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.

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 acid4 (replica plates numbered 1 to 4) with the following results:Identify any colonies that are his⁻, arg⁻, leu⁻.

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.Wild-type polypeptide N . . . Thr–His–Ser–Gly–Leu–Lys–Ala . . . CMutant 1 N . . . Thr–His–Ser–Val–Leu–Lys–Ala . . . CMutant 2 N . . . Thr–His–Ser–CMutant 3 N . . . Thr–Thr–Leu–Asp–CMutant 4 N . . . Thr–Gln–Leu–Trp–Ile–Glu–Gly . . .Identify the mutation that produces each mutant polypeptide.

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.Wild-type polypeptide N . . . Thr–His–Ser–Gly–Leu–Lys–Ala . . . CMutant 1 N . . . Thr–His–Ser–Val–Leu–Lys–Ala . . . CMutant 2 N . . . Thr–His–Ser–CMutant 3 N . . . Thr–Thr–Leu–Asp–CMutant 4 N . . . Thr–Gln–Leu–Trp–Ile–Glu–Gly . . .Determine the wild-type mRNA sequence.

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.Wild-type polypeptide N . . . Thr–His–Ser–Gly–Leu–Lys–Ala . . . CMutant 1 N . . . Thr–His–Ser–Val–Leu–Lys–Ala . . . CMutant 2 N . . . Thr–His–Ser–CMutant 3 N . . . Thr–Thr–Leu–Asp–CMutant 4 N . . . Thr–Gln–Leu–Trp–Ile–Glu–Gly . . .Use the available information to characterize each mutant.

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.