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BIO 1150 Exam 6 Study Guide – DNA, RNA, and Genetic Information

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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 the Hershey-Chase experiment distinguished between DNA and protein as the genetic material using radioactive labeling.

Key Terms and Concepts:

  • Bacteriophage: A virus that infects bacteria.

  • Radioactive labeling: Using radioactive isotopes to trace 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 carried genetic information.

  2. Think about the unique elements found in DNA and proteins: DNA has a sugar-phosphate backbone, while proteins have amino acids, some of which contain sulfur.

  3. Consider which radioactive isotopes would specifically label DNA or protein without labeling the other.

  4. Reflect on why elements like nitrogen, carbon, or oxygen would not be good choices for distinguishing DNA from protein.

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). Sulfur is present in proteins (specifically in cysteine and methionine), but not in DNA. Phosphorus is present in the DNA backbone, but not in 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 each base).

  2. Think about the implications of A = T and C = G for the structure of DNA.

  3. Consider how these ratios suggested a pairing mechanism between bases.

  4. Reflect on how this information, combined with other experimental data, 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 and T are present in equal amounts, as are C and G. This suggested that A pairs with T and C pairs with G, which was crucial for Watson and Crick to deduce 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 tests your understanding of how X-ray crystallography provided evidence for the shape 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.

Step-by-Step Guidance

  1. Recall what kind of data X-ray crystallography provides (patterns of spots that reveal molecular structure).

  2. Think about what Franklin’s famous “Photo 51” showed about DNA’s shape.

  3. Consider what “consistent diameter” means for the arrangement of the two DNA strands.

Try solving on your own before revealing the answer!

Final Answer:

Franklin’s experiment revealed that DNA is a double helix with a consistent diameter, supporting the idea of two strands wound around each other.

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

Background

Topic: DNA Structure and Stability

This question is about the chemical interactions that stabilize the DNA double helix.

Key Terms and Concepts:

  • Base pairs: Pairs of nucleotides on opposite DNA strands.

  • Hydrogen bonds: Weak bonds important for molecular interactions.

Step-by-Step Guidance

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

  2. Think about what kind of bond allows the two strands to separate during replication.

  3. Consider the number of bonds between A-T and C-G pairs.

Try solving on your own before revealing the answer!

Final Answer:

Hydrogen bonds hold the base pairs together: A-T pairs have two hydrogen bonds, and C-G pairs have three.

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 Structure

This question tests your knowledge of nucleotide classification and the structural reasons for specific base pairing.

Key Terms and Concepts:

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

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

  • Base pairing: Purine-pyrimidine pairing maintains uniform DNA width.

Step-by-Step Guidance

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

  2. Recall the structural difference between purines (double ring) and pyrimidines (single ring).

  3. Think about the physical consequences of pairing two purines or two pyrimidines (would the DNA helix be uniform?).

  4. Consider why the DNA double helix needs to maintain a consistent diameter for proper function.

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 be too wide, and pyrimidine-pyrimidine pairs too narrow, disrupting the DNA structure. Purine-pyrimidine pairing keeps the helix uniform.

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

Background

Topic: DNA Structure and Directionality

This question is about the orientation of DNA strands and the numbering of carbons in the sugar molecule.

Key Terms and Concepts:

  • 5’ (five-prime) end: The end of the DNA strand with a phosphate group attached to the 5’ carbon of deoxyribose.

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

  • Directionality: DNA is synthesized and read from 5’ to 3’.

Step-by-Step Guidance

  1. Recall the structure of deoxyribose and how carbons are numbered.

  2. Identify what chemical group is attached to the 5’ and 3’ carbons.

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

Try solving on your own before revealing the answer!

Final Answer:

5’ and 3’ refer to the carbon numbers in the deoxyribose sugar. The 5’ end has a phosphate group, and the 3’ end has a hydroxyl group. This gives DNA 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 has one old and one new strand.

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

Step-by-Step Guidance

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

  2. Think about how growing bacteria in heavy nitrogen (15N) labels their DNA.

  3. Consider what happens when bacteria are switched to light nitrogen (14N) and allowed to replicate.

  4. Reflect on how the distribution of heavy and light DNA after each replication cycle supports the semiconservative model.

Try solving on your own before revealing the answer!

Final Answer:

After one replication in light nitrogen, all DNA molecules were hybrids (one heavy, one light strand), consistent with semiconservative replication. Further rounds produced both hybrid and light DNA, matching predictions for 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 the structural and functional differences between DNA and RNA.

Key Terms and Concepts:

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

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

Step-by-Step Guidance

  1. List the sugar found in each nucleic acid (deoxyribose vs ribose).

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

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

  4. Think about any functional differences (storage vs. transmission of genetic information).

Try solving on your own before revealing the answer!

Final Answer:

DNA contains deoxyribose and is double-stranded with bases A, T, C, G. RNA contains ribose, is usually single-stranded, and uses U instead of T. DNA stores genetic information; RNA is involved in protein synthesis and regulation.

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.

Key Terms and Concepts:

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

  • Base pairing: Ensures accurate replication and transcription.

Step-by-Step Guidance

  1. Define what it means for bases to be complementary.

  2. Recall which bases pair together in DNA and RNA.

  3. Think about the importance of complementary base pairing for genetic fidelity.

Try solving on your own before revealing the answer!

Final Answer:

Complementary bases are those that pair specifically: A with T (or U in RNA), and C with G. This ensures accurate copying of genetic information.

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 containing exons and introns.

  • Mature mRNA: Processed transcript with introns removed, 5’ cap, and poly-A tail.

  • Splicing: Removal of introns and joining of exons.

Step-by-Step Guidance

  1. Recall the steps of RNA processing in eukaryotes (capping, splicing, polyadenylation).

  2. Identify what is present in pre-mRNA but not in mature mRNA.

  3. Think about the functional significance of the 5’ cap and poly-A tail.

Try solving on your own before revealing the answer!

Final Answer:

Pre-mRNA contains both exons and introns; mature mRNA has introns removed, a 5’ cap, and a poly-A tail. These modifications are necessary for stability and translation.

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:

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

  • Degeneracy/redundancy: Multiple codons can specify the same amino acid.

Step-by-Step Guidance

  1. Recall how many codons exist and how many amino acids are encoded.

  2. Think about why more than one codon can code for the same amino acid.

  3. Consider the evolutionary advantage of redundancy in the genetic code.

Try solving on your own before revealing the answer!

Final Answer:

The genetic code is degenerate/redundant because multiple codons can code for the same amino acid, providing a buffer against mutations.

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 tests your understanding of how different mutations affect gene function and protein structure.

Key Terms and Concepts:

  • Nonsense mutation: Changes a codon to a stop codon, truncating the protein.

  • Frameshift mutation: Insertion or deletion that shifts the reading frame.

  • Silent mutation: Does not change the amino acid sequence.

  • Exon: Coding region; Intron: Non-coding region.

Step-by-Step Guidance

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

  2. Think about how a mutation’s location (exon vs intron, early vs late) affects its impact.

  3. Consider which mutations are most likely to cause loss of function and which are likely to be harmless.

Try solving on your own before revealing the answer!

Final Answer:

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

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