Problem set 11

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Problem set 11
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2013-06-15 13:45:33
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14 15
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14.3-14.4 ch 15 page 405-407, 410-421
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  1. Suppose that a mutation occurs in an intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein?
    Explain your answer.
    Because introns are removed prior to translation, an intron mutation would have little effect on a protein’s amino acid sequence unless the mutation occurred within the 5′ splice site, the 3′ splice site, or the branch point. If mutations within these sequences altered splicing, then the mature mRNA would be altered, thus altering the amino acid sequence of the protein. The result could be a protein with additional amino acid sequence. Or, possibly, the altered splicing could introduce a stop codon that stops translation prematurely. If a mutation in the intron induced a frameshift, the reading frame and the amino acid sequence would be altered.
  2. If there were five different types of bases in mRNA instead of four, what would be the minimum codon size (number of nucleotides) required to specify the following numbers of
    different amino acid types:
    (a) 4, (b) 20, (c) 30?
    -determine the number of combinations (codons) possible when there are different numbers of bases and different codon lengths. In general, the number of different codons possible will be equal to: 

    b^lg = number of codons 

    b equals the number of different types of bases and lg equals the number of nucleotides in each codon (codon length). 


    The number of possible codons must be greater than or equal to the number of amino acids specified. Therefore, a codon length of one nucleotide could specify 4 different amino acids, a codon length of two nucleotides could specify 20 different amino acids, and a codon length of three nucleotides could specify 30 different amino acids: (a) one, (b) two, (c) three.
  3. A template strand in bacterial DNA has the following base sequence: 

    5′–AGGTTTAACGTGCAT–3′ 

    What amino acids are encoded by this sequence?
    first work out the mRNA sequence that will be transcribed from this DNA sequence. The mRNA must be antiparallel and complementary to the DNA template strand: 

    • DNA template strand:
    • 5' –AGGTTTAACGTGCAT–3'

    • mRNA copied from DNA:
    • 3' –UCCAAAUUGCACGUA–5'


    An mRNA is translated 5′→3′; so it will be helpful if we turn the RNA molecule around with the 5′ end on the left: 

    mRNA copied from DNA: 5′AUGCACGUUAAACCU–3′ 

  4. The following triplets constitute anticodons found on a series of tRNAs. Name the amino acid carried by each of these tRNAs.

    a. 5′–UUU–3′
    b. 5′–GAC–3′
    c. 5′–UUG–3′
    d. 5′–CAG–3′
  5. What is the one gene, one enzyme hypothesis? Why was this hypothesis an important
    advance in our understanding of genetics?
    • The one gene, one enzyme hypothesis proposed by Beadle and Tatum states that each
    • gene encodes a single, separate protein. Now that we know more about the nature of
    • enzymes and genes, it has been modified to the one gene, one polypeptide hypothesis
    • because many enzymes consist of multiple polypeptides. The original hypothesis helped
    • establish a linear link between genes (DNA) and proteins.
  6. What are isoaccepting tRNAs?
    • Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but
    • accept the same amino acids.
  7. What is the significance of the fact that many synonymous codons differ only in the third nucleotide position?
    Synonymous codons code for the same amino acid, or, in other words, have the same meaning. A nucleotide at the third position of a codon pairs with a nucleotide in the first position of the anticodon. Unlike the other nucleotide positions involved in the codon-anticodon pairing, this pairing is often weak, or “wobbles,” and nonstandard pairings can occur. Because the “wobble,” or nonstandard base-pairing with the anticodons, affects the third nucleotide position, the redundancy of codons ensures that the correct amino acid is inserted in the protein when nonstandard pairing occurs.
  8. Define the following terms as they apply to the genetic code: 

    Reading Frame
    The reading frame refers to how the nucleotides in a nucleic acid molecule are grouped into codons containing three nucleotides. Each sequence of nucleotides has three possible sets of codons, or reading frames.
  9. Define the following terms as they apply to the genetic code: 

    Overlapping code
    If an overlapping code is present, then a single nucleotide is included in more than one codon. The result for a sequence of nucleotides is that more than one type of polypeptide can be encoded within that sequence.
  10. Define the following terms as they apply to the genetic code: 

