Problem Set 1
This problem set covers material from January 10
through January 24
From Griffiths et al.:
Chapter 1, problems 5, 6, 8, 9, 17, 19
Chapter 2, problems 2, 3, 9, 11, 13
Chapter 3, problems 7, 15, 16, 18, 20, 35, 36
Chapter 7, problems 1, 2, 3, 5, 18, 19, 24, 26, 27
1.5) If the DNA is double stranded, A=T and G=C and A+T+C+G = 100%.
If T=15% then C =
[100-15(2)]/2 = 35%.
1.6) If the DNA is double stranded, G=D=24% and A=T=26%.
1.8) Keeping this strand antiparallel to the strand given, the sequence would be:
There are no sequence restrictions in single-stranded DNA.
The percentage of A must
equal the percentage of T only in double stranded DNA.
1.17) a) For the fungus to be orange (as in mutant 1), orange pigment would have to accumulate
and then not be converted into red pigment.
This would occur if enzyme C was defective.
Mutant 2 is yellow.
Using similar logic as in part (a), enzyme B must be defective.
organism defective in both enzyme B and C would still have the phenotype associated with the
block at the earlier step of this biochemical pathway.
In this case, it would be due to lack of
1.19) Protein function can be destroyed by a mutation that causes the substitution of a single
amino acid, even though the protein has the same immunological properties.
enzymes require very specific amino acids in exact positions within their active site.
substitution of one of these key amino acids might have no effect on overall size and shape of the
protein while completely destroying its enzymatic activity.
2.2) PFGE separates DNA molecules by size.
When DNA is carefully isolated from Neurospora
(which has seven different chromosomes) seven bands should be produced using this technique.
Similarly, the pea has seven different chromosomes and will produce seven bands (homologous
chromosomes will co-migrate as a single band).
2.3) There is a total of 4 m of DNA and nine chromosomes per haploid set.
On average, each is
4/9 m long.
At metaphase, their average length is 13
m, so the average packing ratio is 13 x 10
m: 4.4 x10
or roughly 1:34,000.
This remarkable achievement is accomplished through the
interaction of the DNA with proteins.
At its most basic, eukaryotic DNA is associated with
histones in units called nucleosomes and during mitosis, coils into a solenoid.
As loops, it
associates with and winds into a central core of nonhistone protein called the scaffold.