1. Unit 3
MOLECULAR GENETICS

2. I] DNA, RNA and PROTEIN SYNTHESIS
The Analogy
our textbook represents the DNA and there is only one copies.
the textbook never leaves the classroom [nucleus]
in this textbook all of the left hand pages are identical to the right hand pages
if we wish to send a chapter to the shop we transcribe the chapter
that way, if the transcribed chapter is destroyed, we still have the original

3. A] HISTORY of DNA Hammerling, 1930’s using algae 
discovered that the nucleus is source of heredity

4. knew that the nucleus contains proteins and DNA
but DNA did not seem complex enough
4 base pairs only

5. Recall, there are 20 amino acids
while proteins did seem complex enough
This led to the theory that proteins were the molecules of inheritance.

6. when they tried to confirm that proteins control heredity using viruses, Hershey & Chase
actually showed that DNA controls heredity

7. With the help of “borrowed” information from Rosalind Franklin
Watson & Crick determined the structure of DNA in 1953

8. B] DNA STRUCTURE

9. 1] Parts of a nucleotide a nitrogenous base a phosphate pyrimadine
T – thymine
C – cytosine O II
HO—P—OH I OH
a sugar – deoxyribose O OH
HOH2C
purine
A- adenine
G-guanine

10. 2] arrangement of nucleotidesOH
CH2 II
HO—P—OH I
combined to form
a nucleotide
2] arrangement of nucleotides
a] the sugar bonds to the last phosphate at carbon 3 and so sugar end of DNA is called the 3’ end
b] the phosphate bonds to the next sugar at carbon 5 and so the phosphate end of DNA is called the 5’ end

11. c] the two strands are arranged as shown below
5’ end l P
sugar—base……
3’ end l
base—sugar P
5’ end l P
sugar—base…… l P
sugar—base……
3’ end

12. 3. Hydrogen bonding – adenine forms 2 hydrogen bonds with thymine
– cytosine forms 3 hydrogen bonds with guanine
Since A – T and C – G are the usual bondings, if the
amount of A does not equal T or the amount of C does
not equal G,
then the DNA is not double stranded

13. Two equal sized pieces of DNA.
Which is more stable? Why?
G A A T T C T A A T A T C C T T A A G A T T A T A G
A G C C G C C T A G C G T C G G C G G A T C G C
Sections of the DNA with more C-G bonds, are more stable.
[3 H-bonds vs. 2 H-bonds]

14. 4. DNA shape
– the stands are arranged in a double helix

15. 5’ 3’ 3’ 5’ 5. The letter P represents phosphate; S represents sugar
and the letters A, T, C and G are nitrogenous bases 5’
P P P P P P P P P P P P P P
S S S S S S S S S S S S S S
A T A C G C A C T C G A T G
T A T G C G T G A G C T A C
P P P P P P P P P P P P P P 3’ 3’ 5’

16. A __A C G C A C T C G A T G T A T G G T G __ _ C T A C T C A G
6. Normally this is drawn as shown below
A __A C G C A C T C G A T G
T A T G G T G __ _ C T A C T C A G

17. DNA in the cell a. each human cell has about 3 billion base pairs
b. these are arranged into about 25,000 genes
c. in addition, up to 50% of the DNA does not code for any protein
d. in total, the DNA from one cell would be 2m in length
e. double stranded DNA [dsDNA] is very stable, but single stranded DNA [ssDNA] is very easily broken

18. C. DNA Replication 1. Introduction
a. before a cell divides it must copy all of its DNA
b. it must do this while keeping the amount of ssDNA to a minimum
c. also, the DNA will have to be unwound to allow access to the bases

19. DNA Replication
Animations

20. 1. Separating the strands
a] DNA helicase [1]
– breaks H-bonds between the strands at an “origin of replication” and unwinds the DNA
b] DNA gyrase [2]
- releases tension from unwinding
c] replication fork [3]
- an area where the two strands are separated
d] ssB’s [4]
- single-stranded binding proteins keep strands from re-annealing

21. Ignore the Tweed 5’ 3’

22. 2. Building complementary strands a] starting
primase [5] - creates a base strand of RNA
called an RNA-primer [6] at the exposed 3’ end of
the replication fork
[the cell has to use a piece of RNA to start the process because DNA will not stick to DNA as well as RNA will]

23. 3’ 5’ 5’ 3’

24. b] Building -the enzyme DNA polymerase III [7]- adds complimentary
bases that are phosphorylated moving along the old DNA
in the 3’ to 5’ direction
- the energy released by removing 2 Pi’s allows the bonding
the strand that is copied continuously [3’  5’] strand is
called the leading strand [8]

25. c] The other strand DNA polymerase III only moves 3’  5’
- so the lagging strand [9] has to be built in pieces
- these short pieces are called Okazaki fragments [10]

26. 3’ 5’ 5’ 3’ 5’ 3’

27. d] Finishing DNA polymerase I [11] removes the RNA primers
& replaces them with the appropriate DNA bases
DNA ligase [12] attaches Okazaki fragments
by bonding sugar and phosphates

28. C. DNA REPLICATION 3’ 5’ 5’ 3’ 5’ 3’

29. 3’ 5’ 5’ 3’ 5’ 3’

30. 3’ 5’ 5’ 3’ 5’ 3’