    Nonoverlapping Code
    In a nonoverlapping code, a single nucleotide is part of only one codon. This results in the production of a single type of polypeptide from one polynucleotide sequence.
  11. Define the following terms as they apply to the genetic code: 

    Initiation Codon
    An initiation codon establishes the appropriate reading frame and specifies the first amino acid of the protein chain. Typically, the initiation codon is AUG; however, GUG and UUG can also serve as initiation codons although rarely.
  12. Define the following terms as they apply to the genetic code: 

    Termination Codon
    The termination codon signals the termination or end of translation and the end of the protein molecule. There are three termination codons—UAA, UAG, and UGA— which can also be referred to as stop codons or nonsense codons. These codons do not code for amino acids.
  13. Define the following terms as they apply to the genetic code: 


    Sense Codon
    A sense codon is a group of three nucleotides that code for an amino acid. There are 61 sense codons that code for the 20 amino acids commonly found in proteins.
  14. Define the following terms as they apply to the genetic code: 

    Nonsense Codon
    Nonsense codons or termination codons signal the end of translation. These codons do not code for amino acids.
  15. Define the following terms as they apply to the genetic code: 

    Universal Code
    In a universal code, each codon specifies, or codes, for the same amino acid in all organisms. The genetic code is nearly universal, but not completely. Most of the exceptions occur in mitochondrial genes.
  16. Define the following terms as they apply to the genetic code: 


    Nonuniversal Codons
    Most codons are universal (or nearly universal) in that they specify the same amino acids in almost all organisms. However, there are exceptions where a codon has different meanings in different organisms. Most of the known exceptions are the termination codons, which in some organisms do code for amino acids. Occasionally, a sense codon is substituted for another sense codon.
  17. How is the reading frame of a nucleotide sequence set?
    The initiation codon on the mRNA sets the reading frame.
  18. How are tRNAs linked to their corresponding amino acids?
    Each of the 20 different amino acids that are commonly found in proteins has a corresponding aminoacyl-tRNA synthetase that covalently links the amino acid to the correct tRNA molecule.
  19. What events bring about the termination of translation?
    • The process of termination begins when a ribosome encounters a termination codon.
    • Because the termination codon would be located at the “A” site, no corresponding
    • tRNA will enter the ribosome. This allows for the release factors (RF1, RF2, and RF3) to bind the ribosome. RF1 recognizes and interacts with the stop codons UAA and UAG, while RF2 can interact with UAA and UGA. A RF3-GTP complex binds to the ribosome. Termination of protein synthesis is complete when the polypeptide chain is cleaved from the tRNA located at the “P” site. During this process, the GTP is hydrolyzed to GDP.
  20. Assume that the number of different types of bases in RNA is four. What would be the
    minimum codon size (number of nucleotides) required if the number of different types
    of amino acids in proteins were:
    (The number of codons possible must be equal to or greater than the number of different types of amino acids because the codons encode for the different amino acids. To calculate how many possible codons are possible for a given codon size with four different types of bases in the RNA, the following formula can be used: 4n, where n is
    the number of nucleotides within the codon.)
    A. 2 
    B. 8
    C. 17
    6. 45
    e. 75
    A. 1, because 41 = 4 codons, which is more than enough to specify 2 different amino acids.

    b. 2

    c. 3

    d. 3

    e. 4
    (this multiple choice question has been scrambled)
  21. How many codons would be possible in a triplet code if only three bases (A, C, and U)
    were used?
    To calculate the number of possible codons of a triplet code if only three bases are used, the following equation can be used: 3n, where n is the number of nucleotides within the codon. So, the number of possible codons is equal to 33, or 27 possible codons.
  22. Using the genetic code given in Figure 15.10, give the amino acids specified by the bacterial mRNA sequences and indicate the amino and carboxyl ends of the polypeptide produced.

    a. 5 ́–AUGUUUAAAUUUAAAUUUUGA–3 ́
    b. 5 ́–AUGUAUAUAUAUAUAUGA–3 ́ 
    c. 5 ́–AUGGAUGAAAGAUUUCUCGCUUGA–3 ́
    d. 5 ́–AUGGGUUAGGGGACAUCAUUUUGA–3 ́
    Each of the mRNA sequences begins with the three nucleotides AUG. This indicates the start point for translation and allows for a reading frame to be set. In bacteria, the AUG initiation codon codes for N-formyl-methionine. Also, for each of these mRNA sequences, a stop codon is present either at the end of the sequence or within the interior of the sequence.

