Topic 4 Genetic information, variation and relationships between organisms

Question 1

Compare and contrast DNA in eukaryotic cells with DNA in prokaryotic cells.

Answer 1

Similarities:
Both have identical nucleotide structures with deoxyribose attached to phosphate and a base.
Adjacent nucleotides joined by phosphodiester bonds, with complementary bases joined by hydrogen bonds.
DNA in mitochondria and chloroplasts has a similar structure to DNA in prokaryotes (short, circular, not associated with proteins).
Differences:
Eukaryotic DNA is longer and linear, associated with histone proteins, and contains introns.
Prokaryotic DNA is shorter, circular, not associated with histones, and does not contain introns.

Question 2

What is a chromosome, and what is a gene?

Answer 2

Chromosome: Long, linear DNA plus its associated histone proteins found in the nucleus of eukaryotic cells.
Gene: A sequence of DNA (nucleotide) bases that codes for the amino acid sequence of a polypeptide or a functional RNA (e.g., rRNA, tRNA).

Question 3

Describe the nature of the genetic code.

Answer 3

Triplet Code: A sequence of three DNA bases, called a triplet, codes for a specific amino acid.
Universal: The same base triplets code for the same amino acids in all organisms.
Non-overlapping: Each base is part of only one triplet; each triplet is read as a discrete unit.
Degenerate: An amino acid can be coded for by more than one base triplet.

Question 4

What are non-coding base sequences and where are they found?

Answer 4

Non-coding base sequences: DNA that does not code for amino acid sequences/polypeptides.
Found between genes (e.g., non-coding multiple repeats) and within genes (introns).

Question 5

Define ‘genome’ and ‘proteome’.

Answer 5

Genome: The complete set of genes in a cell, including those in mitochondria and/or chloroplasts.
Proteome: The full range of proteins that a cell can produce, coded for by the cell’s DNA/genome.

Question 6

Describe how translation leads to the production of a polypeptide.

Answer 6

mRNA attaches to a ribosome, which moves to a start codon (AUG).
tRNA brings a specific amino acid.
tRNA anticodon binds to the complementary mRNA codon.
Ribosome moves along to the next codon, and another tRNA binds, allowing amino acids to join by condensation, forming a peptide bond.
tRNA is released after the amino acid is added to the polypeptide.
The ribosome continues along the mRNA until a stop codon is reached, completing the polypeptide.

Question 6

Describe the two stages of protein synthesis.

Answer 6

Transcription: Production of messenger RNA (mRNA) from DNA in the nucleus.
Translation: Production of polypeptides from the sequence of codons carried by mRNA at ribosomes.

Question 7

Compare and contrast the structure of tRNA and mRNA.

Answer 7

Similarities: Both are single polynucleotide strands.
Differences:
tRNA is folded into a cloverleaf shape with hydrogen bonds between paired bases; mRNA is linear/straight.
tRNA has an anticodon and an amino acid binding site; mRNA has codons and no amino acid binding site.
tRNA is shorter with a fixed length; mRNA is longer with variable lengths.

Question 8

Describe how mRNA is formed by transcription in eukaryotic cells.

Answer 8

Hydrogen bonds between DNA bases break.
One DNA strand acts as a template.
Free RNA nucleotides align next to their complementary bases on the template strand (uracil replaces thymine).
RNA polymerase joins adjacent RNA nucleotides, forming phosphodiester bonds.
Pre-mRNA is formed and spliced to remove introns, forming mature mRNA.

Question 9

What roles do ATP, tRNA, and ribosomes play in translation?

Answer 9

ATP: Hydrolysis of ATP to ADP + Pi releases energy for amino acid attachment to tRNA and peptide bond formation.
tRNA: Transports a specific amino acid to the ribosome, complementary base pairs with mRNA codons, facilitating peptide bond formation.
Ribosomes: Provide a binding site for mRNA, facilitate tRNA binding, and catalyze peptide bond formation as the ribosome translocates along the mRNA.

Question 10

What is a gene mutation, and how can it affect a protein or enzyme?

