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BIOLOGY 3500: MIDTERM 1 PRACTICE QUESTIONS - DETAILED SOLUTIONS

Biology 3500 Midterm 1 Practice Questions: Detailed Study Notes

DNA Structure and Replication

1. DNA Base Composition

Problem: If thymine (T) makes up 30\% of the bases in a molecule of DNA, what percentage would be made up of cytosine (C)?

Solution:
This problem relies on Chargaff's rules, which state that in a double-stranded DNA molecule:

  1. The amount of adenine (A) is equal to the amount of thymine (T).

  2. The amount of guanine (G) is equal to the amount of cytosine (C).

  3. The total percentage of purines (A+G) equals the total percentage of pyrimidines (T+C).

Given that T = 30\%, according to Chargaff's rules:

  • Adenine (A) = Thymine (T) = 30\%

Now, we know the combined percentage of A and T:

  • A + T = 30\% + 30\% = 60\%

The remaining percentage must be made up of Guanine (G) and Cytosine (C):

  • G + C = 100\% - (A+T) = 100\% - 60\% = 40\%

Since Guanine (G) = Cytosine (C):

  • Cytosine (C) = rac{G+C}{2} = rac{40\%}{2} = 20\%

Answer: Cytosine would make up 20\% of the bases.

3. DNA Replication Fork

Problem: Below is a drawing of a DNA molecule as the replication fork proceeds from left to right. Draw a single long arrow to illustrate synthesis of the leading strand. Draw several short arrows to illustrate synthesis of the lagging strand. Make sure your arrows are pointing in the correct direction.

5’  3’
     5’

(Note: The diagram provided in the transcript is incomplete for full visual representation here. However, the principles of leading and lagging strand synthesis can be explained.)

Explanation of Principles:

  • DNA Polymerase Activity: DNA polymerase can only synthesize new DNA in the 5'
    ightarrow 3' direction.

  • Antiparallel Strands: The two strands of a DNA double helix are antiparallel. If the parental template strand runs 3'
    ightarrow 5', the new strand is synthesized 5'
    ightarrow 3'. If the parental template strand runs 5'
    ightarrow 3', the new strand is synthesized 3'
    ightarrow 5'. However, since DNA polymerase always synthesizes 5'
    ightarrow 3', the template for this new strand must be read 3'
    ightarrow 5'.

  • Leading Strand: Synthesized continuously. The template strand runs 3'
    ightarrow 5' in the direction of the replication fork movement, allowing continuous 5'
    ightarrow 3' synthesis.

  • Lagging Strand: Synthesized discontinuously in short segments called Okazaki fragments. The template strand runs 5'
    ightarrow 3' in the direction of the replication fork movement. To synthesize in the 5'
    ightarrow 3' direction, DNA polymerase must move away from the replication fork, creating fragments that are later ligated.

Diagrammatic Representation (Conceptual):
Assuming the top strand is the template for the lagging strand and the bottom strand is the template for the leading strand, and the fork moves left to right:

Parental DNA
--------------------------------- > (Replication Fork moving right)
5’---Template for Lagging-----------------> 3’ (top strand)
   <---- (short arrows for Okazaki fragments, moving left, 5' to 3')

3’---Template for Leading------------------> 5’ (bottom strand)
   -----------------------------------> (long arrow for continuous synthesis, moving right, 5' to 3')

New DNA strands
---------------------------------

Arrows Direction:

  • Leading Strand: A single long arrow pointing towards the replication fork (in the direction of fork movement), indicating continuous synthesis in the 5'
    ightarrow 3' direction.

  • Lagging Strand: Several short arrows pointing away from the replication fork (opposite to fork movement), each representing an Okazaki fragment synthesized in the 5'
    ightarrow 3' direction.

5. DNA Melting Temperature and G-C Content

Problem: Below are 2 DNA molecules. Which molecule would need to be heated to a higher temperature to separate the strands? Why?

a. Molecule A: 5'-AATAGACTTAATAATATA-3' and 3'-TTATCTGAATTATTATAT-5'
b. Molecule B: 5'-CGGACGGCGCGGGGCGC-3' and 3'-GCCTGCCGCGCCCCGCG-5'

Solution:

To determine which molecule requires a higher temperature for strand separation (denaturation), we need to consider the stability of the base pairing. The stability of a DNA molecule is influenced primarily by its G-C content.

  • Hydrogen Bonds:

    • Adenine (A) pairs with Thymine (T) via two hydrogen bonds.

    • Guanine (G) pairs with Cytosine (C) via three hydrogen bonds.

  • Stability: Because C-G base pairs have three hydrogen bonds compared to two for A-T pairs, C-G rich DNA sequences are more stable and require more energy (higher temperature) to separate their strands.

Analyze Molecule A:

  • Sequence 1: AATAGACTTAATAATATA

  • Sequence 2: TTATCTGAATTATTATAT

  • Total bases: 18 per strand.

  • Count A: 8

  • Count T: 10

  • Count G: 1

  • Count C: 1

  • Total G+C pairs = 1 (or 2 G/C bases). Percentage G-C = (2/36) imes 100\% ext{ or } (1/18) imes 100\% ext{ for one strand} = 5.56\% (very low)

Analyze Molecule B:

  • Sequence 1: CGGACGGCGCGGGGCGC

  • Sequence 2: GCCTGCCGCGCCCCGCG

  • Total bases: 17 per strand.

  • Count A: 1

  • Count T: 0

  • Count G: 10

  • Count C: 7

  • Total G+C pairs = 17 (or 34 G/C bases). Percentage G-C = (34/34) imes 100\% = 100\% (for this particular sequence, or if we count G and C in strand 1: G=10, C=7, no A/T; for strand 2: G=7, C=10, no A/T. The complementary strand is all G's and C's. My initial A/T/G/C counts were incorrect for a paired strand. Let's re-count for strand 1 alone).