    • The amino terminal refers to the end of the protein with a free amino group and will be
    • the first peptide in the chain. The carboxyl terminal refers to the end of the protein with
    • a free carboxyl group and is the last amino acid in the chain. For the following peptide chains reading from left to right, the first amino acid is located at the amino end, while the last amino acid is located at the carboxyl end.

    a. Amino fMet–Phe–Lys–Phe–Lys–Phe Carboxyl

    b. Amino fMet–Tyr–Ile–Tyr–Ile Carboxyl 

    c. Amino fMet–Asp–Glu–Arg–Phe–Leu–AlaCarboxyl 

    • d. Amino fMet–Gly Carboxyl (The stop codon UAG occurs after the codon for
    • glycine.)
  23. A nontemplate strand on DNA has the following base sequence. What amino acid sequence would be encoded by this sequence? 

    5 ́–ATGATACTAAGGCCC–3 ́
    • To determine the amino acid sequence, we need to know the mRNA sequence and the codons present. The nontemplate strand of the DNA has the same sequence as the mRNA, except that thymine containing nucleotides are substituted for the uracil containing nucleotides. So the mRNA sequence would be as follows:
    • 5'–AUGAUACUAAGGCCC–3'.

    • Assuming that the AUG indicates a start codon, then the amino acid sequence would be starting from the amino end of the peptide and ending with the carboxyl end:
    • fMet-Ile–Leu–Arg–Pro.
  24. The following amino acid sequence is found in a tripeptide: Met–Trp–His. Give all possible nucleotide sequences on the mRNA, on the template strand of DNA, and on the nontemplate strand of DNA that could encode this tripeptide.
    • The potential mRNA nucleotide sequences encoding for the tripeptide Met–Trp–His can
    • be determined by using the codon table found in Figure 15.14. From the table, we can see that the amino acid His has two potential codons, while the amino acids Met and Trp each have only one potential codon. Therefore, there are two different mRNA nucleotide sequences that could encode for the tripeptide. Once the potential mRNA nucleotide sequences have been determined, the template and nontemplate DNA strands can be derived from these potential mRNA sequences. 
    • (1) 5'–AUGUGGCAU–3' 
    • DNA template: 3'–TACACCGTA–5' 
    • DNA nontemplate: 5'–ATGTGGCAT–3' 

    • (2) 5'–AUGUGGCAC–3' 
    • DNA template: 3'–TACACCGTG–5'
    • DNA nontemplate: 5'–ATGTGGCAC–3'
  25. An anticodon on a tRNA has the sequence 5 ́–GCA–3 ́. 

    What amino acid is carried by this tRNA?
    • The anticodon 5'–GCA–3' would pair with the codon 5'–CGU–3'. Based on the
    • codon table in Figure 15.14, the amino acid encoded by this codon is cysteine. So,
    • this tRNA is most likely carrying cysteine.
  26. An anticodon on a tRNA has the sequence 5 ́–GCA–3 ́. 

    What would be the effect if the G in the anticodon were mutated to a U?
    The anticodon would now be 3'–ACU–5' and could pair to the codon 5'–UGA–3', a stop codon. The result would be that amino acid cysteine would be placed where the stop codon 5'–UGA–3' was located in the mRNA. Essentially, the stop codon would be suppressed and translation could continue.
  27. Which of the following amino acid changes could result from a mutation that changed a
    single base? For each change that could result from the alteration of a single base,
    determine which position of the codon (first, second, or third nucleotide) in the mRNA
    must be altered for the change to occur.

