Download D = 60% = 390 points

Instructor: Dr. R. S. Zemetra
Office : Ag Biotech 111
Office Hours: M 3:30-4:30, W 2:30 – 4:30 and
F 2:30-3:30 or by appointment
Textbook: Genetics – A Conceptual
Approach by B.A. Pierce
Website:
www.webpages.uidaho.edu/~rzemetra/SWWW
Keys to success in the course:
Read the textbook before class
Read previous lecture’s notes before class
Utilize computer animations
Study previous exams
Ask questions
Exam Schedule - Genetics 314 Sp 2006
1st exam - Monday, Feb. 6th
2nd exam - Monday, March 6th
3rd exam - Monday, April 3nd
4th exam - Friday, April 28nd
Final - Tuesday, May 9th 1-3 pm
Homework
3 written assignments on topics related to
genetics about 2 – 3 pages in length.
Last written assignment will ask you for
an example of how some aspect of
genetics is used in your field of interest
and ask you to come up with a new way
genetics could be used to address a
problem in your field of interest.
There will be one problem set assigned
for the Mendelian genetics section of the
course.
Grading:
Written assignments
3 assignments on current genetics topic
(25 points each)
1 problem set (25 points)
Exams
4 exams (100 points each)
Final exam
Cumulative final (150 points), not
optional
Grades based on your total divided by total
number of points (650 points)
A = 90% = 585 points
B = 80% = 520 points
C = 70% = 455 points
D = 60% = 390 points
Percentages needed for each grade can change
but only downward (i.e. 89% for an A instead
of 90%)
What makes genetics an exciting field to
study?
- A little over 100 years old as a science
- Discovery of how DNA carries genetic
information only occurred about 50
years ago.
- Field of genetics is rapidly changing,
especially in the area of molecular
genetics and the application of genetics
to address problems in many areas.
Examples:
- Genomics
- Proteomics
- Epigenetics
- siRNA
- Others?
All of genetics revolves around the gene
 What is it?
 How is it expressed?
 How is it inherited?
 How does environment influence
expression?
 How does it behave in an:
 individual
 population
To answer these questions we will study several
areas of genetics
 molecular/biochemical genetics
 microbial genetics
 cytogenetics
 qualitative genetics
 quantitative genetics
 population genetics
Since the course revolves around the gene, we
will start with the gene itself and work our
way up.
Molecular

