Lecture 5

Question 1

What is the repeat C of a helical DNA?

Answer 1

• it is the distance on the DNA helix where the DNA exactly repeats itself. This is parallel to the axis

Question 2

What does the value “m” indicate in the x-ray diffraction of DNA experiments?

Answer 2

• it is a values designated as the variable number of polymer residues between a repeat of DNA

Question 3

What does the value “P” pitch mean in the x-ray diffraction of DNA?

Answer 3

• It is the distance parallel to the axis where the helix makes exactly one turn.
• Essentially, the distance between the beginning to the end of one turn is called the pitch,

Question 4

What does the value “h” rise indicate on the x-ray diffraction of DNA?

Answer 4

– it is the distance parallel of the axis from residue to residue.

–Essentially, the “h” is the distance from residue A to residue B on the helix

Question 5

How do you calculate the distance between each residue?

Answer 5

– you use the distance where the dna exactly repeats itself and divide it by the number of integral residues in that repeat.

– I.e if a repeat distance is 10 and there are 5 residues in the repeat, then the distance between each residue will be 2. This means that the distance between every residue is only 2

Question 6

According to the more common structure of right handed b DNA, what are its characteristic repeat, pitch, and rise? What does this indicate about the amount of residues per turn and the correlation between pitch and repeat?

Answer 6

• • DNA has a repeat distance C of exactly 10.5 residues.
• • At the same time, the amount of residues per turn, p, is equal to the repeat residue in the right handed b dna.

– The distance between each residue, h, is 0.34 nm.

– This means the distance of the pitch and repeat is 3.4nm

Question 7

T or F, %A = %T and %G = %C and (%A + %T) = (%G + %C)

Answer 7

False; %A = %T and %G = %C; BUT (%A + %T) =/= (%G + %C)

Question 8

True or false. AT and GC are the same distance apart from their respective C’1 and this ensures a uniform diameter

Answer 8

True

Question 9

Describe chracteristics of B-form DNA right handed helix

Answer 9

• bases are inside helix w/ phosphate backbone and sugars on outside

– pattern of backbone is –> phosphate-sugar-phosphate-sugar

Question 10

T or F, bases aren’t visible in major and minor grooves

Answer 10

False, they are visible, major groove has a larger opening, which makes it more accessible to DNA binding proteins

Question 11

What is the purpose of having the bases stacked close to each other? What does new updates models show about the actual stacking?

Answer 11

• This maximizes the van der waal interactions to stabilize the structure.
• New models show that the base pair are rotated slightly. (36 degrees)

– This is most likely the result to maximize interactions as well as to accommodate the 10 base pairs per turn in the helix of DNA

Question 12

What interaction allows essentially for the secondary structure of DNA to occur?

Answer 12

Hydrogen bonds between nitrogenous bases

Question 13

Based on your knowledge of secondary DNA structure, why is it favorable for the bases to be grouped into the interior of the structure and the backbone towards the outside?

Answer 13

– Since the bases are hydrogen bonded, it would be favorable to remove them from any environment that would resulted in disruption of these H bonds.

– If they weren’t shielded from the aqueous environment, they would probably H bond with something else and denature.

– At the same time, the negative charge of the backbone because of the phosphate is stabilized by the charge-dipole interactions with the aqueous environment, making the structure more stable.

Question 14

What do chargaff’s rule allow you to do?

Answer 14

– It allows you to calculate the amount of the other bases as long a you know one bases percentage since purines: pyrimidine had to have a overall 1:1 ratio.

Question 15

In a collection of DNA, there are found to be 500 bases. Further analysis showed that 20% of these bases were adenine. Based on this information, how many bases are guanine? How can this be?

Answer 15

– 150. Since 20% of the bases are adenine, 20% are T. this equals 40% A+ T. Since bases exist in 1:1, the other 60% of the bases are resulting from G + C.

– Since G+ C exist in equal amount, there is 30% Guanine. 30% of guanine is equal to 150 guanine bases. This seems confusing but this doesn’t violate the purine:pyrimidine ratio of 1:1.

– Purines are A + G and pyrimidines are the C + T. Amount of A is 100 and amount of G is 150. amount of C is 150 and amount of T is 100. This equals 250 purines to 250 pyrimidines, which follows the 1:1 ratio.

Question 16

Which type of DNA is the most common form?

Answer 16

– B form Right handed

Question 17

What is another important finding about the B form right handed helix that allowed watson and crick to make a conclusion ?

Answer 17

– The structure of this secondary structure allowed them to conclude that the helix is antiparallel.

