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BIO 1150 Exam 6 Study Guide – DNA, RNA, and Mutations

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Study Guide - Smart Notes

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Q1. In the Hershey-Chase experiment, what element was radioactively labeled to track the location of proteins? What element was radioactively labeled to track the location of DNA? Why were these specific elements chosen?

Background

Topic: Experimental Evidence for DNA as Genetic Material

This question tests your understanding of how scientists determined whether DNA or protein was the genetic material by using radioactive labeling in bacteriophages.

Key Terms and Concepts:

  • Bacteriophage: A virus that infects bacteria.

  • Radioactive labeling: Using radioactive isotopes to trace the location of molecules.

  • DNA contains phosphorus (P), but not sulfur (S).

  • Proteins contain sulfur (in amino acids like cysteine), but not phosphorus.

Step-by-Step Guidance

  1. Recall that the Hershey-Chase experiment used bacteriophages to infect bacteria and determine which molecule (DNA or protein) entered the cell.

  2. Think about which elements are unique to DNA and which are unique to proteins. DNA has a sugar-phosphate backbone, while proteins have amino acids like cysteine and methionine that contain sulfur.

  3. Consider why radioactive phosphorus and sulfur were chosen instead of other elements like nitrogen or carbon.

  4. Identify which isotope would be used to label DNA and which would be used to label protein, based on their unique elemental composition.

Try solving on your own before revealing the answer!

Final Answer:

Proteins were labeled with radioactive sulfur (S), and DNA was labeled with radioactive phosphorus (P). These elements were chosen because sulfur is found in proteins but not DNA, and phosphorus is found in DNA but not proteins. This allowed the researchers to track each molecule separately.

Q2. How did Chargaff’s experimental results help other scientists figure out the structure of DNA?

Background

Topic: DNA Base Pairing and Structure

This question is about how Chargaff’s findings on nucleotide composition contributed to the discovery of DNA’s double helix structure.

Key Terms and Concepts:

  • Chargaff’s rules: In DNA, the amount of adenine (A) equals thymine (T), and the amount of cytosine (C) equals guanine (G).

  • Base pairing: The specific hydrogen bonding between A-T and C-G.

Step-by-Step Guidance

  1. Recall what Chargaff measured in DNA samples from different organisms (the relative amounts of A, T, C, and G).

  2. Think about how the observation that A = T and C = G could suggest a pairing mechanism.

  3. Consider how this information, combined with structural data (like Franklin’s X-ray images), helped Watson and Crick propose the double helix model.

Try solving on your own before revealing the answer!

Final Answer:

Chargaff’s results showed that A pairs with T and C pairs with G, which helped Watson and Crick deduce the base-pairing rules and the double helix structure of DNA.

Q3. What was the key conclusion from Rosalind Franklin’s X-ray crystallography experiment?

Background

Topic: DNA Structure Determination

This question focuses on the structural insights gained from X-ray diffraction studies of DNA.

Key Terms and Concepts:

  • X-ray crystallography: A technique to determine the 3D structure of molecules.

  • Double helix: The spiral structure of DNA.

  • Consistent diameter: The uniform width of the DNA molecule.

Step-by-Step Guidance

  1. Recall what X-ray crystallography reveals about molecular structure (e.g., repeating patterns, helical shapes).

  2. Think about what Franklin’s famous “Photo 51” showed regarding the shape and dimensions of DNA.

  3. Consider how this evidence supported the double helix model and the idea of a consistent diameter between the two strands.

Try solving on your own before revealing the answer!

Final Answer:

Franklin’s X-ray crystallography showed that DNA is a double helix with a consistent diameter, which was crucial evidence for the double helix model.

Q4. What type of bond keeps opposing DNA bases connected in base pairs?

Background

Topic: DNA Structure and Stability

This question tests your knowledge of the chemical interactions that stabilize the DNA double helix.

Key Terms and Concepts:

  • Base pairs: Pairs of nucleotides (A-T, C-G) held together in the DNA double helix.

  • Hydrogen bonds: Weak bonds important for the specificity and stability of base pairing.

Step-by-Step Guidance

  1. Recall the types of chemical bonds found in DNA (covalent bonds in the backbone, hydrogen bonds between bases).

  2. Identify which bond type is responsible for holding the two strands together at the base pairs.

  3. Think about why this type of bond is suitable for DNA replication and function.

Try solving on your own before revealing the answer!