Cytological

Individual

Population
So the first questions that need to be
answered are:
 What is a gene?
 What is it made of?
 How does it replicate?
 How is it expressed?
Gene = The fundamental physical unit of
heredity whose existence can be
confirmed by allelic variants and
which occupies a specific gene locus.
A gene is a DNA sequence coding for
a single polypeptide, t-RNA or r-RNA.
Characteristics needed of the carrier of
genetic information:
 Highly accurate replication
 storage
 transmission
 Large carrying capacity
 Be capable of variation
In a cell there were two candidates for
carriers:
 proteins
 nucleic acids - deoxyribonucleic acid
ribonucleic acid
First indirect evidence
Griffith (1928) Transformation of
Pneumococcus
Two types of Pneumococcus bacteria
 R - rough - nonvirulent
 S - smooth - virulent
Bacteria were injected into mice to observe
the reaction.
Expected
R
S
S
heat killed
R+S
heat killed
Observed
What happened?
Heat-killed S cells transformed R cells
Griffith speculated that the transforming
factor was DNA
First direct evidence
Avery, MacLeod and McCarty (1944)
Repeated Griffith’s research but separated
the heat-killed S cells into component parts
by removing one component at a time
enzymatically:
 protein removed by protease
 RNA removed by ribonuclease
 DNA removed by deoxyribonuclease
Then transformed R cells with each extract
and tested for virulence.
Treatment
expected
observed
protease
R
S
R
S
RNA ase
R
S
R
S
DNA ase
R
S
R
S
Only after treatment with the
deoxyribonuclease did transformation fail to
occur.
Confirming evidence
Hershey and Chase (1952)
Radioactive labeled components of phage
P32 - labels only DNA
S35 - labels only protein
Bacteria were then infected with the P32 or S35
labeled phage, after infection the solution was
put in a blender to separate the bacteria and
phage, and then the solution was tested to
determine where the P32 or S35 was located.
Label
P32
S35
cells
solution
X
X
So it was finally accepted that DNA carried
the genetic information in a cell.
So DNA carried the genetic information.
Question: How did it carry the information?
The key is understanding the structure of
DNA
A. Components of DNA
1) a phosphate group
2) a five carbon sugar
2-deoxyribose
3) four cyclic nitrogen containing bases
a. pyrimidines
Thymine
Cytosine
b. purines
Adenine
Guanine
The components come together to form
nucleotides
nucleotide
example: deoxythymine monophosphate
Other nucleic acid:
Ribonucleic acid (RNA)
Differences:
1) sugar - ribose instead of deoxyribose
2) bases - Uracil replaces Thymine the
other bases stay the same
Uracil
Thymine
3) number of strands - one
B. Structure of DNA
Watson and Crick (1953)
Solved the structure of DNA and in so doing
determined how the genetic information was
stored.
Information used to solve the puzzle:
1. Chargaff’s rules
2. X-ray crystallography of Wilkens and
Franklin
1. Chargaff’s rules
a) amount of Adenine = Thymine
amount of Guanine = Cytosine
b) amount of A + G = T + C
amount of A + T  G + C
2. X ray crystallography pattern indicated
that DNA was in a helix form.
Based on this information Watson and Crick
determined that:
 DNA was a double stranded helix
 each strand consisted of nucleotides held
together by phosphodiester bonds
 the stands were held together by hydrogen
bonds between the bases
AT
2 hydrogen bonds
GC
3 hydrogen bonds
 the strands run anti- parallel
 strands are complementary
Types of DNA
Right-handed Helixes
1.  - DNA:
 23o A diameter
 11 bases/turn
 base pairs tilted in relationship to the axis
2.  - DNA:
 20o A diameter
 10 bases/turn
 base pairs perpendicular in relationship to
the axis
Left-handed Helix
1.  - DNA:
 18o A diameter
 12 bases/turn
 base pairs not parallel
Denaturation/renaturation of DNA
Slow heating and slow cooling of DNA can
give information on the base composition and
the level of unique base sequences in DNA
Denaturation
Slowly heating double-stranded DNA will
break the hydrogen bonds holding the
strands together resulting in two single
strands.
The temperature needed to separate the
strands depends on the base composition of
the DNA. Why?
Renaturation
Slow cooling of the denatured DNA would
allow double-stranded DNA to reform.
The rate of reannealing is related to the
amount of DNA and the percentage of unique
base sequences .
The greater the amount of DNA and/or the
higher the percentage of unique base
sequences the slower the rate of reannealing.
DNA renaturation is graphed using a Cot
plot. The graph is based on:
Co - initial concentration of single-stranded
DNA
C - concentration of single -stranded DNA
after time (t) has elapsed
t - time of the reaction
The Watson and Crick model of DNA with its
double strands and base complementarity
provided the critical information needed to
explain how the genetic information could be
replicated and expressed.
Crick came up with the ‘Central Dogma’ of
genetics
Replication
DNA
RNA
Protein
transcription
translation
Question: How does DNA replicate?
DNA replication
Three theories
1) conservative
2) semi-conservative
3) dispersive
Meselson and Stahl (1958)
Proved DNA replicated in a semiconservative manner.
They used a heavy isotope of nitrogen (N15) to
label DNA in Escherichia coli (E. coli).
N14 normal
N15 heavy
experiment:
1) grow E. coli in N15 media.
2) transfer E. coli to N14 media.
3) extract DNA after various lengths of time.
4) determine DNA densities
Test densities using a cesium chloride (CsCl)
density gradient.
increasing
density
Centrifuge and the DNA will band where its
density is equal to a region in the gradient.