    Leu -> Gln
    • Of the six codons that encode for Leu, only two could be mutated by the alteration
    • of a single base to produce the codons for Gln:

    -CUA (Leu)—Change the second position to A to produce CAA (Gln).

    -CUG (Leu)—Change the second position to A to produce CAG (Gln).
  28. Which of the following amino acid changes could result from a mutation that changed asingle base? For each change that could result from the alteration of a single base,determine which position of the codon (first, second, or third nucleotide) in the mRNAmust be altered for the change to occur.

    Phe-> Ser
    • Both Phe codons (UUU and UUC) could be mutated at the second position to
    • produce Ser codons:

    -UUU (Phe)—Change the second position to C to produce UCU (Ser).

    -UUC (Phe)—Change the second postion to C to produce UCC (Ser).
  29. Which of the following amino acid changes could result from a mutation that changed asingle base? For each change that could result from the alteration of a single base,determine which position of the codon (first, second, or third nucleotide) in the mRNAmust be altered for the change to occur.

    Phe->Ile
    • Both Phe codons (UUU and UUC) could be mutated at the first position to produce
    • Ile codons:

    -UUU (Phe)—Change the first position to A to produce AUU (Ile).

    -UUC (Phe)—Change the first position to A to produce AUC (Ile).
  30. Which of the following amino acid changes could result from a mutation that changed asingle base? For each change that could result from the alteration of a single base,determine which position of the codon (first, second, or third nucleotide) in the mRNAmust be altered for the change to occur.

    Pro->Ala
    All four codons for Pro can be mutated at the first position to produce Ala codons: 

    • -CCU (Pro)—Change the first position to G to produce GCU (Ala).
    • -CCC (Pro)—Change the first position to G to produce GCC (Ala).
    • -CCA (Pro)—Change the first position to G to produce GCA (Ala).
    • -CCG (Pro)—Change the first position to G to produce GCG (Ala).
  31. Which of the following amino acid changes could result from a mutation that changed asingle base? For each change that could result from the alteration of a single base,determine which position of the codon (first, second, or third nucleotide) in the mRNAmust be altered for the change to occur.

    Asn->Lys
    Both codons for Asn can be mutated at a single position to produce Lys codons:

    • -AAU (Asn)—Change the third position to A to produce AAA (Lys).
    • -AAU (Asn)—Change the third position to G to produce AAG (Lys).
    • -AAC (Asn)—Change the third postion to A to produce AAA (Lys).
    • -AAC (Asn)—Change the third position to G to produce AAG (Lys).
  32. Which of the following amino acid changes could result from a mutation that changed asingle base? For each change that could result from the alteration of a single base,determine which position of the codon (first, second, or third nucleotide) in the mRNAmust be altered for the change to occur.

    Ile->Asn
    • Only two of the three Ile codons can be mutated at a single position to produce Asn
    • codons:

    • -AUU (Ile)—Change the second position to A to produce AAU (Asn).
    • -AUC (Ile)—Change the second position to A to produce AAC (Asn).
  33. Arrange the following components of translation in the approximate order in which they would appear or be used during protein synthesis.
    • The components are in order according to when they are used or play a key role in
    • translation. The potential exception is initiation factor 3. Initiation factor 3 could
    • possibly be listed first because it is necessary to prevent the 30s ribosome from
    • associating with the 50s ribosome. It binds to the 30s subunit prior to the formation
    • of the 30s initiation complex. However, during translation events the release of
    • initiation factor 3, allows the 70s initiation complex to form, a key step in
    • translation.

