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BIO 1150 Exam 6 Study Guide – Step-by-Step Guidance for Key Questions

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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, used in the experiment.

  • Radioactive labeling: Attaching a radioactive isotope to a molecule to track its location.

  • DNA vs. Protein composition: DNA contains phosphorus but not sulfur; proteins often contain sulfur (in cysteine and methionine) 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 bacterial cells.

  2. Think about the unique elements found in DNA and proteins. DNA contains a sugar-phosphate backbone, while proteins contain certain amino acids with sulfur.

  3. Consider which radioactive isotopes would specifically label only DNA or only protein, based on their unique elements.

  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 (because some amino acids contain sulfur, and DNA does not). DNA was labeled with radioactive phosphorus (because the sugar-phosphate backbone contains phosphorus, and proteins do not).

Nitrogen, carbon, or oxygen would not have been good choices since both proteins and DNA are rich in these elements.

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 contributed to the discovery of the double helix structure of DNA.

Key Terms and Concepts:

  • Chargaff’s rules: In any DNA sample, 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 finding that A = T and C = G in all DNA samples.

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

  4. Reflect on how base pairing explains Chargaff’s observations.

Try solving on your own before revealing the answer!

Final Answer:

Chargaff’s experiments 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 X-ray diffraction).

  2. Think about what Franklin’s famous “Photo 51” revealed about the symmetry and dimensions of DNA.

  3. Consider what structural features (e.g., helical shape, consistent diameter) were inferred from the X-ray data.

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 between the two strands.

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

Background

Topic: DNA structure and chemical bonding

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

Key Terms and Concepts:

  • Hydrogen bond: A weak bond between a hydrogen atom and an electronegative atom (like N or O), important in base pairing.

  • Base pairs: Pairs of nucleotides (A-T, C-G) held together by hydrogen bonds.

Step-by-Step Guidance

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

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

  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 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 structure and pairing

This question tests your knowledge of the chemical structure of DNA bases and the importance of base pairing rules for DNA 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 uniform width of the DNA double 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 the spatial arrangement of the DNA double helix and why pairing a double-ring with a single-ring base is necessary.

  4. Consider what would happen to the DNA structure if two purines or two pyrimidines paired together.

Try solving on your own before revealing the answer!

Final Answer:

Purines: Adenine (A) and Guanine (G). Pyrimidines: Cytosine (C) and Thymine (T). Purine-purine pairs would be too large, causing bulges; pyrimidine-pyrimidine pairs would be too small, making the helix unstable. 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 free phosphate group attached to the 5’ carbon of the sugar.

  • 3’ end: The end with a free 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 deoxyribose (in DNA) and ribose (in RNA) and how the carbons are numbered.

  2. Identify which end of the nucleotide chain has a phosphate group and which has a hydroxyl group.

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

Try solving on your own before revealing the answer!

Final Answer:

The 5’ and 3’ refer to the numbering of carbons in the sugar. 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.

Step-by-Step Guidance

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

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

  3. Consider what pattern of DNA densities would be expected after one and two rounds of replication for each model.

  4. Reflect on how the observed results matched the predictions of the semiconservative model.

Try solving on your own before revealing the answer!

Final Answer:

After growing bacteria in heavy nitrogen and then switching to light nitrogen, Meselson and Stahl observed hybrid DNA after one replication and a mix of hybrid and light DNA after two, consistent with semiconservative replication.

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 is about comparing the chemical and structural differences between DNA and RNA.

Key Terms and Concepts:

  • Deoxyribose vs. Ribose: DNA contains deoxyribose; RNA contains ribose.

  • Bases: DNA uses thymine (T); RNA uses uracil (U).

  • Strandedness: DNA is usually double-stranded; RNA is usually single-stranded.

Step-by-Step Guidance

  1. List the sugar found in DNA and the sugar found in RNA.

  2. Identify the four bases in DNA and the four in RNA.

  3. Consider the typical structure (double helix vs. single strand) of each molecule.

  4. Think about any other differences, such as stability or function.

Try solving on your own before revealing the answer!

Final Answer:

DNA contains deoxyribose and uses A, T, C, G; RNA contains ribose and uses 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 in nucleic acids

This question is about the concept of complementary base pairing and which bases pair together in DNA and RNA.

Key Terms and Concepts:

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

  • Base pairing rules: A pairs with T (or U in RNA), C pairs with G.

Step-by-Step Guidance

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

  2. List the base pairs in DNA and in RNA.

  3. Consider the importance of complementary base pairing for replication and transcription.

Try solving on your own before revealing the answer!

Final Answer:

Complementary bases can bind together via hydrogen bonds: A pairs with T (or U in RNA), and C pairs with G.

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

Background

Topic: RNA processing in eukaryotes

This question is about the modifications that occur to pre-mRNA to produce mature mRNA.

Key Terms and Concepts:

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

  • Splicing: Removal of introns to produce mature mRNA.

  • 5’ cap and poly-A tail: Modifications added to mature mRNA for stability and export.

Step-by-Step Guidance

  1. Recall what is present in the initial RNA transcript (pre-mRNA) after transcription.

  2. List the processing steps that occur to produce mature mRNA (splicing, capping, polyadenylation).

  3. Think about the functional significance of these modifications.

Try solving on your own before revealing the answer!

Final Answer:

Mature mRNA has had introns removed (splicing) and contains a 5’ cap and poly-A tail, while pre-mRNA contains both exons and introns.

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 nature of the genetic code and how multiple codons can specify the same amino acid.

Key Terms and Concepts:

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

  • Degeneracy/redundancy: More than one codon can code for 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 some amino acids are specified by more than one codon.

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

Try solving on your own before revealing the answer!

Final Answer:

Degenerate/redundant means that 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: Types and effects of mutations

This question is about how different mutations affect gene function, depending on their type and location.

Key Terms and Concepts:

  • Nonsense mutation: A mutation that introduces a premature stop codon.

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

  • Silent mutation: A mutation that does not change the amino acid sequence.

  • Exon vs. intron: Exons code for protein; introns are non-coding regions.

Step-by-Step Guidance

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

  2. Think about which mutations are most likely to disrupt protein function (e.g., those that introduce stop codons or shift the reading frame).

  3. Consider the impact of mutation location (exon vs. intron, early vs. late in the gene).

  4. Reflect on which mutations are least likely to have an effect (e.g., silent mutations, mutations in introns).

Try solving on your own before revealing the answer!

Final Answer:

Most disruptive: Nonsense or frameshift mutations in early exons. Least disruptive: Silent mutations or point mutations in introns.

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