Biol 405: Lecture 3 (9/2/15)

Front 1

DNA size is expressed in what 3 ways?

Back 1

1. Number of base pairs

2. Molecular eight ~660 is molecular weight of 1 base pair

3. Length ~34angstrom per helical turn of 10.4 base pairs

Front 2

What methods are used to measure DNA size?

Back 2

1. Electron Microscopy

2. Gel electrophoresis

Front 3

What is the most popular way to express DNA size?

Back 3

Number of base pairs

Front 4

What is the formula used to determine molecular weight?

Back 4

Molecular weight = # of Base Pairs * 660

Front 5

What is the formula used to determine Length

Back 5

Length = # of base pairs * H

Front 6

What is the formula used to determine H?

Back 6

Rise/Residue

Front 7

What is H traditionally?

Back 7

3.4 angstroms OR 3.4x10^-4nanometers

Front 8

Given that SV40 virus has 5226 base pairs. What is it's molecular weight?

Back 8

Molecular weight = #bp * 660 -->

MW= 5226*660 -->

MW= 3.5x10^6

Front 9

Given that SV40 virus has 5226 base pairs. what is the length?

Back 9

Length = #bp * H -->

Length = 5226 * 3.4x10^-4

Length = 1.7 nanometers

Front 10

Explain how the same DNA molecule could assume two to be different sizes?

Back 10

The same DNA molecule could assume two different conformation (Relaxed and Supercoiled)

Front 11

Between "relaxed" and "supercoiled" DNA. which moves faster through a CsCl gradient and there for has greater density?

Back 11

Supercoiled DNA

Front 12

Based on the image, which band represents a relaxed DNA?

Back 12

A

Front 13

Based on the image, which band represents supercoiled DNA?

Back 13

B

Front 14

What reasons would supercoiled DNA have to have a higher density than relaxed DNA?

Back 14

1. Supercoiled DNA binds to more CsCl

2. It has a smaller molecular volume, hence more molecules packed in a given volume, leading to a higher density.

Front 15

The number of supercoils in a DNA molecule

Back 15

Writhe

Front 16

The positive # indicating the total number of turns and supercoils regardless of whether the supercoils are negative

Back 16

Twist

Front 17

The total number of turns and supercoils taking into account positive and negative integers

Back 17

Linking number

Front 18

Given the following information. What would happen if a DNA molecule were nicked and unwound 2x.

Twist= 25

writhe= 0

Linking number = 25

Back 18

If DNA were nicked and unwound two times then it will spontaneously negatively supercoil right-handedly. This will cause the following

Twist=25

Writhe= -2

Linking number=23

Front 19

Given the following information. What would happen if a DNA molecule were nicked, unwound 2x, and the molecule doesn't spontaneously supercoil

Twist= 25

writhe= 0

Linking number = 25

Back 19

If the DNA were left as is and not allowed to supercoil then the following would occur.

Twist= 23

Writhe= 0

Linking number 23

Front 20

What would happen to the Writhe if DNA was overwound

Back 20

If DNA was overwound then it would result in a left-handed turn and a positive supercoil resulting in positive writhe number

Front 21

What would happen to the Writhe if DNA was underwound

Back 21

If DNA was underwound then it would result in a right-handed turn and a negative supercoil resulting in negative writhe number

Front 22

Which of the following must be integers?

1. Writh

2. Linking number

3. Twist

Back 22

Linking number

Front 23

The free energy is minimal when about ______ of the change in L is expressed in T and ______ in W

Back 23

30% ; 70%

Front 24

Given that Linking Number change is 2 what is the most stable (geometrical) isomer if T0=25 and W=0.

Back 24

25-2*0.3 = 24.4

T=24.4

0-2*0.7 = -1.4

W=-1.4

Front 25

What is the formula to determine Supercoiling Density?

Back 25

(L-L0)/L0

Front 26

What happens to the DNA molecule when histones are removed

Back 26

30-nm fibers and nucleosomes disappear

Front 27

Can nicks be used to relax supercoiled DNA?

Back 27

yes

Front 28

The lagging strand is also known as the ______________ and the leading strand is also known as the ________________

Back 28

Discontinuous strand ; continuous strand

Front 29

Label the 5' and 3' end of each DNA fragment shown

Back 29

Front 30

How many replication forks are shown here?

Back 30

2

Front 31

Describe the experiment performed by Matthew Meselson and Franklin Stahl perform to detemine if DNA replicates conservitively, semiconservatively, or through random replicates

Back 31

Meselson and Stahl ran a DNA molecule amongst N15 heavy isotopes for 3 generations. This ensured that all of the DNA in the last generation had the N15 then They moved these DNA molecules in an environment of light N14 isotopes. They then allowed the DNA to replicate for 2 generations. After each generation they measured the DNA through a CsCl gradient and photographed their results under UV light. If the DNA replicated conservitively then after the first generation of DNA molecules would show one two distinct bands one with heavy and the other with light. If the they replicated dispersively then after the first and second generation there would only be one band indicating that the DNA molecules contained both the heavy and light isotopes in the CsCl gradient. If they replicated semiconservatively then after the first generation there would be one band and then after the second generation there would be 2 bands. They found that the DNA molecules replicate semiconservatively.

Front 32

Matthew Mesleson and Franklin Stahl determined that DNA replicates semiconservatively. What isotopes did they use and why? Were they radioactive?

Back 32

N15 (heavy) and N14 (light) were used in order to be used in a CsCl gradient which separates molecules based on density. These are not radioactive isotopes.