    • fMet-tRNAfMet
    • 30S initiation complex
    • initiation factor 3
    • 70S initiation complex
    • elongation factor Tu
    • peptidyl transferase
    • elongation factor G
    • release factor 1
  34. The following diagram illustrates a step in the process of translation. Sketch the diagram and identify the following elements on it.

    a. 5′ and 3′ ends of the mRNA
    b. A,P,andEsites
    c. Start codon
    d. Stop codon
    e. Amino and carboxyl ends of the newly synthesized polypeptide chain
    f. Approximate location of the next peptide bond that will be formed
    g. Place on the ribosome where release factor 1 will bind


  35. a. What will be the anticodon of the next tRNA added to the A site of the ribosome? 

    b. What will be the next amino acid added to the growing polypeptide chain?
    a. The anticodon 3' UGC 5' is complementary to the codon 5' ACG 3', which is located at the A site of the ribosome. Notice the anticodon and codon are antiparallel. 

    b. The codon 5' ACG 3' encodes the amino acid threonine.
  36. For each of the following sequences , place a check mark in the appropriate space to indicate the process most immediately affected by deleting the sequence. Choose only one
  37. Several experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that codes for methionine initiator tRNA (tRNAiMet) was located and changed. The nucleotides that specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was 5′–CCA–3′ instead of 5′–CAU–3′. When this mutated gene was placed into a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of the proteins produced contained extra amino acids, and others contained fewer amino acids.

    a. What do these results indicate about how the ribosome recognizes the starting point
    for translation in eukaryotic cells? Explain your reasoning.
    By mutating the anticodon to 5'–CCA–3' from 5'–CAU–3' on tRNAiMet, the initiator tRNA will now recognize the codon 5'–UGG–3', which normally would code only for Trp. If translation initiation by the ribosome in eukaroytes occurs by binding the 5' cap of the mRNA followed by scanning, then the first 5'–UGG–3' codon recognized by the mutated tRNAiMet will be the start site for translation. If the first 5'–UGG–3' codon occurs prior to the normal 5'–AUG–3' codon, then a protein containing extra amino acids could be produced. If the first 5'–UGG–3' codon occurs after the normal 5'–AUG–3', then a shorter protein will be produced. Finally, truncated proteins could also be produced by the first 5'–UGG–3' being out of frame of the normal coding sequence. If this happens, then most likely a stop codon will be encountered before the end of the normal coding sequence and will terminate translation. The data suggest that translation initiation takes place by scanning of the ribosome for the appropriate start sequence.
  38. Several experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that codes for methionine initiator tRNA (tRNAiMet) was located and changed. The nucleotides that specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was 5′–CCA–3′ instead of 5′–CAU–3′. When this mutated gene was placed into a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of the proteins produced contained extra amino acids, and others contained fewer amino acids.

    If the same experiment had been conducted on bacterial cells, what results would you expect?
    Very little or no protein synthesis would be expected. Translation initiation in bacteria requires the 16S RNA of the small ribosomal subunit to interact with the Shine-Dalgarno sequence. This interaction serves to line up the ribosome over the start codon. If the anticodon has been changed such that the start codon cannot be recognized, then protein synthesis is not likely to take place.
  39. Explain the wobble hypothesis? In what sense might this be beneficial?
    The 5’ nucleotide of the anticodon in tRNA can pair with more than one nucleotide in the mRNA. Thus, although each tRNA can be charged with only one specific amino acid, the anticodon can recognize more than one codon for that particular amino acid. This reduces the number of tRNAs needed for translation.
  40. Contrast the effects of frameshift, missense and nonsense mutations on a protein amino acid sequence.
    A frameshift mutation changes the reading frame and causes the wrong amino acids to be inserted into the polypeptide after the change in the DNA. Missense mutations result in a single amino acid substitution in the protein (the total number of amino acids remains unchanged). Nonsense mutations replace a codon specifying an amino acid with a stop codon. The protein is shortened or truncated.
  41. List one major piece of evidence that DNA does not serve directly as the template for protein
    synthesis in eukaryotes.
    Translation in eukaryotes occurs in the cytoplasm; DNA is confined to the nucleus.
  42. If a tRNA had the anticodon sequence 5’ CGA 3’, what would be the sequence and polarity of the corresponding codon in the mRNA? In the DNA on the non-template strand?
    5’ UCG 3’ in the mRNA; 5’ TCG 3’ in the non-template strand of the DNA.
  43. What is the minimum number of nucleotides in an mRNA molecule necessary to encode a protein of 141 amino acids.
    426 (don’t forget the termination codon)

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