Let's re-count for Molecule B's first strand (5'-CGGACGGCGCGGGGCGC-3'):

  • G: 10

  • C: 7

  • A: 0

  • T: 0

The entire molecule B is composed solely of G and C bases. Therefore, its G-C content is 100\%.

Comparing the G-C content:

  • Molecule A has a very low G-C content (predominantly A-T pairs).

  • Molecule B has a 100\% G-C content (composed entirely of G-C pairs).

Answer: Molecule B would need to be heated to a higher temperature to separate the strands. This is because Molecule B has a much higher percentage of Guanine-Cytosine (G-C) base pairs compared to Molecule A. G-C pairs are held together by three hydrogen bonds, whereas A-T pairs are held by only two hydrogen bonds. More hydrogen bonds require more thermal energy to break, thus increasing the melting temperature.

Gene Expression: Transcription, Splicing, and Translation

2. Gene Sequence Analysis (Eukaryotic vs. Prokaryotic)

Problem: Below is the coding strand of a gene A:
5'–GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA-3'

a. Write the sequence of the complementary strand below the coding strand, indicating the 5’ and 3’ ends.

Solution:
The complementary strand will be antiparallel to the coding strand and follow the base pairing rules (A with T, G with C).

  • Coding Strand: 5'-GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA-3'

  • Complementary Strand: 3'-CTGGTGTTACTGTACCATGCTGAAATGCTTG GGGCGTCCTTATCGATTGTTGGCTGT-5'

b. In eukaryotes, GU and AG mark the beginning and end of introns, respectively. Assuming that the entire sequence is initially transcribed, write the sequence of the mature mRNA that would result from this gene.

Solution:

  1. Transcription: First, transcribe the coding strand into a pre-mRNA sequence by replacing T with U. The mRNA sequence will be identical to the coding strand, with T's replaced by U's, and will also run 5'
    ightarrow 3'.
    Pre-mRNA: 5'-GACACCAUGAC AUGGUACGACUUUACG AA CCCGCAGGAAUAGCU AAC AACCGACA-3'

  2. Intron Splicing: Identify introns based on the GU (start) and AG (end) splice sites. We are looking for GU and AG motifs within the transcribed pre-mRNA sequence.
    Let's re-examine the full coding strand and find the potential intron sites:
    Coding: 5'-GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA-3'
    Pre-mRNA: 5'-GACACCAUGACAUGGUACGACUUUACGAA CCCGCAGGA AUAGCUAA CAACCGA CA-3'
    Searching for GU…AG pairs:

    • The sequence contains CCGCAGGA and AAACGGACA. GU and AG are specific splice signals embedded in the intron.
      Let's mark potential splice sites, considering the mature mRNA will contain only exons.
      Looking at the transcribed mRNA:
      5'–GACACCAUGAC AUGGUACGACUUUACGAACCCGCAGGAAUAGCU AAC AACCGACA-3'

    • The sequence segment CGAAA CCCGCAGGA contains GU and AG in the form of G...AG (common 5' and 3' splice sites) or GU...AG.
      Let's find the first GU and the closest subsequent AG (as introns are typically longer than a few bases):
      Original gene sequence (coding strand):
      5'-GACACCATGACAT GGTA CGACT TTACGAACCCGCAGGAATAGCTAACAACCGACA-3'
      Pre-mRNA (T -> U):
      5'-GACACCAUGACAUGGUACGACUUUACGAACCCGCAGGAAUAGCU AAC AACCGACA-3'

    The problem statement says GU and AG mark the beginning and end of introns in eukaryotes. This typically implies GU at the 5' splice site (intron starts with GU) and AG at the 3' splice site (intron ends with AG).

    Let's look for GU followed by ...AG:
    GACACCAUGACAUGGUACGACUUUACG**AA CCCGCAGGA**AUAGCU AAC AACCGACA
    This sequence: ACGAA CCCGCAGGA AUAGCU doesn't immediately jump out with a clear GU…AG that is typically short for an example. Let's assume the question implies identifying actual intron sequences within the provided string that would be removed.

    If we carefully scan the pre-mRNA for GU…AG:
    GACACCAUGACAUGGUACGACUUUACGAACCCGCAGGAAUAGCU AAC AACCGACA

    • First GU is at index 12 (counting from 0): GACACCAUGAC**AU**GGUAC... -> no. GU is not a splice signal here. It needs to be G**U**.

    • The sequence GACACC**AUG**AC**AUGG**UACGACUUUACG**AAC**CCGC**AGG**AAUAGCU AAC AACCGACA

    Let's assume the question implicitly defines the intron boundaries carefully, and we are looking for the first intron. The problem is a bit ambiguous without clearer intron demarcation in the example. However, if we assume the entire sequence is initially transcribed, and we must find the introns using GUstart and AGend:

    Scanning the mRNA for GU:
    5'–GACACCAUGAC AUGGUACGACUUUACGAA CCCGCAGGAAUAGCU AAC AACCGACA-3'

    • AUG (start codon) is not an intron start.

    • UGG (tryptophan) is not an intron start.

    • The problem likely expects us to find a GU that functions as a 5' splice site and AG as a 3' splice site for an intron.

    Let's consider the sequence GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA again. Without explicit markings or a convention for small examples, it's hard to precisely pinpoint the intron locations based solely on the short coding strand.

    However, often practice problems will hint at or explicitly mark sequences as introns. If we look for GU as 5' splice site and AG as 3' splice site:
    ...ACGAACCCGC**AGGA**ATAGCTAACAACC**GACA
    No clear GU followed by an AG forming an intron within this fairly short sequence.

    Let's re-evaluate the problem statement: