Page 153 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

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

36. The human gene for ß2 lens crystallin has the compo­ normal protein and 20 are unrelated junk. No muta­
nents listed below. The numbers represent nucleotide tion is involved in the production of this 114­amino
pairs that make up the particular component. Assume acid protein.
for simplicity that no alternative splicing is involved. p. Outline a hypothesis for the process that would
produce this protein. Your hypothesis should ex­
5′ UTR 174 plain why 94 amino acids are the same as in the
1st exon 119 normal ß2 lens crystallin.
1st intron 532
2nd exon 337 q. Explain why you would expect that a polypeptide
2nd intron 1431 such as the 114­mer described above would on av­
3rd exon 208 erage have 20 amino acids of junk.
3rd intron 380 Section 8.4
4th exon 444
4th intron 99 37. In prokaryotes, a search for genes in a DNA sequence
5th exon 546 involves scanning the DNA sequence for long open
3′ UTR 715 reading frames (that is, reading frames uninterrupted
by stop codons). What problem can you see with this
Answer the following questions about the ß2 lens approach in eukaryotes?
crystallin gene, primary transcript, and gene product. 38. a. The genetic code table shown in Fig. 8.2 applies
Questions asking where should be answered with one both to humans and to E. coli. Suppose that you
of the 11 components from the list or with None. have purified a piece of DNA from the human
Assume poly­A tails contain 150 As. genome containing the entire gene encoding the
a. How large is the ß2 lens crystallin gene in bp (base hormone insulin. You now transform this piece of
pairs)? DNA into E. coli. Why can’t E. coli cells contain­
b. How large is the primary transcript for ß2 lens ing the human insulin gene actually make insulin?
crystallin in bases? b. Pharmaceutical companies have actually been able
c. How large is the mature mRNA for ß2 lens crystal­ to obtain E. coli cells that make human insulin;
lin in bases? such insulin can be purified from the bacterial cells
d. Where would you find the base pairs encoding the and used to treat diabetic patients. How were the
initiation codon? pharmaceutical companies able to create such
e. Where would you find the base pairs encoding the bacterial factories for making insulin?
stop codon? 39. a. Very few if any eukaryotic genes contain tracts
f. Where would you find the base pairs encoding the with more than 25 As or Ts in a row, yet almost all
5′ cap? eukaryotic mRNAs have a tract with more than
g. Where would you find the base pairs that constitute 100 As in a row. How is this possible?
the promoter? b. Scientists know the nucleotide sequences that direct
h. Which intron interrupts the 3′ UTR? the termination of bacterial gene transcription, but
i. Where would you find the sequences encoding the they generally have little idea about the nature of
the nucleotide sequences that direct transcription
C terminus? termination in eukaryotic cells. Explain the basis
j. Where would you find the sequence encoding the of this statement.
poly­A tail?
k. How large is the coding region of the gene in bp 40. Explain how differences in the initiation of translation
dictate that eukaryotic mRNAs are monocistronic
(base pairs)? while prokaryotic mRNAs may be polycistronic.
l. How many amino acids are in the ß2 lens crystallin
protein? Section 8.5
m. Which intron interrupts a codon? 41. Do you think each of the following types of mutations
n. Which intron is located between codons? would have very severe effects, mild effects, or no effect
o. Where would you be likely to find the site specifying at all?
poly­A addition? a. Nonsense mutations occurring in the sequences en­
You find in lens­forming cells from several different coding amino acids near the N terminus of the protein
people small amounts of a polypeptide that has the b. Nonsense mutations occurring in the sequences en­
same N terminus as the normal ß lens crystallin, but it coding amino acids near the C terminus of the protein
has a different C terminus. The polypeptide is c. Frameshift mutations occurring in the sequences en­
114 amino acids long, of which 94 are shared with the coding amino acids near the N terminus of the protein