Week of 08/28

Chapter One:
Question #4: What are the 3 features of DNA that suit it for its role as a hereditary
Its ability to replicate/duplicate
It stores informational content
It is relatively stable while being able to mutate
Chapter 2:
Question #14: In normal double-helical DNA is it true thata.) A plus C always equals G plus T?Yes, A base pairs with T and G base pairs with
C. So, there will always be an equal number of A’s as there are T’s and an equal
number of G’s as there are C’s.
b.) A plus G always equals C plus T?Yes. A and G are purines, while C and T are
pyrimidines. The number of purines always equals the number of pyrimidines since a
base pair always occurs between a purine and a pyrimidine.
Question # 17: The genome of the bacterium Hemophilus influenzae is 1830kb in size,
and sequencing has shown that it has 1703 genes.
a.) What is the average spacing between the central points of genes?
1830kb=1,830,000 base pairs/1703 genes ~1075 base pairs/ gene. This
estimate does not take into account where the gene begins or ends or whether there
are any spaces in this estimate. It calculates an AVERAGE spacing.
b.) If the average size of a coding region of a gene is assumed to be 1kb, what is the
average distance between the genes?
1075 base pairs/gene – 1000 base pairs/geneAn average distance of ~75 base
pairs between genes
c.) What is in this region?
What is in this spacer region? The space between genes often contains DNA of
unknown function. Regulatory? Structural? Parasitic? Junk?
d.) What proportion of the genome is expected to be introns?
We are dealing with a prokaryote here, and generally they do not have introns
So, the proportion would be 0See Figure 2-4 in text for good comparison!
e.) What is the average space between coding regions?
--Exact same question as letter b!! 