequences of all possible anticodons that were originally transcribed with A in the wobble position? (assume A is always modified to I)
b. In human tRNAs, what are the sequences of all possible anticodons that were originally transcribed with U in the wobble position (any single type of tRNA with a U at the wobble position can be modified only in a single way)
c. how might the wobble Us in each of the anticodons in b be modified and still be consistent?
d. what is the theoretical minimal number of different tRNA genes that must exist in the human genome? (assume that xo^5U pairs with A, G or U only)
a. the anticodons that can have I as the wobble base are: 5′ ICG, 5′ ICC, 5′ IGG, 5′ IGA, 5′ IGU, 5′ IGC, 5′ IAG, 5′ IAC, and 5′ IAU.
b. the only anticodon seqence that could never be allowed to have a 5’ U is 5’ UUA
c. xm^5U can be the wobble base for all anticodons except for 5′ UUA. anticodons that would be able to have a xo5U as the wobble base are: 5′ UAG, 5′ UAC, 5′ UGA, 5′ UGG, 5′ UGU, 5′ UGC, 5′ UCG, and 5′ UCC.
d. the minimal number of tRNAs required in humans is 32.
the human genome contains about 500 genes for tRNAs
a. do you think that each one of these tRNA genes has a different function?
b. can you explain why the human genome might have evolved so as to house so many different tRNA genes/
a. many of the 500 human tRNA genes are functionally redundant
b. having many different genes that encode the same tRNA is one way to produce the large quantities of tRNA molecules that cells need. Also, multiple genes for each tRNA molecules protect the organism from the effects of mutation of tRNA genes
the yeast gene encoding a protein found in the mitotic single was cloned by a laboratory studying mitosis. the gene encodes a protein of 477 amino acids
a. what is the minimum length in nucleotides of the protein coding part of this yeast gene?
b. a partial sequence of one DNA strand in an exon containing the middle of the coding region of the yeast gene is given here what is the sequence of nucleotides of the mRNA in this region of the gene? she the 5’ and 3’ directionality of your strand
5’ GTAAGTTAACTTTCGACTAGTCCAGGT 3’
c. what is the sequence of amino acids in this part of the yeast mitotic spindle
a. the minimum length of the coding region is 477 Amino acids x 3 bases/codon = 1431 base pairs
b. XXX
template strand 5′ GTAAGTTAACTTTCGACTAGTCCAGGGT 3′
RNA-like strand 3′ CATTCAATTGAAAGCTGATCAGGTCCCA 5′
XOX
mRNA 5′ ACCCUGGACUAGUCGAAAGUUAACUUAC 3′ c. N…Pro Trp Thr Ser Arg Lys Leu Thr Tyr...C
The sequence of a complete eukaryotic gene encoding the small protein Met Tyr Gly Ala is shown here. All of the written sequences on the template strand are transcribed into RNA.
5’
CCCCTATGCCCCCCTGGGGGGAGGATCAAAACACTTACCTGACATGGC
3’
3’
GGGGATACGGGGGGACCCCCCTCCTAGTTTTGTAATGGACATGTACCG
5’
a. which strand is the template strand? In which direction does RNA polymerase move along the template as it transcribes this gene?
b. what is the sequence of the nucleotides in the processes mRNA molecule for this gene? Indicate the 5’ and 3’ polarity of this mRNA
c. A single base mutation in the gene results in synthesis of the peptide Met Tyr Thr. what is the sequence of nucleotides making up the mRNA produced by this mutant gene?
a. The bottom strand is the RNA-like strand, so the top strand is the template. The RNA polymerase moves 3′-to-5′ along the template; that is, in the right-to-left direction with respect to the written sequence.
b.5′ GCC AUG UAC AG|G GGG GCA UAG GGG 3′
c. Thr at the C terminus of the truncated protein could occur if the G base on the bottom strand that just precedes the junction between the intron and the first exon The sequence of the mutant mRNA is:
5′ GCC AUG UAC ACG UAA GUG UUU UGA UCC UCC CCC AGG GGG GCA UAG GGG 3′
Do you think each of the following types of mutations would have very severe effects, mild effects, or no effect at all?