Final Answer:

Hydrogen bonds hold the opposing DNA bases together in base pairs.

Q5. Which bases (A, T, C, or G) are purines and which ones are pyrimidines? Why must a purine always bond with a pyrimidine? How would a purine-purine pair or a pyrimidine-pyrimidine pair obstruct the structure of the DNA molecule?

Background

Topic: DNA Base Pairing and Molecular Structure

This question explores the chemical structure of DNA bases and the importance of base pairing for the double helix’s uniform structure.

Key Terms and Concepts:

  • Purines: Double-ring bases (adenine and guanine).

  • Pyrimidines: Single-ring bases (cytosine and thymine).

  • Base pairing: Purine-pyrimidine pairing maintains the consistent width of the DNA helix.

Step-by-Step Guidance

  1. Identify which bases are purines (A, G) and which are pyrimidines (C, T).

  2. Recall the structural difference: purines have two rings, pyrimidines have one.

  3. Think about what would happen to the DNA helix if two purines or two pyrimidines paired together (would the helix be too wide or too narrow?).

  4. Explain why purine-pyrimidine pairing is necessary for the uniform structure of DNA.

Try solving on your own before revealing the answer!

Final Answer:

Adenine (A) and guanine (G) are purines; cytosine (C) and thymine (T) are pyrimidines. Purine-purine pairs would make the helix too wide, and pyrimidine-pyrimidine pairs would make it too narrow. Purine-pyrimidine pairing keeps the DNA structure uniform.

Q6. What do the terms 5’ and 3’ refer to?

Background

Topic: DNA and RNA Structure

This question is about the directionality of nucleic acids and how the numbering of carbons in the sugar determines the 5’ and 3’ ends.

Key Terms and Concepts:

  • 5’ end: The end of a nucleic acid strand with a phosphate group attached to the 5’ carbon of the sugar.

  • 3’ end: The end with a hydroxyl (-OH) group on the 3’ carbon of the sugar.

  • Directionality: Nucleic acids are synthesized and read from 5’ to 3’.

Step-by-Step Guidance

  1. Recall the structure of the deoxyribose (or ribose) sugar in nucleotides and how the carbons are numbered.

  2. Identify what functional group is attached to the 5’ carbon (phosphate) and the 3’ carbon (hydroxyl group).

  3. Think about why this directionality is important for DNA replication and transcription.

Try solving on your own before revealing the answer!

Final Answer:

The 5’ and 3’ refer to the carbons in the sugar of the nucleotide. The 5’ end has a phosphate group, and the 3’ end has a hydroxyl group. This gives DNA and RNA strands directionality.

Q7. How did the Meselson-Stahl experiment show that DNA is replicated in a semiconservative manner?

Background

Topic: DNA Replication Models

This question tests your understanding of how experimental evidence supported the semiconservative model of DNA replication.

Key Terms and Concepts:

  • Semiconservative replication: Each new DNA molecule consists of one old strand and one new strand.

  • Isotopic labeling: Using heavy and light nitrogen to distinguish old and new DNA strands.

  • Density gradient centrifugation: Technique to separate DNA based on density.

Step-by-Step Guidance

  1. Recall the three models of DNA replication: conservative, semiconservative, and dispersive.

  2. Understand how Meselson and Stahl used heavy () and light () nitrogen to label DNA strands.

  3. Think about what pattern you would expect to see after one and two rounds of replication if DNA replication is semiconservative.

  4. Consider how the experimental results matched the predictions of the semiconservative model.

Try solving on your own before revealing the answer!

Final Answer:

The experiment showed that after one round of replication, all DNA was of intermediate density (hybrid), and after more rounds, both hybrid and light DNA appeared, matching the semiconservative model.

Q8. What are the differences between RNA and DNA? List as many as you can think of.

Background

Topic: Nucleic Acid Structure and Function

This question asks you to compare and contrast the two main types of nucleic acids in cells.

Key Terms and Concepts:

  • DNA: Deoxyribonucleic acid, double-stranded, contains deoxyribose, bases A, T, C, G.

  • RNA: Ribonucleic acid, single-stranded, contains ribose, bases A, U, C, G.

  • Structural and functional differences (e.g., stability, roles in the cell).

