exam 1 Flashcards

1
Q

Define “chemistry”

A

study of matter and how it changes

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2
Q

Define what makes a thing “living”

A
  1. Replication
  2. Complex
  3. Organized
  4. Carries out energy transformations
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3
Q

List properties of biomolecules that make them ideal for supporting life

A
  1. most of them are organic compounds
  2. specific shapes and dimensions
    3.functional groups determines their chemical properties
    4 asymmetric
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4
Q

Distinguish between prokaryotic and eukaryotic cells

A

Prokaryotic( archaea and bacteria)
cells lack a true nucleus, have a simpler structure, and often have a single circular DNA molecule, while eukaryotic cells have a distinct nucleus, are more complex with membrane-bound organelles, linear chromosomes, and a larger size (plant, animals, fungi, protists)

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5
Q

Qualitatively describe enthalpy and entropy

A

enthalpy relates to the total energy content of a system and the heat exchanged during processes, while entropy relates to the degree of disorder or randomness within a system.

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6
Q

Use the relationship between enthalpy, entropy, and Gibbs free energy to determine whether or not a particular
chemical reaction is favorable

A

These relationships help us determine whether a particular process is spontaneous (favorable) or non-spontaneous (unfavorable)

The Gibbs free energy equation is as follows:

∆G = ∆H - T∆S

Where:

∆G is the change in Gibbs free energy.
∆H is the change in enthalpy.
∆S is the change in entropy.
T is the absolute temperature in Kelvin.

Spontaneous at given temp if:
If ∆G < 0 (negative), the process is spontaneous in the forward direction.
If ∆G > 0 (positive), the process is non-spontaneous in the forward direction
If ∆G = 0, the system is at equilibrium

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7
Q

Relate standard state Gibbs free energy and equilibrium constant of a chemical reaction

A

∆G° = -RT ln(K)

Where:

∆G° is the standard state Gibbs free energy change for the reaction.
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K)).
T is the absolute temperature in Kelvin.
ln represents the natural logarithm.
K is the equilibrium constant for the reaction.

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8
Q

Calculate Gibbs free energy for a system not at equilibrium

A

∆G = ∆G° + RT ln(Q)

Where:

∆G is the Gibbs free energy change for the non-equilibrium state.
∆G° is the standard state Gibbs free energy change.
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K)).
T is the absolute temperature in Kelvin.
ln represents the natural logarithm.
Q is the reaction quotient, which is a measure of the ratio of the concentrations (or activities) of products to reactants in the non-equilibrium state.

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9
Q

Describe how the structure and chemical properties of water give rise to solvation, dialysis, and clathrate and
micelle formation

A

Solvation: Water’s ability to form hydrogen bonds with polar or charged substances allows it to disperse and surround solute particles, leading to the dissolution of solutes in aqueous solutions.

Dialysis: Water’s small size and ability to form hydrogen bonds enable it to pass through semipermeable membranes during dialysis, separating solute molecules from a solvent.

Clathrate Formation: Water molecules can trap small molecules or gases within their lattice-like structure, forming clathrate hydrates. This is significant in natural environments like deep-sea sediments where methane hydrates are found.

Micelle Formation: Amphiphilic molecules self-assemble in aqueous solutions, with hydrophobic regions clustering in the core and hydrophilic regions facing the surrounding water molecules. This property is crucial in processes like emulsification and the absorption of dietary fats.

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10
Q

Illustrate how nonpolar molecules spontaneously separate from water and how this represents the state of
greatest entropy

A

The spontaneous separation of nonpolar molecules from water is driven by the principle of increasing entropy. When nonpolar molecules are introduced into a polar solvent like water, they disrupt the structured arrangement of water molecules due to their lack of polar groups. To maximize disorder and entropy, nonpolar molecules tend to spontaneously aggregate or separate from water. This separation, known as the hydrophobic effect, reduces unfavorable interactions, releases water molecules from their ordered state, and leads to a system with greater entropy. In this separated state, both water and nonpolar molecules adopt a more disordered arrangement, representing a state of maximum entropy.

