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Ch. 11 - Gene Mutation, DNA Repair, and Homologous Recombination
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
All textbooksSanders 3rd EditionCh. 11 - Gene Mutation, DNA Repair, and Homologous RecombinationProblem 30b
Chapter 11, Problem 30b

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.
Table comparing wild-type and mutant polypeptide sequences for amino acid analysis.
Determine the wild-type mRNA sequence.

Verified step by step guidance
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Step 1: Understand the relationship between amino acids and codons. Each amino acid is encoded by a specific codon in the mRNA sequence. Use the genetic code table to identify the codons corresponding to the amino acids in the wild-type polypeptide sequence.
Step 2: Analyze the wild-type polypeptide sequence (Thr–His–Ser–Gly–Leu–Lys–Ala) and determine the codons for each amino acid. For example, Thr is encoded by codons such as ACU, ACC, ACA, or ACG. Repeat this process for all amino acids in the sequence.
Step 3: Compare the wild-type sequence with the mutant sequences to identify the changes. For example, Mutant 1 has Val instead of Gly, indicating a single nucleotide change in the codon for Gly. Mutant 2 is truncated after Ser, suggesting a premature stop codon. Mutant 3 and Mutant 4 show more extensive changes, which may involve frameshift mutations or multiple nucleotide substitutions.
Step 4: Use the information from the mutants to infer the possible nucleotide sequence of the wild-type mRNA. For example, if a mutation changes Gly to Val, determine the codon for Gly in the wild-type and the codon for Val in the mutant to identify the nucleotide change.
Step 5: Assemble the wild-type mRNA sequence by combining the codons for each amino acid in the wild-type polypeptide. Ensure that the sequence aligns with the genetic code and accounts for the observed mutations in the mutants.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Genetic Code

The genetic code is a set of rules that defines how the sequence of nucleotides in mRNA is translated into the sequence of amino acids in a polypeptide. Each amino acid is specified by a codon, which is a sequence of three nucleotides. Understanding the genetic code is essential for determining the mRNA sequence that corresponds to a given polypeptide sequence.
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The Genetic Code

Amino Acid Structure

Amino acids are the building blocks of proteins, each consisting of a central carbon atom, an amino group, a carboxyl group, a hydrogen atom, and a variable R group that determines the specific properties of the amino acid. The sequence of amino acids in a polypeptide is critical for its function, and knowing the amino acid sequence helps in deducing the corresponding mRNA sequence through codon mapping.
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Ribosome Structure

Mutations

Mutations are changes in the nucleotide sequence of DNA that can lead to alterations in the amino acid sequence of proteins. Analyzing the differences between the wild-type and mutant polypeptide sequences can provide insights into which nucleotides have changed, allowing for the reconstruction of the wild-type mRNA sequence by identifying the original codons that correspond to the wild-type amino acids.
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Mutations and Phenotypes
Related Practice
Textbook Question

A wild-type culture of haploid yeast is exposed to ethyl methanesulfonate (EMS). Yeast cells are plated on a complete medium, and 6 colonies (colonies numbered 1 to 6) are transferred to a new complete medium plate for further study. Four replica plates are made from the complete medium plate to plates containing minimal medium or minimal medium plus one amino acid (replica plates numbered 1 to 4) with the following results:

For colonies 1, 3, and 5, write '+' for the wild-type synthesis and '−' for the mutant synthesis of histidine and leucine.

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Textbook Question

A wild-type culture of haploid yeast is exposed to ethyl methanesulfonate (EMS). Yeast cells are plated on a complete medium, and 6 colonies (colonies numbered 1 to 6) are transferred to a new complete medium plate for further study. Four replica plates are made from the complete medium plate to plates containing minimal medium or minimal medium plus one amino acid (replica plates numbered 1 to 4) with the following results:

Are there any colonies for which genotype information cannot be determined? If so, which colony or colonies?

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Textbook Question

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.

Use the available information to characterize each mutant.

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Textbook Question

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.

Identify the mutation that produces each mutant polypeptide.

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Textbook Question

Experiments by Charles Yanofsky in the 1950s and 1960s helped characterize the nature of tryptophan synthesis in E. coli. In one of Yanofsky's experiments, he identified glycine (Gly) as the wild-type amino acid in position 211 of tryptophan synthetase, the product of the trpA gene. He identified two independent missense mutants with defective tryptophan synthetase at these positions that resulted from base-pair substitutions. One mutant encoded arginine (Arg) and another encoded glutamic acid (Glu). At position 235, wild-type tryptophan synthetase contains serine (Ser) but a base-pair substitution mutant encodes leucine (Leu). At position 243, the wild-type polypeptide contains glutamine and a base-pair substitution mutant encodes a stop codon. Identify the most likely wild-type codons for positions 211, 235, and 243. Justify your answer in each case.

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Textbook Question

Alkaptonuria is a human autosomal recessive disorder caused by mutation of the HAO gene that encodes the enzyme homogentisic acid oxidase. A map of the HAO gene region reveals four BamHI restriction sites (B1 to B4) in the wild-type allele and three BamHI restriction sites in the mutant allele. BamHI utilizes the restriction sequence 5′-GGATCC-3′. The BamHI restriction sequence identified as B3 is altered to 5′-GGAACC-3′ in the mutant allele. The mutation results in a Ser-to-Thr missense mutation. Restriction maps of the two alleles are shown below, and the binding sites of two molecular probes (probe A and probe B) are identified.

DNA samples taken from a mother (M), father (F), and two children (C1 and C2) are analyzed by Southern blotting of BamHI-digested DNA. The gel electrophoresis results are illustrated.

Using A to represent the wild-type allele and a for the mutant allele, identify the genotype of each family member. Identify any family member who is alkaptonuric.

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