Genetics Study Guide: DNA Structure, Gene Expression, and the Lac Operon

Q1. You are analyzing a double-stranded DNA molecule. You separate the two single strands and find that one of them is 38% adenine and thymine. What percentage of guanine must the entire original double-stranded molecule contain? Show your work.

Background

Topic: DNA Base Composition (Chargaff's Rules)

This question tests your understanding of base pairing rules in double-stranded DNA and how to calculate the percentage of each nucleotide based on given information.

Key Terms and Formulas

• Chargaff's Rules: In double-stranded DNA, the amount of adenine (A) equals thymine (T), and the amount of guanine (G) equals cytosine (C).

• Base Pairing: A pairs with T, G pairs with C.

• Percentages: The total percentage of all four bases (A, T, G, C) in double-stranded DNA adds up to 100%.

Step-by-Step Guidance

1. Recognize that the percentage given (38% A+T) is for one single strand. In double-stranded DNA, the total percentage of A+T is the same for both strands combined.

2. Since A pairs with T, and G pairs with C, the percentage of A+T in the double-stranded molecule is the same as in one strand.

3. Subtract the percentage of A+T from 100% to find the combined percentage of G+C in the double-stranded DNA:

   $100\% - 38\% = \text{G+C}\%$

4. Since G and C are present in equal amounts, divide the G+C percentage by 2 to find the percentage of guanine:

   $\text{Guanine}\% = \frac{\text{G+C}\%}{2}$

Try solving on your own before revealing the answer!

Final Answer: 31% guanine

$100\% - 38\% = 62\%$ is G+C. Half of this is guanine: $\frac{62\%}{2} = 31\%$.

This follows from Chargaff's rules: in double-stranded DNA, G = C and A = T.

Q2. The DNA sequence below comes from the coding strand of a eukaryote. The sequence includes two exons and one intron. The sequence is broken up into ten nucleotide blocks for ease of reference. The numbers above each block of ten nucleotides indicate what number within the given sequence the first nucleotide of the block represents. [In your calculations, you must reference the specific number of the nucleotides (e.g. 26) and not just the block where it resides (e.g. 21)].

Background

Topic: Eukaryotic Gene Structure and mRNA Processing

This question tests your ability to identify regulatory and coding regions in a gene, including the TATA box, poly-adenylation signal, transcriptional start site, and splice sites. It also asks you to calculate the length of the raw transcript.

Key Terms and Sequences

• TATA box: A promoter sequence (TATAAA) important for transcription initiation.

• Poly-adenylation signal: Sequence (AAUAAA) signaling the end of transcription.

• Transcriptional start site: The nucleotide where transcription begins, typically 20–25 bases downstream of the TATA box.

• 5' splice site: Consensus sequence at the exon-intron boundary (e.g., [A/C]GGU[Pu]AGUA).

• 3' splice site: Consensus sequence at the intron-exon boundary (e.g., [Py]12[N][Py]AGG).

Step-by-Step Guidance

1. Locate the TATA box (TATAAA) in the sequence and label it.

2. Find the poly-adenylation signal (AAUAAA) and label it.

3. Identify the transcriptional start site, which is about 20–25 nucleotides downstream from the TATA box. Underline and label this region.

4. Highlight the 5' splice site and 3' splice site sequences using the consensus sequences provided.

5. To calculate the number of nucleotides in the raw transcript, subtract the position of the first nucleotide in the transcript from the position at the end of the poly-adenylation site, then add 1 (if both endpoints are included):

   $\text{Transcript length} = \text{End position} - \text{Start position} + 1$

Try solving on your own before revealing the answer!

Final Answer: 311 nucleotides

$346 - 35 = 311$ nucleotides (assuming the transcript starts at position 36 and ends at 346).

This calculation gives the length of the raw transcript from the transcriptional start site to the end of the poly-adenylation signal.

Q3. The lac operon of E. coli controls the expression of genes for lactose metabolism. Mutations to various regions of the lac operon affect gene expression and function. Given different mutations (I, I', Oc, P', Z), determine the functionality of the lac operon (or partial diploid operon system) under the following genetic and cellular conditions. For each genotype, indicate whether the operon is repressed, expressed constitutively, or inducible.

Background

Topic: Prokaryotic Gene Regulation (Lac Operon)

This question tests your understanding of how mutations in the lac operon affect its regulation and expression, including the concepts of repression, constitutive expression, and inducibility.

Key Terms and Concepts

• I (repressor): Wild-type repressor protein that binds the operator to block transcription.

• Is (superrepressor): Repressor that cannot be removed from the operator, always represses.

• Oc (operator constitutive): Operator cannot bind repressor, so genes are always on.

• P' (promoter mutant): Promoter not recognized by RNA polymerase, so no transcription.

• Z (structural gene): Codes for β-galactosidase; Z- is nonfunctional.

• Partial diploid: A cell with two copies of the operon; if either is expressed, the system is not repressed.

• Constitutive expression: Genes are always on, regardless of lactose presence.

• Inducible expression: Genes are off unless lactose is present.

Step-by-Step Guidance

1. For each genotype, identify which mutation(s) are present and recall their effects on the operon's regulation.

2. Determine if the repressor can bind the operator (I, Is, Oc), and whether the promoter allows transcription (P').

3. For partial diploids, remember that if either operon copy is expressed, the system is not repressed.

4. Use the definitions of 'repressed', 'constitutive', and 'inducible' to fill in the table for each genotype.

5. Check if the Z gene is functional; if not, β-galactosidase will not be produced even if the operon is expressed.

Try solving on your own before revealing the answer!

Final Answer: See Table Below

Genotype	Repressed	Constitutive	Inducible
ISB*G SZ	no	yes	no
ISB OZ*	yes	no	no
TkB*O82*	no	no	yes
ISP*0°Z*	yes	yes	no
PP" oe2*	no	no	yes
I°p*oe2z*	yes	no	yes
I°P*0*Z*/	yes	yes	no
1 P*0°Z*/	no	yes	no

Each row reflects the effect of the specific mutations on lac operon expression.