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11
Q

Define amphiphilic molecules and describe their behavior in water

A

Micelle Formation: Amphiphilic molecules spontaneously organize into micelles, with hydrophilic heads facing outward in contact with water and hydrophobic tails clustering in the core to minimize contact with water. This structure stabilizes the system.

Emulsification: Amphiphilic molecules are used to create stable emulsions by surrounding and dispersing hydrophobic substances (like oils) in water. The hydrophilic heads interact with water, while the hydrophobic tails interact with the hydrophobic substances, preventing them from coalescing.

Biological Significance: In biological systems, amphiphilic molecules like phospholipids form the basis of cell membranes, with hydrophilic head groups interacting with the aqueous environment and hydrophobic tails creating the lipid bilayer.

Detergent Action: Many detergents and surfactants are amphiphilic molecules that can solubilize hydrophobic substances, allowing them to be removed in water due to their interaction with the hydrophilic heads of the detergent molecules.

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12
Q

Describe and rank the relative strengths of non-covalent (electrostatic, hydrogen bond, van der Waals)
interactions in water

A

Ranking the relative strengths of these interactions in water:

Electrostatic Interactions (Strongest)
Hydrogen Bonds
Van der Waals Interactions (Weakest)

Electrostatic Interactions: These are the strongest non-covalent interactions. They arise from the attraction between oppositely charged ions or molecules. In water, they play a vital role in dissolving salts and stabilizing charged biomolecules.

Hydrogen Bonds: Hydrogen bonds are weaker than electrostatic interactions but stronger than van der Waals forces. They form between hydrogen atoms covalently bonded to electronegative atoms (e.g., oxygen) and neighboring electronegative atoms. In water, hydrogen bonds contribute to its unique properties and are essential for biological molecule stability.

Van der Waals Interactions: These are the weakest of the three non-covalent interactions. They result from temporary fluctuations in electron distribution, inducing temporary dipoles in molecules. In water, van der Waals interactions are involved in the attraction between nonpolar molecules and are significant in hydrophobic interactions.

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13
Q

Relate hydrogen ion concentration to a pH scale

A

pH = -log[H⁺]

A pH value less than 7 indicates acidity, with lower values indicating stronger acidity (higher [H⁺] concentration).
A pH value greater than 7 indicates alkalinity, with higher values indicating stronger alkalinity (lower [H⁺] concentration).

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14
Q

Use the Henderson-Hasselbalch equation to determine the pH of buffered solutions

A

pH = pKa + log([A⁻]/[HA])

Where:

pH is the pH of the buffered solution.
pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka) of the weak acid (or base) in the buffer.
[A⁻] is the concentration of the conjugate base.
[HA] is the concentration of the weak acid.

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15
Q

Describe how weak acid/conjugate base pairs function as buffers

A

eak acid/conjugate base pairs function as buffers by maintaining a stable pH in a solution. They achieve this by shifting their equilibrium to absorb or release H⁺ ions in response to changes in acidity or alkalinity, thereby resisting large pH changes.

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16
Q

Draw the structures of five nitrogenous bases: adenine, cytosine, guanine, thymine, and uracil

A
17
Q

Build nucleosides and nucleotides from nitrogenous bases, ribose sugars, and phosphates

A
18
Q

Describe in detail the B-form structure of DNA first proposed by Watson and Crick

A

Helical Nature: DNA forms a right-handed helix, with two antiparallel strands twisted around each other.

Base Pairing: Complementary base pairs (A-T and C-G) are held together by hydrogen bonds, providing stability.

Sugar-Phosphate Backbone: A repeating sugar-phosphate backbone runs along the outside of each DNA strand.

Antiparallel Strands: DNA strands run in opposite 5’ to 3’ and 3’ to 5’ directions.

Major and Minor Grooves: The helix creates major and minor grooves, important for protein binding and DNA recognition.

Base Pairing Rules: Adenine pairs with thymine (A-T) via two hydrogen bonds, and cytosine pairs with guanine (C-G) via three hydrogen bonds.