Step-by-Step Guidance

  1. List the sugar found in each nucleic acid (deoxyribose in DNA, ribose in RNA).

  2. Identify the nitrogenous bases present in each (T in DNA, U in RNA).

  3. Note the typical structure (double helix for DNA, single strand for RNA).

  4. Consider other differences, such as stability and cellular roles.

Try solving on your own before revealing the answer!

Final Answer:

DNA contains deoxyribose and the bases A, T, C, G; RNA contains ribose and the bases A, U, C, G. DNA is usually double-stranded, RNA is usually single-stranded.

Q9. When we talk about DNA or RNA bases being “complementary”, what does this mean? What bases are complementary to each other?

Background

Topic: Base Pairing Rules

This question is about the specificity of base pairing in nucleic acids and how it ensures accurate replication and transcription.

Key Terms and Concepts:

  • Complementary bases: Bases that pair specifically via hydrogen bonds (A-T/U, C-G).

  • Base pairing ensures accurate copying of genetic information.

Step-by-Step Guidance

  1. Define what it means for bases to be complementary (ability to form stable hydrogen bonds).

  2. List the complementary pairs in DNA (A-T, C-G) and in RNA (A-U, C-G).

  3. Think about why complementarity is important for DNA replication and transcription.

Try solving on your own before revealing the answer!

Final Answer:

Complementary bases are those that pair together: A with T (or U in RNA), and C with G.

Q10. What are the differences between pre-mRNA and a mature mRNA transcript?

Background

Topic: RNA Processing

This question tests your understanding of how eukaryotic mRNA is processed before translation.

Key Terms and Concepts:

  • Pre-mRNA: The initial transcript that includes both exons and introns.

  • Mature mRNA: The processed transcript with introns removed, and a 5’ cap and poly-A tail added.

  • Splicing: Removal of introns and joining of exons.

Step-by-Step Guidance

  1. Recall the steps of RNA processing: capping, splicing, and polyadenylation.

  2. Identify what is present in pre-mRNA that is not in mature mRNA (introns).

  3. List the modifications that occur to produce mature mRNA (5’ cap, poly-A tail, splicing).

Try solving on your own before revealing the answer!

Final Answer:

Mature mRNA has had its introns removed and has a 5’ cap and poly-A tail, while pre-mRNA contains introns and may not be fully processed.

Q11. What does it mean that the genetic code is described as “degenerate” or “redundant”?

Background

Topic: Genetic Code Properties

This question is about the relationship between codons and amino acids in the genetic code.

Key Terms and Concepts:

  • Degenerate/redundant: More than one codon can specify the same amino acid.

  • Codon: A sequence of three nucleotides that codes for an amino acid.

Step-by-Step Guidance

  1. Recall that there are 64 possible codons but only 20 amino acids.

  2. Think about how multiple codons can code for the same amino acid (e.g., GGU, GGC, GGA, GGG all code for glycine).

  3. Consider why this redundancy might be beneficial for the cell (e.g., minimizing the effects of mutations).

Try solving on your own before revealing the answer!

Final Answer:

The genetic code is degenerate/redundant because multiple codons can specify the same amino acid.

Q12. What type of mutation is the most likely to disrupt the function of a gene? What type of mutation is the least likely to disrupt gene function? Think about the type of mutation as well as the location of the mutation (exon vs intron, early in the gene vs late in the gene).

Background

Topic: Mutations and Their Effects

This question asks you to evaluate the impact of different types of mutations on gene function, considering both the mutation type and its location.

Key Terms and Concepts:

  • Nonsense mutation: Changes a codon to a stop codon, leading to a truncated protein.

  • Frameshift mutation: Insertion or deletion that shifts the reading frame, altering downstream amino acids.

  • Silent mutation: Changes a codon but does not change the amino acid.

  • Exon: Coding region of a gene.

  • Intron: Non-coding region, usually removed during RNA processing.

Step-by-Step Guidance

  1. Recall the definitions of different mutation types (nonsense, missense, silent, frameshift).

  2. Think about which mutations would have the greatest effect on the protein product (e.g., early stop codons, frameshifts).

  3. Consider the location: mutations in exons vs introns, early vs late in the gene.

  4. Identify which mutations are least likely to affect gene function (e.g., silent mutations, mutations in introns).

Try solving on your own before revealing the answer!

Final Answer:

Nonsense or frameshift mutations in early exons are most disruptive; silent mutations or mutations in introns are least likely to affect gene function.

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