Question 18

A certain sequence of DNA is 5'AATCTG3'. Which of the following is its complementary sequence?
A) 5'TTAGAC'3
B) 5'CAGATT3'
C)5'ATTTGC'3
D)5'GATCGC'3

Answer 18

B –> written 5’ to 3’

Question 19

What are other implications of chargaffs rules?

Answer 19

– Since A binds with T and G binds with C, it allowed for inferences to be made about how replication occurs.

– Since each strand was complementary, the sequence of one defines the sequence of the expected replicated strand

Question 20

What were the 3 models of DNA replication proposed by Watson and Crick based on conclusions made from chargaffs rules ?

Answer 20

• conservative
• dispersive
• semiconservative

Question 21

How does conservative replication differ from semi conservative and dispersive?

Answer 21

– in conservative, the parental DNA is left as is and it is collectively replicated to form an entirely new duplex helix. There is no unwinding of the parental .

– In semi-conservative, each parent strand is “ripped off” and replicated via chargaff rule and two new strands different from the original parent are observed where one strand is parental and one strand is daughter.

– In dispersive, there is just a mix of parental and daughter in each strand

Question 22

What occurred during the Meselsohn and Stahl experiment?

Answer 22

– used density gradient centrifugation in CsCl to separate DNA species of different densities

– E-coli was grown in heavy nitrogen initially so that the parental dna consisted of only heavy nitrogen.

– They switched the medium to a light nitrogen mixture and saw that the after replication of the F1 generation, the density of the DNA exited in an intermediate between the heavy and light nitrogen.

– This provided evidence for semi-conservative since the intermediate density showed that one strand was from the parental DNA (heavy) and one strand a daughter strand (the light nitrogen used to replicate).

–Subsequent generations showed that daughters resulted in only light nitrogen, again indicated more evidence for the semiconservative method.

Question 23

What are other secondary structures of nucleic acids? What is its characteristics and what may induce this conformation?

Answer 23

– There are also right handed A DNA. This DNA type has 11bp/turn and has a rise of 0.255nm compared to a 0.34 nm between each nucleotide.

– This indicates nucleotides are closer together and the helix has wider turns.

– This is seen in RNA RNA and DNA RNA hybrids. This conformation can be induced by low-humidity environment

Question 24

What is the tertiary structure of DNA?

Answer 24

• • supercoil
• • important when DNA needs to be packaged into a nucleus
• • supercoiling can allow for compaction in a cell

Question 25

What is the difference between negative and positive supercoils. Whats their purpose?

Answer 25

– One is results from undertwisting (neg.) and the other from over twisting. (pos.)

– IT is done to relieve helical stress from the under/overtwisting that was at a higher energy state and very stressed

Question 26

What are methods for secondary structures of single-stranded nucleic acids?

Answer 26

In summary: they can be in a random coil, stacked bases (h bonding) or even partially double-stranded by self complementary sequences (H - bonding)

Question 27

True or false. Random coiled strands of DNA have higher energy than native DNA?

Answer 27

True

Question 28

What is the purpose of tRNA?

Answer 28

• involved in protein synthesis, have extensive regions of double-stranded structures that create an overall 3^0 structure
• important to its function

Question 29

On an absorbance temp graph, what does an abrupt change in absorbance indicate?

Answer 29

– It is seen as the melting point temperature of the native DNA

Question 30

True or false. free energy is positive at low temps for the formation of random coils of DNA from a helix but negative at high temperature.

Answer 30

True

Question 31

T or F, A/T pairs are stronger than G/C pairs as they have 3 H bonds instead of 2

Answer 31

False, G/C pairs need more heat to be broken

Question 32

Based on the gibbs free energy equation, if free energy is positive at low temps and negative at high temps, what could the signs of enthalpy and entropy be?

Answer 32

– The only way for free energy to become negative as temperature increases is for the change in enthalpy to be positive and entropy to be positive.

– This would mean that going from helix to random coil requires an addition of heat to break bonds which makes sense.

– As temperature increases, the significance of the T* entropy increases, eventually overcoming the unfavorability of the enthalpy, making the free energy negative.

– This would make the formation of the random coil favorable.

– Having any other sign would not allow this to occur. A negative enthalpy and negative entropy values would go from negative free energy to positive (the reverse).

– Positive enthalpy but negative entropy will never be favorable.

– Negative enthalpy and positive entropy will result always result in a favorable reaction and would be very bad for DNA since it will want to spontaneously denature at any temperature

Question 33

What occurs at the Tm for a DNA molecule? (on graph)

Answer 33

– At the Tm, the molecule becomes denatured and is split from its alpha helix form and a sudden abrupt change in absorbance of the structure occurs.

– A higher absorbance ( ability to absorb stuff) is experienced at a given wavelength of energy for a denatured product.