This structure efficiently stores and transmits genetic information, ensuring accurate replication and transcription of DNA.

19
Q

Draw a schematic diagram of the Central Dogma of Molecular Biology

A
20
Q

Define the following terms: replication, transcription, and translation

A

Replication: This process involves making an exact copy of DNA. It occurs during cell division and ensures that each daughter cell receives a complete set of genetic information in the form of DNA.

Transcription: Transcription is the synthesis of an RNA molecule using a DNA template. During transcription, an RNA molecule, known as mRNA, is created, carrying genetic information from DNA to the ribosome for protein synthesis.

Translation: Translation is the process of converting the information in mRNA into a sequence of amino acids to form a protein. Ribosomes read the mRNA codons, and transfer RNA molecules bring the corresponding amino acids to build the protein.

21
Q

Describe the roles of three types of RNA (mRNA, rRNA, and tRNA) during translation

A

Messenger RNA (mRNA):

Role: Carries genetic information from DNA to the ribosome.
Function: mRNA provides the template for protein synthesis by specifying the sequence of amino acids through codons. It acts as the “message” for protein production.
Ribosomal RNA (rRNA):

Role: Serves as a structural component of ribosomes.
Function: rRNA forms the framework of ribosomes, facilitating the interaction between mRNA, tRNA, and amino acids. It provides the physical platform for protein synthesis.
Transfer RNA (tRNA):

Role: Delivers amino acids to the ribosome during translation.
Function: tRNA molecules are specific to particular amino acids and contain anticodons that complement mRNA codons. They ensure the accurate incorporation of amino acids into the growing protein chain.

22
Q

Draw the chemical reaction catalyzed by nucleic acid polymerase enzymes

A
23
Q

Illustrate how dideoxy-NTPs can be used to sequence DNA by the Sanger method

A

Template DNA Preparation: Start with the DNA sample you want to sequence.

Primer Annealing: Attach a known DNA primer to the template.

DNA Polymerase and dNTPs: Add DNA polymerase and regular deoxyribonucleotide triphosphates (dNTPs) for DNA synthesis.

Dideoxy-NTPs: Include labeled dideoxyribonucleotide triphosphates (ddNTPs) that lack the 3’-OH group needed for chain elongation.

DNA Synthesis: DNA polymerase extends the primer, but when a ddNTP is incorporated, chain termination occurs.

Multiple Reactions: Set up separate reactions for each ddNTP (ddA, ddT, ddG, ddC) alongside regular dNTPs.

Fragment Separation: Run terminated DNA fragments through electrophoresis to separate them by size.

Fluorescent Detection: Detect fluorescence emitted by labeled ddNTPs as fragments move through the gel or capillary.

Data Collection: Record fluorescence data, with each peak representing a terminated DNA fragment.

Base Calling: Analyze data to determine the DNA sequence based on the order of peaks.

Sequence Reconstruction: Combine data from each ddNTP reaction to obtain the full DNA sequence.

24
Q

List key differences between DNA and RNA Polymerases

A

Function:

DNA Polymerase: Synthesizes new DNA strands during DNA replication and repair.
RNA Polymerase: Synthesizes RNA molecules from a DNA template during transcription.
Product:

DNA Polymerase: Produces a complementary DNA strand using deoxyribonucleotide triphosphates (dNTPs).
RNA Polymerase: Produces an RNA molecule using ribonucleotide triphosphates (NTPs).
Nucleotides Used:

DNA Polymerase: Uses dNTPs containing deoxyribose sugar and thymine (T).
RNA Polymerase: Uses NTPs containing ribose sugar and uracil (U) instead of thymine (T).
Initiation Site:

DNA Polymerase: Requires a primer to initiate synthesis.
RNA Polymerase: Can initiate transcription without a primer, recognizing promoter sequences.
Editing and Proofreading:

DNA Polymerase: Typically has proofreading capabilities to correct errors in DNA synthesis.
RNA Polymerase: Generally lacks proofreading mechanisms, resulting in a higher error rate.
Final Product Length:

DNA Polymerase: Synthesizes long DNA strands, potentially entire chromosomes.
RNA Polymerase: Produces shorter RNA molecules, such as mRNA, rRNA, and tRNA.
Role in Genetic Information Flow:

DNA Polymerase: Mainly involved in DNA replication and repair.
RNA Polymerase: Involved in the transcription of DNA into various types of RNA for protein synthesis and cellular processes.

