Page 147 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

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

pairs distributed among 23 chromosomes containing an of manageable size, making many copies of those pieces to
average of 130 million nucleotide pairs apiece. obtain enough material for study, and characterizing the
In Chapters 9, 10 and 11, we describe how researchers pieces down to the level of nucleotide sequence. The scien­
analyze the mass of genetic information in the chromo­ tists then try to reconstruct the DNA sequence of an entire
somes of a genome as they try to discover what parts of the genome by determining the spatial relationship between
DNA are genes and how those genes influence phenotype. the many pieces. Finally, they use this knowledge to exam­
They begin their analysis by breaking the DNA into pieces ine the genomic variations that make individuals unique.

SOLVED PROBLEMS

I. A geneticist examined the amino acid sequence of a at position 2. Mutants 2, 4, and 6 affect a base pair different
particular protein in a variety of E. coli mutants. The from that affected by mutant 1, so they could recombine
amino acid in position 40 in the normal enzyme is with mutant 1.
glycine. The following table shows the substitutions In summary, the sequence of nucleotides on the RNA­
the geneticist found at amino acid position 40 in six like strand of the wild­type and mutant genes at this posi­
mutant forms of the enzyme. tion must be:

mutant 1 cysteine wild type 5′ G G T/C 3′
mutant 2 valine mutant 1 5′ T G T/C 3′
mutant 3 serine mutant 2 5′ G T T/C 3′
mutant 4 aspartic acid mutant 3 5′ A G T/C 3′
mutant 5 arginine mutant 4 5′ G A T/C 3′
mutant 6 alanine mutant 5 5′ C G T/C 3′
mutant 6 5′ G C T/C 3′
Determine the nature of the base substitution that
must have occurred in the DNA in each case. Which II. The double­stranded circular DNA molecule that
of these mutants would be capable of recombination forms the genome of the SV40 tumor virus can be de­
with mutant 1 to form a wild­type gene? natured into single­stranded DNA molecules. Because
the base composition of the two strands differs, the
strands can be separated on the basis of their density
Answer into two strands designated W(atson) and C(rick).
To determine the base substitutions, use the genetic code When each of the purified preparations of the sin­
table (see Fig. 8.2). The original amino acid was glycine, gle strands was mixed with mRNA from cells infected
which can be encoded by GGU, GGC, GGA, or GGC. Mu­ with the virus, hybrids were formed between the RNA
tant 1 results in a cysteine at position 40; Cys codons are and DNA. Closer analysis of these hybridizations
either UGU or UGC. A change in the base pair in the DNA showed that RNAs that hybridized with the W prepa­
encoding the first position in the codon (a G–C to T–A ration were different from RNAs that hybridized with
transversion) must have occurred, and the original glycine the C preparation. What does this tell you about the
codon must therefore have been either GGU or GGC. Valine transcription templates for the different classes of
(in mutant 2) is encoded by GUN (with N representing any RNAs?
one of the four bases), but assuming that the mutation is a
single base change, the Val codon must be either GUU or Answer
GUC. The change must have been a G–C to T–A transver­ An understanding of transcription and the polarity of DNA
sion in the DNA for the second position of the codon. To strands in the double helix are needed to answer this ques­
get from glycine to serine (mutant 3) with only one base tion. Some genes use one strand of the DNA as a template;
change, the GGU or GGC would be changed to AGU or others use the opposite strand as a template. Because of the
AGC, respectively. A transition occurred (G–C to A–T) at different polarities of the DNA strands, one set of genes
the first position. Aspartic acid (mutant 4) is encoded by would be transcribed in a clockwise direction on the circu­
GAU or GAC, so the DNA of mutant 4 is the result of a lar DNA (using, say, the W strand as the template), and the
G–C to A–T transition at position 2. Arginine (mutant 5) is other set would be transcribed in a counterclockwise direc­
encoded by CGN, so the DNA of mutant 5 must have un­ tion (with the C strand as template).
dergone a G–C to C–G transversion at position 1. Finally,
alanine (mutant 6) is encoded by GCN, so the DNA of III. Geneticists interested in human hemoglobins have
mutant 6 must have undergone a G–C to C–G transversion found a very large number of mutant forms. Some of