Front 33

Why did Meselson and Stahl wait 3 generations before introducing the DNA molecules to the N14 solution?

Back 33

By waiting 3 generations, Meselson and Stahl ensured that all of the DNA molecule were N15 labeled.

Front 34

Using the image, which of generation shows that the DNA is H/L

Back 34

1.0

Front 35

Based on the image at what generation does DNA show a L/L and H/L band

Back 35

Begining from 1.9-2.5

Front 36

Using the image explain why H/H are no longer seen in later generations?

Back 36

H/H is no longer seen because the DNA is not exposed to the N15 solution (heavy) and is only in the N14 (light) solution. Thus any replication that goes on will only progress with N14.

Front 37

T/F: The following image taken by Meselson and Stahl was taken by a radiograph

Back 37

False. The isotopes used were not radioactive and therefore no radiograph was used. The image shown was taken using a camera under UV light.

Front 38

T/F: For many years "labeled" has been synonymous with "radioactive"

Back 38

True

Front 39

Why are radioactive tracers used as opposed to nonradioactive tracers?

Back 39

Radioactive tracers allow vanishingly small quantities of substances to be detected

Front 40

T/F: All labels are radioactive

Back 40

False. Such examples of unradioactive tracers include N15

Front 41

Radioactive emissions from a sample create photons of visible light detected by a photomultiplier tube

Back 41

Liquid Scinitllation Counting

Front 42

Describe the process of Liquid Scintillation Counting

Back 42

1. Radioactive material is put into a vial containing scintillation fluid.

2. Fluid containing fluor fluoresces when hit with radioactive emissions.

3. The Liquid Scintillation Counter acts to convert invisible radioactivity into visible light when radioactive emissions hit the Fluor

4. The unit is counted per minute or CPM

Front 43

If Liquid Scintilation Counting material uses Fluor (a fluorescent material) and should then be visible to be counted, why would radioactive emissions be required?

Back 43

The radioactive emissions hit the Fluor which emit fluorescence. This process converts radioactivity into visible light which is then counted.

Front 44

Knowing that DNA replication is semi-conservative. The next question is to determine if DNA proceeds through replication Continuously, Discontinuously, or Semi-Discontinuously. What experiment was performed to determine if DNA replication proceeded Continuously?

Back 44

In order to determine if DNA replication proceeded continuously.

Front 45

Using the image, which of the following (A, B, or C) would not be a place where to label thymidine with tritium?

Back 45

C. Tritium can't label this -OH because this H+ is lost when it is built on the DNA strand.

Front 46

What organism was used when demonstrating the semidiscontinuous mode of DNA replication?

Back 46

T4

Front 47

How were thymidine labeled in order to demonstrate the semidiscontinuous mode of DNA replication?

Back 47

It was labeled with Tritium.

Front 48

Explain how Okazaki fragments were used to demonstrate the semidiscontinuous mode of DNA replication

Back 48

Okazaki fragments are used to demonstrate that DNA replication does not occur continuously.

Front 49

Describe the experiment performed based on the image

Back 49

The image is the results of an experiment used to demosntrate the semidiscontiuous mode of DNA replication. T4 bacteriophage was the organism used. Thymidine were labeled with tritium. As thymidine is added to the growing DNA strands, they will label the strand. The label allowed experimenters to determine if DNA replication produced okazaki fragments. If okazaki fragments are formed and then later ligated during DNA replication then it will be shown that DNA replication proceeds through a continuous mode. In order to determine which is true, experimenters eliminated ligating enzymes responsible for the joining of okazaki fragments. The graph onthe left shows normal DNA replication. The longer the Thymidine labels were present, the longer the strands formed. The graph on the right showed demonstrated disabled ligase. The results of the graph on the right show that a great majority of replicated DNA were actually short pieces. These implied that DNA replicates in small chunks and are then ligated later. This showed that DNA does not replicate continuously. However at thi point it is not known whether it proceeds discontinuous or semi-discontinously.

Front 50

What does the dut gene do?

Back 50

Dut gene codes for the dUTPase which converts dUTP to dUMP, hence preventing dUTP (as dUMP) from being incorporated into the DNA.

Front 51

What is the ung gene responsible for?

Back 51

Ung gene codes for Uracil N-glycosylase. It removes uraicl molecule that have been incorporated into DNA.

Front 52

What two mechanisms keep DNA "dUMP-free"?

Back 52

1. The dUTPase (encoded by dut gene) keeps the ratio of dUTP/dTTP low in the cell so that few dUTP molecules can be incorporated into DNA

2. The uracil-N-glycosylase (encoded by the ung gene) removes the uracil group from DNA if any dUMP has been incorporated into DNA

Front 53

How did Okazaki determine that DNA replicates semidiscontinuously?

Back 53

Okazaki developed a dut+ung- mutant which minimized the incorporation of dUMP (presence of dUTPase) and formation of abasic sites (no uracil-N-glycosylase). He found that 50% o newly labeled DNA in these cells was still in short pieces. Hence, the conclusion is that DNA replication is semidiscontinuous

Front 54

Why did Okazaki develop mutants to determine if DNA replicated semidiscontinuously

Back 54

Ung produceds Uracil-N-glycosylase which is responsible for removing uracil from growing DNA strands which would create abasic sites and nicks. WIthout ung, abasic sites won't form and neither will nicks. With that he found that 50% of newly formed DNA were still in short pieces while the other 50% were in long continuous strands.