25
Q

List the reagents necessary to carry out the polymerase chain reaction (PCR) and describe how it works

A

Denaturation: The reaction starts with heating DNA to a high temperature (94-98°C), causing the DNA strands to separate (denature).

Annealing: The temperature is lowered (50-68°C), allowing DNA primers to anneal (bind) to their complementary sequences on the single-stranded DNA template.

Extension: The temperature is raised (68-72°C), and DNA polymerase extends the primers by adding complementary nucleotides to the template strand, synthesizing new DNA strands.

Cycle Repetition: These steps (denaturation, annealing, extension) are repeated 20-40 times, doubling the amount of DNA each cycle.

Final Extension: After the final cycle, there is an extension step to ensure complete strand synthesis.

26
Q

Describe separation of DNA fragments by agarose gel electrophoresis

A

Agarose Gel Preparation: Agarose is mixed with electrophoresis buffer, poured into a gel mold, and allowed to solidify.

DNA Sample Preparation: DNA samples are mixed with a loading buffer and loaded into wells on the gel, often with a DNA ladder for size reference.

Electrophoresis Setup: The gel is placed in an electrophoresis chamber filled with buffer. Electrodes are connected to create an electric field from negative to positive.

Electrophoresis: DNA fragments, negatively charged, migrate through the gel towards the positive electrode. Smaller fragments move faster and separate from larger ones.

Monitoring: Progress is monitored by tracking dye or DNA ladder movement. Electrophoresis is stopped when fragments have separated adequately.

Visualization: The gel is stained with a fluorescent dye and visualized under UV light, revealing a pattern of bands corresponding to DNA fragments.

Analysis: DNA fragment sizes are determined by comparing their migration distances to standards. The gel image provides information about fragment sizes and quantities.

26
Q

Illustrate the function of restriction endonuclease and DNA ligase enzymes

A

Restriction Endonucleases:

Function: Recognize specific DNA sequences and cleave DNA at or near these recognition sites.
Illustration: They cut DNA, creating fragments with complementary ends.
Applications: Used in DNA cloning and genetic engineering to create recombinant DNA molecules.
DNA Ligase:

Function: Catalyze the formation of phosphodiester bonds between adjacent nucleotides in DNA strands, sealing breaks or nicks in the DNA backbone.
Illustration: Joins DNA fragments by sealing single-stranded overhangs created by restriction enzymes.
Applications: Essential for DNA replication, repair, and the creation of recombinant DNA molecules.

27
Q

Show how plasmid vector DNA can be used for molecular cloning

A

Select a Plasmid Vector: Choose a suitable circular DNA plasmid as the cloning vehicle.

Prepare Plasmid DNA: Isolate the plasmid vector DNA from bacteria.

Select DNA Fragment: Identify the DNA fragment to be cloned, obtained through PCR, restriction enzyme digestion, or synthesis.

Prepare DNA Fragment: Modify the DNA fragment if necessary, ensuring it has compatible ends for ligation.

DNA Fragment Ligation: Mix the plasmid vector and DNA fragment, using DNA ligase to join them.

Transformation: Introduce the ligated DNA into bacterial cells via transformation.

Select Transformants: Plate the transformed bacteria on selective medium; only those with the plasmid will grow.

Screen for Clones: Identify bacterial colonies containing the desired DNA fragment through screening methods.

Expansion and Isolation: Grow and isolate bacteria with the cloned DNA fragment.

Verification: Confirm the identity and integrity of the cloned DNA fragment, often by sequencing.

