Section 3 Flashcards

1
Q

What is the difference in structure between purines and pyrimidines? How might this affect the properties of the nucleic acids that they make up?

A

Purines have two rings, pyrimidines have one. These rings have alternating single and double bonds, showing resonance.

The resonance imparts partial double-bond character to most of the bonds within the ring structures. As a result, pyrimidines have a planar structure, while purines have a nearly planar structure with a slight curve.

Because of this resonance, the electrons in the rings can absorb UV light at around 260 nm. This is how we detect and measure the concentration of nucleic acids.

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2
Q

What does the molar extinction coefficient (ε) measure, and what are its units?

A

The molar extinction coefficient (ε) measures the amount of light absorbed by a 1M solution with a light path length of 1 cm. Its units can be expressed as per molar per centimeter (M⁻¹ · cm⁻¹), liters per Molar per centimeter (L · mol⁻¹ · cm⁻¹), or litres per gram per centimeter (g⁻¹ · cm⁻¹ · L). This property is specific to the molecule being measured.

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3
Q

What does the ability of bases to absorb UV light depend on?

A

It varies depending on whether or not they exist in their free form or part of a nucleic acid

The ability of nitrogenous bases to absorb UV (ultraviolet) light is influenced by their molecular environment. When nitrogenous bases are in their free form, not bound within a nucleic acid, they are more capable of absorbing UV light. This is because, in a free state, each individual base has the freedom to interact with incoming UV photons and absorb the energy.

On the other hand, when these bases are incorporated into a nucleic acid (such as DNA or RNA), they are arranged in a stacked and close-packed manner. This close interaction between adjacent bases within the nucleic acid structure alters their ability to absorb UV light. In this context, the stacked bases collectively absorb less UV light compared to the same concentration of free nucleotides.

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4
Q

What does Beer’s Law state about the relationship between absorbance and concentration in a solution?

A

Beer’s Law states that the absorbance of light at a certain wavelength is directly proportional to the concentration of the solution. In other words, the darker a solution (higher absorbance), the more concentrated it is as it absorbs more light.

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5
Q

The equation for Beer’s Law is A260 = ε260 * c * l.

What do each of the variables represent?

A

A260 is the absorbance of UV light (optical density) at 260 nm. It has no units.

ε260 is the extinction coefficient [g⁻¹ · cm⁻¹ · L]. This is the light absorbed by a 1 M solution over a 1 cm path-length.

c represents the concentration of the solution. (mol/L)

l is the path length, or the distance that light travels through a sample in an analytical cell. (cm)

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6
Q

The absorbance value for a nucleic acid solution, as measured by a spectrophotometer, is 1. The
path length is a standard 1 cm, and the extinction coefficient is 0.020 g⁻¹ · cm⁻¹ · L.
What is the concentration of the nucleic acid solution?

A

ANSWER: 50 g/L

A₂₆₀ = ε₂₆₀ * c * l
c = A₂₆₀ / (ε₂₆₀ * l)

Given:
ε₂₆₀ = 0.020 g⁻¹·cm⁻¹·L
A₂₆₀ = 1 (as per your calculations)
l = 1 cm

Now, plug in the values:
c = 1 / (0.020 g⁻¹·cm⁻¹·L * 1 cm)

c = 1 / 0.020 g⁻¹·L

c = 50 g/L

So, the concentration (c) of the solution is 50 grams per liter (g/L).

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7
Q

How does the absorption of UV light differ between single-stranded (ss) and double-stranded (ds) DNA, and how does it compare to free nucleotides?

A

The absorption of UV light varies depending on whether the DNA is single-stranded (ss) or double-stranded (ds).

The order of the amount of UV absorption is as follows: dsDNA < ssDNA &laquo_space;free nucleotides.

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8
Q

What does the hyperchromic effect in DNA refer to, and when does it occur?

A

The hyperchromic effect refers to the large increase in light absorption at 260 nm that occurs as double-helical DNA unwinds (or melts). This happens when DNA denatures, and the base pairs are disrupted, leading to the separation of the two strands, forming single-stranded DNA (ssDNA).

