Concept explainers
You learned in Problem 21 in Chapter 7 that the neurodegenerative disease ALS can be caused by expansion of a hexanucleotide repeat region (5′-GGGGCC-3′) outside of the open reading frame (but within the first intron) of the gene called C9ORF72. While a normal C9ORF72 allele has 2–23 copies of the hexanucleotide repeat unit, dominant disease-causing alleles have hundreds or even thousands of copies.
Researchers observed that the first intron of the C9ORF72 disease allele is transcribed not only from the normal template strand of DNA, but also from the nontemplate strand. Even more unusual, both types of repeatregion transcripts are translated in all six reading frames in an AUG-independent manner—a process called repeat-associated non-ATG translation, or RAN translation. These discoveries led to the hypothesis that the proteins made from the repeats might contribute to ALS.
a. | What polypeptides are made from the repeat-region transcripts? |
b. | According to the RAN translation hypothesis, why are disease-causing C9ORF72 alleles dominant to normal alleles? |
c. | How would you classify the mutant alleles? Do they cause a loss of function or a gain of function? Are they amorphic, hypomorphic, hypermorphic, neomorphic, or antimorphic? (Note: More than one answer might be possible.) |
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Genetics: From Genes to Genomes
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- The D1S80 locus is located on human chromosome 1 and is characterized by a repeating 16 base pair (bp) sequence. Alleles for this locus vary depending on the number of repeats present, thus affecting the size of the locus. The D1S80 locus also contains two conserved sequences, a 32bp sequence at one end and a 113bp sequence at the other end. If the DNA of an individual is targeted for D1S80 amplification, and one of the resulting amplicons is approximately 785bp in size, how many repeats would be present in this D1S80 allele? The amplicon of interest is indicated by a red arrow in the diagram below.arrow_forwardThe human genome contains thousands of sequences known as small open reading frames, some of which encode proteins of about 30 amino acids. What is the minimum number of nucleotides required to encode such a protein?arrow_forwardThe DNA of a deletion mutant of λ bacteriophage has a length of 15.4383 μm instead of 19.6356 μm. How many base pairs are missing from this mutant? *arrow_forward
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