Applications: Use the cloned DNA for various molecular biology applications, such as gene expression studies or protein production.

28
Q

Name the twenty common amino acids, draw their structures, and give their three- and one-letter code names

A
29
Q

Determine the formal charge on amino acids at different pH

A

Low pH (Acidic Conditions):

Amino acids have a net positive formal charge.
The amino group (NH2) is protonated (NH₃⁺), and the carboxyl group (COOH) remains in its acidic form (COOH).
Intermediate pH (Neutral Conditions):

Amino acids may have a formal charge close to zero.
Both the amino group (NH2) and the carboxyl group (COOH) may be partially protonated or deprotonated.
High pH (Basic Conditions):

Amino acids have a net negative formal charge.
The amino group (NH2) remains deprotonated (NH2), and the carboxyl group (COOH) becomes fully deprotonated (COO⁻).

30
Q

Classify amino acids according to nonpolar, polar uncharged, and polar charged

A

Nonpolar Amino Acids (Hydrophobic):

Alanine (Ala, A)
Valine (Val, V)
Leucine (Leu, L)
Isoleucine (Ile, I)
Methionine (Met, M)
Phenylalanine (Phe, F)
Tryptophan (Trp, W)
Proline (Pro, P)
Polar Uncharged Amino Acids:

Serine (Ser, S)
Threonine (Thr, T)
Cysteine (Cys, C)
Asparagine (Asn, N)
Glutamine (Gln, Q)
Tyrosine (Tyr, Y)
Polar Charged Amino Acids:

Aspartic Acid (Asp, D) - Negatively charged (acidic)
Glutamic Acid (Glu, E) - Negatively charged (acidic)
Lysine (Lys, K) - Positively charged (basic)
Arginine (Arg, R) - Positively charged (basic)
Histidine (His, H) - Positively charged (basic)

31
Q

Describe the amino acids in terms of their stereochemistry

A

L-Amino Acids:

Most amino acids in proteins are L-amino acids.
In L-amino acids, the amino group (NH2) is on the left side when the carboxyl group is at the top of the α-carbon.
L-amino acids are essential for protein synthesis and constitute the building blocks of proteins in living organisms.
D-Amino Acids:

D-amino acids are mirror images of L-amino acids.
In D-amino acids, the amino group (NH2) is on the right side when the carboxyl group is at the top of the α-carbon.
D-amino acids are relatively rare in proteins but can be found in specific contexts, such as certain bacterial cell walls.

32
Q

Define a protein assay and appreciate its significance in isolation of proteins

A

a laboratory technique used to quantitatively determine the concentration of proteins in biological samples. These assays are of great significance in the isolation and analysis of proteins for various purposes, including research, diagnostics, biotechnology, and quality control. Protein assays provide precise measurements, allow the analysis of diverse sample types, aid in protein purification, and ensure the quality and accuracy of bioproducts.

33
Q

Use the phenomenon of “salting out” as a means to fractionate protein mixtures

A

to fractionate protein mixtures based on differences in protein solubility in the presence of salts, typically ammonium sulfate ((NH4)2SO4). The process involves gradually adding a concentrated salt solution to the protein mixture, leading to the precipitation of proteins at different salt concentrations due to their varying solubilities. This method allows for the selective isolation of proteins while leaving other molecules in the supernatant.

34
Q

Describe protein purification by ion exchange, size exclusion, and affinity chromatography

A

Ion Exchange Chromatography: This method separates proteins based on their net charge differences. Proteins with opposite charges to the stationary phase resin bind, while those with similar charges flow through. Elution is achieved by changing salt concentration or pH.

Size Exclusion Chromatography (SEC): SEC separates proteins based on their size or molecular weight. Larger proteins are eluted earlier, while smaller proteins enter the column pores and are delayed. It is commonly used for desalting and buffer exchange.

Affinity Chromatography: Affinity chromatography relies on specific interactions between a protein and an immobilized ligand. The target protein binds selectively to the ligand, allowing for highly specific purification. Elution is achieved by disrupting the specific interaction.