Problem Set 1 Flashcards
(25 cards)
List two experiments that demonstrated DNA as the genetic material
Avery and Hershey-Chase experiments
Describe the Avery experiment
The Avery experiment used the same R and S type Strep from the Griffith Experiment. Living S type cells were lysed and filtered to remove carbohydrates and lipids. The remaining liquid (containing DNA, RNA, and protein) was treated with one of three enzymes…. Either DNase, RNase, or protease. These enzymes degrade one specific biomolecule but no other. After enzymatic treatment, the solution was exposed to living R type cells and plated. They then looked to see if transformation occurred, indicated by the presence of both R and S type cells. If RNase and protease were used, transformation still occurred, indicating that the genetic material was still present. Application of DNase wiped out the ability to transform the cells (only R grew) indicating that the enzyme had destroyed the genetic material.
Describe the Hershey-Chase Experiment
Hershey and Chase used radioactive isotopes to lab proteins and DNA in a virus. Viruses are limited to
only two biological molecules. By growing the virus in the presence of radioactive phosphorous or
radioactive sulfur, they could make the DNA or the protein of the virus radioactive. After labelling they
then exposed the virus to new bacteria, allowed infection to occur, and then agitated the solution to knock any viral particles of the surface of the cell. Cells were collected by centrifugation and measured for the presence of radioactive isotopes. Any radiation present should be inside the cells if it were part of the molecule that left the virus and invaded the bacteria. Viruses treated with radioactive phosphorous inserted the radiation into the cell, indicating that DNA made it inside the cell. Radioactive sulfur did not enter the cell… what part of the protein that made of up the capsule left outside the cell.
List the Building Blocks of a Nucleotide. With regard to the 5’ and 3’ positions on a sugar molecule, how are nucleotides linked together to form a strand of DNA?
The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and
a phosphate group. In a nucleotide, the phosphate is already linked to the 5’-position on the sugar.
When two nucleotides are linked together, a phosphate on one nucleotide forms a covalent bond with the 3’ hyrdroxyl group on another nucleotide.
What was the significance of the Griffith experiment?
This experiment demonstrated that some chemical could be released from bacteria and make its way inside other bacteria, therefore changing their physical characteristics (R –> S, no disease to disease). Whatever this chemical substance was it must be the genetic material.
Describe the bonding mechanism of the DNA molecule.
Phosphodiester bonds (covalent bonds) hold the sides of the ladder (sugar phosphate backbone) together. Hydrogen bonds between opposite complementary bases hold one strand to the opposite strand.
Draw the structure of DNA
Base pairing in DNA consists of purine-pyrimidine pairs, so why can’t A-C and G-T pairs form?
Pairing between A-C and G-T can occur, but the electronegative atoms do not line up as readily,
making the thermodynamics of these base pairs less stable for the entire DNA molecule. A-T and G-C
pairing is more thermodynamically stable and therefore preferred in nature.
How many of each of the following does this DNA molecule have?
AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 3' hydroxyls b. Hydrogen bonds c. Purines d. Ribose sugars
Two 3’ hycroxyls, one at the end of each strand. 2 hydrogen bonds between AT and 3
between each CG… so a total of 58 hydrogen bonds. G and A are the purines… 23 total. There
must be one purine in one strand or the other at each position. This is DNA so 0 ribose sugars,
46 deoxyribonse.
Which parts of a nucleotide (e.g., phosphate, sugar, and/or bases) are found in the major and minor grooves of double-stranded DNA and which parts are found in the DNA backbone?
The nucleotide bases occupy the major and minor grooves. Phosphate and sugar are found in
the backbone.
If you knew that a DNA-binding protein does not recognize a specific base sequence, would you expect that it recognizes the major groove, the minor groove, or the DNA backbone? Explain your reasoning
If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not binding in the grooves, but instead is binding to the sugar-phosphate backbone. DNA-binding proteins that recognize a base sequence must bind into the major or minor groove of DNA, which is where the bases are accessible to such proteins. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNA-binding proteins such as histones, which do not recognize a base sequence, bind to the DNA backbone.
Let’s suppose an organism has a G + C content of 64% in its DNA. What are the percentages of A, T, G, and C?
G = 32%, C = 32%, A = 18%, T = 18%.
In humans, an average-sized chromosome contains about 100 million bp of DNA. If the DNA in such a chromosome were stretched out in a linear manner, how long would it be?
There are 108 base pairs in this chromosome. In a double helix, a single base pair traverses
about 0.34 nm, which equals 0.34 x 10^–9 meters. If we multiply the two values together:
10^8 (0.34 x 10^–9) = 0.34 x 10^–1 m, or 0.034 m, or 3.4 cm. The answer is 3.4 cm, which equals 1.3 inches!
That is enormously long considering that a typical human cell is only 10 to 100 μm in diameter. The DNA
has to be greatly compacted to fit into a living cell.
Let’s suppose that a DNA molecule is 1 cm long and contains 15% adenine. How many cytosine’s would this DNA molecule contain?
The first thing you need to do is to determine how many base pairs are in this DNA molecule.
10^-2
- ————- = 2.9x10^-7 base pairs
0. 34x10^-92.9x10^-7 base pairs which equals 5.8 x 10^7 nucleotides.
