Clear and orderly summary of Chapter 9 of the book "Genetics: Analysis and Principles, 6th Edition by Robert Brooker". Together with all my other summaries of Genetics I got an 8,5 for this course.
Genetic Summary Chapter 9 – Molecular Structure of DNA and RNA
Molecular genetics: the study of DNA structure and function at the molecular level.
9.1 Identification of DNA as the Genetic Material
Genetic material must meet four criteria to fulfill its role
1. Information; genetic material must contain information to construct an
organism.
2. Transmission; DNA is transmitted (overgedragen) from one cell to the next.
From parent to offspring through the egg and sperm.
3. Replication; genetic material must be copied during division (deling). Mistakes
are made so the genes you become from your mother are not exactly the
same as the genes she gives to you.
4. Variation; genetic material must vary so they can account (verklaren) for the
phenotypic differences. Think of mutations
Chromosomes contain DNA and proteins. RNA is found in the vicinity (nabijheid) of
chromosomes. Further research was needed to precisely identify the genetic
material.
Experiments to identify the genetic material
Griffith; transformation (1) Plasmid or segment of
chromosomal DNA is introduced into a bacterial cell (2) a normal
cell is converted (omgezet) into a malignant cell (kwaadaardig).
Transforming principle substance that cause this (this was
unknown)
What it means to genetic terms?
1. Information; the transformed bacteria acquired (verwerft)
the information to make a capsule
2. Variation; this exist in the ability to create a capsule and cause mortality
(sterfte) in mice.
3. Replicated; the genetic material that is necessary to create a capsule
4. Transmitted; transmission from mother to daughter cell during cell
division(deling)
Avery, MacLeod and McCarty; They asked what the transforming principle was?
DNA, RNA, proteins and carbohydrates (koolhydraten) are in living cells. These
people separate these components and look if any of them was the genetic material.
Conclusion, material in the DNA might be the genetic material.
To verify that DNA was responsible for the transformation they treated samples of
DNA with enzymes that digest (verteren)
- DNase; enzyme that cuts the sugar-phosphate backbone in DNA
- RNase; enzyme that cuts the sugar-phosphate backbone in RNA
- Portease; enzyme that digest (verteren) the polypeptide backbone in a protein
Conclusion, DNA is the transforming principle. Because DNA with DNase prevented
conversion (omzetting) of type R to type S
Hershey and Chase;
Bacteriophage or phage is a virus that infects bacterial cells. T2 is a virus, so the T2
phage consist of genetic material. It consists only of DNA and proteins. During
, infection the phage attached on the outside of the bacterium and does not enter the
cell. Only the genetic material enters the bacterial cell.
They asked? What is in the genetic material?
They used radioisotopes to distinguish (onderscheiden) proteins from DNA.
Sulfar atoms are only found in proteins
Phosphorus atoms are only found in DNA.
Radioactive S and P were used. Most of the radioactive P had entered the bacterial
cells whereas most of the radioactive S remained outside the cells.
Conclusion; the genetic material is DNA
9.2 Overview of DNA and RNA Structure
DNA and RNA are nucleic acid repeating nucleotide units.
They release (vrijlaten) hydrogen ions (H+) (= acid) in solution and have a negative
charge at neutral pH. So, that is why they are called nucleic acid.
There are four levels of complexity of DNA:
1. Nucleotides sugar + one to three phosphates + base
2. Strand nucleotides linked together
3. Double helix two strands interact
4. Three-dimensional structure folding(vouwen) and bending (buigen) the
double helix
Proteins influence the structure of DNA
9.3 Nucleotide Structure
Nucleotide
- Sugar; deoxyribose (DNA) or ribose (RNA)
- Phosphate 5’; they are attached to suger via an ester bond.
- Base 1’; (A+T 2 hydrogen bonds and C+G 3 hydrogen bond, in RNA U instead
of T)
o Purines: this base has a double ring structure. A and G.
o Pyrimidines: this base has a
single ring structure. C, T and
U.
o Purines always bond with
pyrimidines, this keep the
width of the double helix
relatively constant.
The nitrogen and carbon atoms in bases
are given numbers like normal
- purines 1 through 9
- pyrimidines 1 through 6.
The carbon atoms in sugars are given numbers like 1’
-OH group of the sugars is on the 3’. The group is
important to form covalent linkages with each other.
Sugar + Base = nucleoside (example. Ribose +
adenine = adenosine AND deoxyribose + adenine =
deoxyadenosine)
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