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9
Q

Why does single-stranded DNA (ssDNA) exhibit higher UV light absorption than double-stranded DNA (dsDNA) at the same concentration during the hyperchromic effect?

A

During the hyperchromic effect, when DNA transitions from dsDNA to ssDNA, the resonance within the bases in each strand is no longer constrained. As a result, the UV light absorption of ssDNA becomes higher than that of dsDNA at the same concentration.

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10
Q

How can the transition of DNA from double-stranded to single-stranded (or vice versa) be detected?

A

The transition of DNA from double-stranded to single-stranded (or vice versa) can be detected by monitoring the change in absorption of UV light, specifically at 260 nm. The hyperchromic effect, with increased UV absorption during denaturation, is a key indicator of this transition.

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11
Q

What is the hypochromic effect in DNA, and when does it occur?

A

The hypochromic effect refers to the significant decrease in light absorption at 260 nm that occurs as single strands of DNA anneal to form double-helical DNA. This effect is observed when the forces stabilizing the DNA double helix, such as hydrogen bonding and base stacking, limit the amount of resonance that can occur within the aromatic rings of the bases. As a result, there is a reduced amount of UV light absorbed as single-stranded DNA (ssDNA) anneals to form double-stranded DNA (dsDNA).

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12
Q

Identify the correct relative order of UV absorption (from most to least)
Response recorded
Free dNTPs > dsDNA > ssDNA
Free dNTPs > ssDNA > dsDNA
ssDNA > dsDNA > Free dNTPs
dsDNA > ssDNA > Free dNTPs
None of the above

A

Free dNTPs > ssDNA > dsDNA

most absorbance occurs with the most electron freedom, which means least constraint

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13
Q

What is the melting temperature (Tm) of DNA, and how is it defined?

A

The melting temperature (Tm) is the temperature at which 50% of the DNA is in the single-stranded form (ssDNA), and the other 50% is in the double-stranded form (dsDNA). It’s a critical parameter in DNA analysis and can be calculated from the absorbance data, often measured at 260 nm.

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14
Q

What happens to the ability of DNA to resonate as it transitions from a double-stranded to a single-stranded state, and how does this change affect its interaction with light?

A

As DNA shifts from a double-stranded to a single-stranded configuration, it gains a greater capacity to resonate. This heightened resonance and enhanced freedom of electrons lead to a greater absorption of light.

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15
Q

Does DNA absorb more UV light in its single stranded for or double stranded form?

A

DNA absorbs more light in its single stranded form, as the resonance within the nitrogenous bases are no longer constrained by the forces that stabilize the DNA double helix.

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16
Q

How can the process of DNA denaturation and the determination of the melting temperature be used to estimate the base composition of a specific DNA sequence?

A

DNA denaturation and the determination of the melting temperature can help estimate the base composition of a DNA sequence. GC base pairs have three hydrogen bonds and are stronger than AT base pairs with two hydrogen bonds. Consequently, DNA with a higher content of GC base pairs has a higher melting point. By carefully controlling denaturation conditions, regions rich in AT base pairs denature first, creating denatured regions or bubbles that can be visualized using electron microscopy. This information can provide insights into the DNA sequence’s base composition.

17
Q

Why is it sometimes necessary to control the denaturation of DNA in a laboratory setting?

A

DNA denaturation is controlled in the laboratory for various purposes. Two key reasons include:

  • Analyzing the structure and function of DNA, specifically examining the interactions that hold the two DNA strands together.
  • Estimating the amount of G and C bases within a DNA sequence. GC base pairs are stronger than AT base pairs, so sequences with high GC content have higher melting points. DNA denaturation is also utilized in techniques like polymerase chain reaction (PCR), covered in Section 04 of this module.
18
Q

How can you calculate the melting temperature (Tm) of a DNA strand based on its GC content, and what is the formula for this calculation?

