Skip to main content
Pearson+ LogoPearson+ Logo
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 32a

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 maps of wild-type and mutant alleles with BamHI sites and molecular 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.
Southern blot results showing DNA bands for mother (M), father (F), and children (C1, C2) with sizes in kilobases.
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.

Verified step by step guidance
1
Step 1: Understand the genetic basis of alkaptonuria. It is an autosomal recessive disorder caused by mutations in the HAO gene. This means that individuals with two copies of the mutant allele (aa) will express the disorder, while individuals with one wild-type allele (A) and one mutant allele (Aa) are carriers and do not express the disorder.
Step 2: Analyze the restriction maps provided. The wild-type allele (A) has four BamHI restriction sites, while the mutant allele (a) has three BamHI restriction sites due to the alteration of the B3 site from 5′-GGATCC-3′ to 5′-GGAACC-3′. This change affects the digestion pattern observed in Southern blotting.
Step 3: Examine the Southern blot results for each family member (M, F, C1, and C2). The banding patterns on the gel correspond to the DNA fragments generated by BamHI digestion. Compare these patterns to the expected digestion patterns for the wild-type allele (A) and mutant allele (a).
Step 4: Assign genotypes based on the banding patterns. For example, individuals with bands corresponding to both wild-type and mutant alleles are heterozygous (Aa), while individuals with bands corresponding only to the mutant allele are homozygous recessive (aa) and are affected by alkaptonuria.
Step 5: Identify any family member who is alkaptonuric. Homozygous recessive individuals (aa) will express the disorder. Use the genotype assignments from Step 4 to determine which family member(s) are affected.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

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, individuals must inherit two copies of the mutant HAO gene to exhibit symptoms. Carriers, who possess one normal and one mutant allele, do not show symptoms but can pass the mutant allele to their offspring.
Recommended video:
Guided course
09:08
Autosomal Pedigrees

Restriction Enzymes and Restriction Mapping

Restriction enzymes, like BamHI, cut DNA at specific sequences, allowing researchers to create restriction maps that illustrate the locations of these cut sites. In this scenario, the presence of different numbers of BamHI sites in the wild-type and mutant alleles helps identify genetic differences. Analyzing these patterns through techniques like Southern blotting can reveal the genotypes of individuals based on the presence or absence of specific bands in the gel.
Recommended video:
Guided course
07:11
Mapping with Markers

Missense Mutation

A missense mutation occurs when a single nucleotide change results in the substitution of one amino acid for another in a protein. In the context of alkaptonuria, the Ser-to-Thr missense mutation in the HAO gene alters the enzyme's function, leading to the accumulation of homogentisic acid. Understanding this mutation is crucial for determining the biochemical basis of the disorder and its inheritance pattern.
Recommended video:
Guided course
09:49
Point Mutations
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.

Determine the wild-type mRNA sequence.

577
views
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.

413
views
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
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.

In a separate figure, draw the gel electrophoresis band patterns for all the genotypes that could be found in children of this couple.

419
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.

534
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.

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

633
views