Page 155 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

314 Chapter 8 Gene Expression: The Flow of Information from DNA to RNA to Protein

Tyr
898 amino acids long; in Family 2, deletion of a single 49. Why is a nonsense suppressor tRNA , even though it
base results in a 766­amino acid polypeptide; and in has a mutant anticodon that cannot recognize a tyrosine
Family 3, a C­to­G transversion in exon 10 yields a codon, charged with tyrosine by Tyr tRNA synthetase?
458­amino acid protein. (Hint: Refer to Fig. 8.19.)
a. Hypothesize as to how each of the three SCN9A 50. A mutant B. adonis bacterium has a nonsense sup­
mutations affects gene structure: Why are truncated pressor tRNA that inserts glutamine (Gln) to match
proteins made in each case? UAG (but not other nonsense) codons. [This species
2
b. How would you classify the mutant alleles? Do does not modify wobble position C residues to k C and
Pyl
these cause loss of function or gain of function? does not have tRNA (see Problem 57).]
Are they amorphs, hypomorphs, hypermorphs, a. What is the anticodon of the suppressing tRNA?
neomorphs, or antimorphs? Indicate the 5′ and 3′ ends.
c. Explain in molecular terms why CPA/anosmia is a b. What is the sequence of the template strand of the
Gln
recessive condition. wild­type tRNA encoding gene that was altered
47. You learned in Problem 21 in Chapter 7 that the to produce the suppressor, assuming that only a
neurodegenerative disease ALS can be caused by expan­ single­base­pair alteration was involved?
sion of a hexanucleotide repeat region (5′­GGGGCC­3′) c. What is the minimum number of tRNA genes
Gln
outside of the open reading frame (but within the first that could be present in a wild­type B. adonis cell?
intron) of the gene called C9ORF72. While a normal Describe the corresponding anticodons.
C9ORF72 allele has 2–23 copies of the hexanucleotide 51. You are studying mutations in a bacterial gene that
repeat unit, dominant disease­causing alleles have hun­ codes for an enzyme whose amino acid sequence is
dreds or even thousands of copies. known. In the wild­type protein, proline is the fifth
Researchers observed that the first intron of the amino acid from the amino terminal end. In one of
C9ORF72 disease allele is transcribed not only from your mutants with nonfunctional enzyme, you find a
the normal template strand of DNA, but also from the serine at position number 5. You subject this mutant to
nontemplate strand. Even more unusual, both types of further mutagenesis and recover three different strains.
repeat­region transcripts are translated in all six reading Strain A has a proline at position number 5 and acts
frames in an AUG­independent manner—a process just like a wild­type strain. Strain B has tryptophan
called repeat-associated non-ATG translation, or RAN at position number 5 and also acts like wild type.
translation. These discoveries led to the hypothesis Strain C has no detectable enzyme function at any
that the proteins made from the repeats might temperature, and you can’t recover any protein that
contribute to ALS. resembles the enzyme. You mutagenize strain C and
a. What polypeptides are made from the repeat­region recover a strain (C­1) that has enzyme function. The
transcripts? second mutation in C­1 that is responsible for the
b. According to the RAN translation hypothesis, why recovery of enzyme function does not map at the
are disease­causing C9ORF72 alleles dominant to enzyme locus.
normal alleles? a. What is the nucleotide sequence in both strands of
c. How would you classify the mutant alleles? Do the wild­type gene at this location?
they cause a loss of function or a gain of function? b. Why does strain B have a wild­type phenotype? Why
Are they amorphic, hypomorphic, hypermorphic, does the original mutant with serine at position
neomorphic, or antimorphic? (Note: More than 5 lack function?
one answer might be possible.) c. What is the nature of the mutation in strain C?
48. When 1 million cells of a culture of haploid yeast d. What is the second mutation that arose in C­1?
−
carrying a met auxotrophic mutation were plated on 52. Another class of suppressor mutations, not described
petri plates lacking methionine (Met), five colonies in the chapter, are mutations that suppress missense
grew. You would expect cells in which the original mutations.
−
met mutation was reversed (by a base change back to
the original sequence) would grow on the media lacking a. Why would bacterial strains carrying such mis­
methionine, but some of these apparent reversions could sense suppressor mutations generally grow more
be due to a mutation in a different gene that somehow slowly than strains carrying nonsense suppressor
−
suppresses the original met mutations. How would mutations?
you be able to determine if the mutations in your five b. What other kinds of mutations can you imagine in
colonies were due either to a precise reversion of the genes encoding components needed for gene ex­
−
original met mutation or to the generation of a suppres­ pression that would suppress a missense mutation
sor mutation in a gene on another chromosome? in a protein­coding gene?