Tm = 0.41 (%G+C) + 69.3°C

A

The melting temperature (Tm) of a DNA strand can be determined based on its GC content using the formula: Tm = 0.41 (%G+C) + 69.3°C. Under standard conditions, Tm increases with increasing GC content in a linear relationship.

Example:
To calculate the Tm for a DNA strand with 50% GC content, you can use the formula: Tm = 0.41 (50) + 69.3°C, which equals approximately 90°C.

19
Q

What is important to keep in mind about the melting temperature equation (Tm)

Tm = 0.41 (%G+C) + 69.3°C

A

That the equation is true for long DNA segments and a concentration of NaCL of 0.2 M

It is important to keep the salt concentration constant because more salt means more stability, therefore more heat you have to add to make it melt.

20
Q

What are some external factors that influence Tm?

A

High pH (>9) can cause the deprotonation of nitrogenous bases, reducing H-bonding

21
Q

How can nucleic acids from different animal species form hybrid duplexes, and what is the process of hybridization in molecular biology?

A

In hybridization experiments, two samples of isolated DNA or RNA are denatured by heating and then mixed. As the sample cools, complementary strands can form a duplex, whether it’s a DNA-DNA, RNA-DNA, or RNA-RNA hybrid. It’s worth noting that RNA duplexes are more stable than DNA duplexes.

The stability of an RNA-DNA hybrid falls between that of double-stranded RNA and double-stranded DNA. Due to this increased stability, denaturing a double-helical RNA at neutral pH often requires higher temperatures, typically 20°C or more, than those needed to denature a DNA molecule with a similar sequence.

22
Q

Predict the effect of Salt Concentration on the melting temperature and/or absorbance of DNA.

A

Salt in the sample buffer stabilizes the duplex by stabilizing the negative charge on the backbone - this has the effect of raising the melting temperature in a linear fashion proportionally to the salt concentration.

23
Q

Predict the effect of organic solvents on the melting temperature and/or absorbance of DNA.

A

Increasing organic solvents (eg formamide) in a solution lowers the melting temperature by destabilizing the duplex

24
Q

Predict the effect of DNA concentration on the melting temperature and/or absorbance of DNA.

A

DNA concentration affects the absorbance value, but not the melting temperature, since this is a property of the GC content.

25
Q

What are the three factors that influence DNA hybridization, and how do they affect the hybridization process?

A

DNA Concentration: A more concentrated DNA sample results in more frequent intermolecular collisions, leading to a faster hybridization rate.

Base Pair Mismatch: Mismatched nucleotides in hybridizing nucleic acid segments that are supposed to be completely complementary can disrupt base pairing, stacking interactions, and DNA stability.

pH: In the pH range of 5 to 9, there is only a minor effect on DNA hybridization. However, more extreme pH conditions, whether strongly alkaline or acidic, have a more significant impact on the efficiency of DNA hybridization.
- pH < 5 causes the liberation of all bases from the helix
- pH > 9 causes the conversion of dsDNA to ssDNA

26
Q

What is stringency in the context of nucleic acid hybridization, and how can it be adjusted to influence the hybridization process?

A

Stringency in nucleic acid hybridization refers to the degree of complementarity required between two strands for them to hybridize. High stringency requires high compatibility, while low stringency allows for hybridization even in the presence of some base mismatches. Stringency can be adjusted using factors that affect DNA duplex stability, including temperature, salt, and organic solvents.

Increased stringency is achieved by high temperatures, low-salt concentrations, and the presence of organic solvents. Decreased stringency is achieved with low temperatures, high-salt concentrations, and the absence of organic solvents. These adjustments control the favorability of hybridization in specific conditions.

27
Q

How can stringency be increased in hybridization experiments?

A

Stringency can be increased by using high temperatures close to the melting temperature (Tm) of the hybrid, low-salt concentrations, and the presence of organic solvents.

28
Q

What are the conditions that decrease stringency in hybridization?

A

Stringency is decreased by using low temperatures (below Tm), high-salt concentrations, and the absence of organic solvents, making hybrid formation more favorable.