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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
Chapter 11, Problem 32b

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.
Restriction map showing BamHI sites in wild-type and mutant alleles, with probes A and B indicated.
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.
Gel electrophoresis results showing DNA band patterns for mother, father, and two children, with size markers indicated.
In a separate figure, draw the gel electrophoresis band patterns for all the genotypes that could be found in children of this couple.

Verified step by step guidance
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Step 1: Understand the genetic basis of the problem. Alkaptonuria is an autosomal recessive disorder caused by a mutation in the HAO gene. The mutation alters a BamHI restriction site, changing the sequence from 5′-GGATCC-3′ to 5′-GGAACC-3′, which results in a Ser-to-Thr missense mutation. This change affects the restriction map of the mutant allele.
Step 2: Analyze the restriction maps provided for the wild-type and mutant alleles. The wild-type allele has four BamHI restriction sites (B1 to B4), while the mutant allele has three BamHI restriction sites due to the alteration at B3. This difference will result in distinct fragment sizes when the DNA is digested with BamHI.
Step 3: Review the Southern blotting technique. Southern blotting involves digesting DNA with a restriction enzyme (in this case, BamHI), separating the fragments by gel electrophoresis, transferring them to a membrane, and hybridizing them with specific probes (probe A and probe B). The probes will bind to complementary sequences, allowing visualization of specific DNA fragments.
Step 4: Determine the possible genotypes of the children based on the parents' genotypes. Since the disorder is autosomal recessive, the parents are likely heterozygous (carriers) for the mutant allele. The children could inherit one of three genotypes: homozygous wild-type (WT/WT), heterozygous (WT/Mut), or homozygous mutant (Mut/Mut). Each genotype will produce a distinct band pattern on the gel.
Step 5: Draw the gel electrophoresis band patterns for each genotype. For homozygous wild-type (WT/WT), the band pattern will correspond to the fragment sizes of the wild-type allele. For heterozygous (WT/Mut), the band pattern will show fragments from both the wild-type and mutant alleles. For homozygous mutant (Mut/Mut), the band pattern will correspond to the fragment sizes of the mutant allele. Use the restriction maps and probe binding sites to determine the exact fragment sizes and their positions on the gel.

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Key Concepts

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

Autosomal Recessive Inheritance

Autosomal recessive inheritance refers to a pattern where two copies of a mutated gene (one from each parent) are necessary for an individual to express a trait or disorder. In the case of alkaptonuria, both parents must carry at least one copy of the mutated HAO gene for their children to potentially inherit the disorder. This means that unaffected parents can still pass on the mutation to their offspring.
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Restriction Enzymes and Restriction Mapping

Restriction enzymes, like BamHI, are proteins that cut DNA at specific sequences, allowing researchers to analyze genetic material. Restriction mapping involves determining the locations of these cut sites within a DNA molecule, which can help identify differences between alleles, such as the presence of mutations. In this scenario, the alteration of the BamHI site in the mutant allele is crucial for understanding how the mutation affects the gene.
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Southern Blotting

Southern blotting is a technique used to detect specific DNA sequences within a complex mixture. It involves the digestion of DNA with restriction enzymes, followed by gel electrophoresis to separate the fragments, and then transferring the fragments to a membrane for hybridization with labeled probes. This method is essential for analyzing the genotypes of the family members in the context of alkaptonuria, as it allows for the visualization of the presence or absence of specific alleles.
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Related Practice
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.

959
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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.

568
views
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.

Explain how the DNA sequence change results in a Ser-to-Thr missense mutation.

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Textbook Question

In an experiment employing the methods of the Ames test, two strains of Salmonella are used. Strain A contains a base-substitution mutation, and Strain B contains a frameshift mutation. Four plates are prepared to test the mutagenicity of the compound ethyl methanesulfonate (EMS). Plate 1 is a control plate with Strain A and S9 extract but no EMS. Plate 2 is also a control plate and contains Strain B and S9 extract but no EMS. Plate 3 contains Strain A along with S9 extract and EMS, and Plate 4 contains Strain B, S9 extract, and EMS.

Characterize the expected distribution of colony growth on the four plates. Defend your growth prediction for each plate.

633
views
Textbook Question

In an experiment employing the methods of the Ames test, two strains of Salmonella are used. Strain A contains a base-substitution mutation, and Strain B contains a frameshift mutation. Four plates are prepared to test the mutagenicity of the compound ethyl methanesulfonate (EMS). Plate 1 is a control plate with Strain A and S9 extract but no EMS. Plate 2 is also a control plate and contains Strain B and S9 extract but no EMS. Plate 3 contains Strain A along with S9 extract and EMS, and Plate 4 contains Strain B, S9 extract, and EMS.

What event is being detected by growth of a colony on any of the four plates?

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