a. nonsense mutations occurring in the sequences encoding amino acids near the N terminus of the protein
b. nonsense mutations occurring in the sequences encoding amino acids near the C terminus
c. frameshift mutations occurring in the sequences encoding amino acids near the N terminus of the protein
d. frameshift mutations occurring in the sequences encoding Amino acids near the C terminus of the protein
e. silent mutations
f. conservative missense mutations
g. nonconservative missense mutations affecting the active sites of the protein
h. nonconservative missense mutations not in the active site of the protein
a. very severe effect as there will be no functional protein
b. probably mild effect if none of the last few amino acids are important for function
c. very severe effect as most of the amino acids in the protein will be incorrect
d. probably mild effect as only the last few amino acids will be affected
e. no effect because by definition, a silent mutation maintains the same amino acid
f. mild to no effect as the replacement is with an amino acid with similar chemical properties
g. severe effect as the mutation is likely to destroy the protein′s activity
h. could be severe if it affects the protein structure enough to hinder the action of the protein, or could be mild if the protein′s function can tolerate the substitution
A mutant B. adonis bacterium has a nonsense suppressor tRNA that inserts glutamine to match UAG codons.
a. what is the anticodon of the suppressing tRNA? Indicate the 5’ and 3’ ends
b. what is the sequence of the template strand of the wild type tRNA^Gln encoding gene that was altered to produce the suppressor, assuming that only a single base pair alteration was involved?
c. what is the minimum number of tRNA^Gln genes that could be present in a wild type B. adonis cell? Describe the corresponding anticodons
a. The anticodon in this nonsense suppressor tRNA is either 5′ CUA 3′ or 5′ xm5UUA 3
b. the sequence of the template strand of DNA for the wild-type tRNAGln gene is either 5′ CAG 3′ or 5′ CAA 3′.
c. at least two tRNAGln genes must exist in a wild-type B. adonis cell. One would code for the tRNAGln (anticodon 5′ CUG 3′ or 5′ xm5UUG 3′) that was changed into a nonsense suppressor, and the other gene would code for the tRNAGln (anticodon of 5′ xm5s2UUG 3′) that recognizes both normal Gln codons.
You are studying mutations in a bacterial gene that codes for an enzyme whose amino acid sequence is known. In the wild type protein, proline is the fifth amino acid from the amino terminal end. In one of your mutants with nonfunctional enzyme, you find a serine at position number 5. You subject this mutant to further mutagenises and recover three different strains. Strain A has a proline at position number 5and acts just like a wild type strain. strain B has tryptophan at position number 5 and acts like a wild type. Strain c has no detectable enzyme function at any temp, and you can’t recover any protein that resembles the enzyme. You mutagenize strain C and recover a strain C-1 that has enzyme function. The second mutation in C1 that is responsible for the recovery of enzyme function does not map at the enzyme locus.
a. what is the nucleotide sequence in both strand of the wild type gene at this location
b. why does strain B have a wild type phenotype? why does the original mutant with serine at position 5 lack function?
c. what is the nature of the mutation in strain C?
d. what is the second mutation that arose in C1?
a. the original wild-type proline codon must have been 5′ CCG, so the sequence of the DNA in this region must have been: 5′ CCG 3′ 3′ GGC 5′
b. Trp at position 5 is compatible with the function of the enzyme encoded by the gene Ser at position 5 is not compatible with enzyme function - it is a nonconservative substitution
c. Strain C does not have any detectable protein, so it is likely to be a nonsense mutation that stops translation after only a few amino acids have been added.
d. Because the reversion mutation does not map to the enzyme locus, it must be a mutation in a tRNA gene that would encode a nonsense suppressing tRNA.
At least one nonsense suppressing tRNA is known that can suppress more than one type of nonsense codon.
a. what is the anticodon of such a suppressing tRNA?
b. what stop codons would it suppress
c. could this tRNA possibly also function as a missense suppressor
d. what are the amino acids most likely to be carried by this suppressing tRNA?
a. . A tRNA with the anticodon 5′ UUA 3′ (U modified to xm5s2U) could recognize both nonsense codons
b. This nonsense suppressing tRNA would suppress 5′ UAA 3′ and 5′ UAG 3′.
c. no
d. Gln, Glu, Lys, Leu, Ser, Tyr, and possibly Trp.
Brenner’s m mutant phages (m^1-M^6) were suppressed when grown in suppressor (su-) mutant bacteria; they produced full length M proteins that functioned like wild type M protein
a. what gene do you think was mutant in the su- bacteria
b. when the m- phages were prorated in the su- bacterial strain, not all of the proteins made by the mutant am alleles were identical to wild type M protein. How did some of them differ
a. The mutant gene was a tRNA whose anticodon could recognize the stop codon 5′ UAG 3′.
b. Each M protein had a single amino acid change in its polypeptide sequence; the amino acid with which the nonsense suppressor tRNA was charged was inserted in place of the amino acid specified by the normal codon that was changed to UAG by the nonsense mutation