If 15% are adenine, then 15% must also be thymine. This leaves 70% for cytosine and guanine. Because
cytosine and guanine bind to each other, there must be 35% cytosine and 35% guanine. If you multiply
5.8 x 10^7 times 0.35, you get (5.8 x 10^7 )(0.35) = 2.0 x 10^7 cytosines (about 20 million cytosines)
The bases that make up DNA are able to absorb ultraviolet light at 260 nm. This absorbance is highest when bases are not paired with a complementary partner. Two different DNA samples are exposed to different temperatures and their ability to absorb UV light is measured. Explain the following results:
When the strands of DNA are tightly paired, the bases are close together and the absorbance is low. As they pull apart the absorbance goes up until it reaches the maximum. The red sample is able to separate into separate strands at a lower temperature than the blue sample. It takes more energy to break the bonds holding the two strands together in the blue sample, implying that there are more hydrogen bonds in the blue sample than the red, since it takes more energy to separate the strands.
This is due to differences in makeup, the blue sample has more CG base pairs with 3 hydrogen bonds
each, while the red sample probably has more AT base pairs with only 2 H bonds.
List the structural differences between DNA and RNA.
The sugar in DNA is deoxyribose; in RNA it is ribose.
The difference between the 2 sugars comes
down to the presence or absence of OH at carbon 2’. DNA contains the base thymine, while RNA has
uracil.
DNA is a double helical structure. RNA is single stranded, although parts of it may form double-
stranded regions when internal base pairing occurs within a strand of RNA.
Write out a sequence of an RNA molecule that could form a stem-loop with 24 nucleotides in the stem and 16 nucleotides in the loop.
Though many sequences are possible, here is an example of an RNA molecule that could form a
hairpin that contains 24 nucleotides in the stem and 16 nucleotides in the loop.
5’-(GAUCCCUAAACG) GAUCCCAGGACUCCC (ACGUUUAGGGAUC)–3’ The complementary stem
regions are in parenthesis
For each of the following nucleic acid molecules deduce whether it is DNA or RNA, double or single stranded
Molecule
% A % G % T %C % U
A 33 17 33 17 0
B 33 33 17 17 0
C 26 24 0 24 26
D 21 40 21 18 0
E 15 40 0 30 15
F 30 20 15 20 15contains thymine so DNA, Chargaff’s rules so probably double stranded.
B is DNA but C does not = G and A does not = T so single stranded
C. RNA because of uracile presence. Probably double stranded
D. DNA single stranded
E. RNA single stranded
F. One strand of DNA paired with one strand of RNA (like during transcription)
Discuss the advantages and disadvantages of having a compact DNA structure.
Organisms must make their DNA compact so it can fit inside of their cells.
Why do eukaryotic cells make their DNA more compact when they are preparing to divide?
Making it compact allows organisms to carry a bigger genome, more genes, and thereby become more complex. The two main disadvantages are that compact DNA is (1) more difficult to replicate and (2) more difficult to express genes. It is necessary to make it very compact in preparation for cell division to prevent the chromosomes from being tangled up.
Describe in steps how the double helix of DNA, which is 2nm in width, gives rise to a chromosome that is 700 nm in width
The DNA double helix is 2 nm wide. It wraps around a bundle of histone proteins, doing two full revolutions, to form a nucleosome. These series of nucleosomes (beads on a string) is 11 nm wide, providing more than 5X compaction. Each nucleosome includes the protein H1, which is attracted to other H1 nearby. The interaction between the individual nucleosomes
folds into a fiber that is 30 nm wide. The 30 nm fiber then interacts with a group of scaffold proteins, forming large loops that come off a central axis made of the scaffold. The scaffold is then able to fold over on itself multiple times, making a more compact structure until it is 700 nm
wide.
Describe in steps how the double helix of DNA, which is 2nm in width, gives rise to a chromosome that is 700 nm in width
There are 146 bp around the core histones. If the linker region is 54 bp, you expect 200 bp of DNA (i.e., 146 + 54) for each nucleosome and linker region. If you divide 46,000 bp by 200 bp, you get 230. Because there are two molecules of H2A for each nucleosome, there would be 460 molecules of H2A in a 46,000-bp sample of DNA.
In a particular eukaryotic species, the linker DNA (from one nucleosome to the next) averages 54 bp in length. Remember that each nucleosome has 146 bp of DNA wrapped around a histone. How many molecules of H2A would you expect to find in a DNA sample that is 46,000 bp in length?
There are 146 bp around the core histones. If the linker region is 54 bp, you expect 200 bp of
DNA (i.e., 146 + 54) for each nucleosome and linker region. If you divide 46,000 bp by 200 bp, you get
230. Because there are two molecules of H2A for each nucleosome, there would be 460 molecules of
H2A in a 46,000-bp sample of DNA.
Take a look at the figure below. Is this showing an 11 nm fiber, a 30 nm fiber, or a 300 nm fiber? Is this DNA from a cell this is in M phase or interphase?
You are looking at a 30 nm fiber, which is the predominant form of DNA found in the radial
loops of a cell that is in interphase.