Answer 10

Gene Mutation: A change in the base sequence of DNA on chromosomes that can arise spontaneously during DNA replication.
Effects on Protein/Enzyme:
1.Mutation changes the sequence of base triplets in DNA and the codons on mRNA.
2.Alters the sequence of amino acids in the polypeptide.
3.Changes the position of hydrogen, ionic, or disulfide bonds in the protein.
4.Alters the protein’s tertiary structure, possibly making an enzyme’s active site incompatible with its substrate, preventing enzyme-substrate complex formation.

Question 11

Explain the effects of substitution and deletion mutations.

Answer 11

Substitution Mutation:
1.A base/nucleotide in DNA is replaced by a different one, changing one mRNA codon.
2.This may change one amino acid in the polypeptide, altering the protein’s tertiary structure.
3.However, due to the degenerate nature of the genetic code, the amino acid may remain unchanged, especially if the mutation occurs in an intron.
Deletion Mutation:
1.One nucleotide/base is removed from the DNA sequence, causing a frameshift mutation.
2.Alters the sequence of mRNA codons from the mutation point onward.
3.Changes the sequence of amino acids, potentially altering the tertiary structure of the protein.

Question 11

Describe the three types of adaptations and provide examples.

Answer 11

Anatomical: Structural features (e.g., thick fur for insulation).
Physiological: Chemical processes (e.g., enzymes for digestion).
Behavioral: Actions that increase survival (e.g., birds migrating to warmer climates).

Question 12

Describe the process of meiosis and how it leads to genetic variation.

Answer 12

Meiosis Process:
1.Interphase: DNA replicates, forming sister chromatids joined by a centromere.
2.Meiosis I: Homologous chromosomes separate after crossing over and independent segregation, resulting in two haploid cells.
3.Meiosis II: Sister chromatids separate, resulting in four genetically varied haploid daughter cells.
Genetic Variation:
Crossing Over: Homologous chromosomes exchange alleles at chiasmata, creating new allele combinations.
Independent Segregation: Random alignment of homologous chromosomes creates different combinations of maternal and paternal chromosomes in daughter cells.

Question 12

Describe how mutations in chromosome number can arise.

Answer 12

Process: Chromosome non-disjunction during meiosis, where homologous chromosomes (meiosis I) or sister chromatids (meiosis II) fail to separate.
Outcome: Results in gametes with an extra chromosome (n+1) or missing a chromosome (n-1).

Question 12

Explain the importance of meiosis.

Answer 12

Purpose:
Produces haploid gametes, ensuring that diploid chromosome number is restored at fertilization.
Maintains chromosome number across generations.
Introduces genetic variation through independent segregation and crossing over.

Question 13

What is genetic diversity, and how does it relate to natural selection?

Answer 13

Genetic Diversity: The number of different alleles of genes in a population.
Relation to Natural Selection:
Mutations and genetic diversity allow natural selection to occur.
Certain alleles may provide a selective advantage in specific environments, leading to increased reproductive success and a change in allele frequency over generations.

Question 14

Describe aseptic techniques and their importance in microbial experiments.

Answer 14

Aseptic Techniques:
Wash hands, disinfect surfaces, and sterilize equipment to prevent contamination.
Flame the neck of bacterial containers, minimize exposure to air, and work near a Bunsen burner to avoid airborne contamination.
Seal Petri dishes partially to allow oxygen in and prevent anaerobic bacteria growth, which could be pathogenic.
Importance: Prevents contamination and ensures the validity of results when investigating the effects of antimicrobial substances on microbial growth.

Question 14

Describe a method to investigate the effect of antimicrobial substances on microbial growth.

Answer 14

1.Use aseptic techniques to prepare the working area and inoculate agar plates with bacteria.
2.Place discs soaked in different antimicrobial agents on the agar.
3.Incubate the plates at 25°C for 48 hours, then measure the diameter of the inhibition zones around each disc.
4.Calculate the area of inhibition using the formula πr².

Question 14

Explain directional and stabilizing selection with examples.

Answer 14

Directional Selection:
Example: Antibiotic resistance in bacteria.
Mechanism: Favors individuals with extreme traits that are advantageous in a changing environment.
Outcome: Shift in the population’s trait distribution toward the extreme.
Stabilizing Selection:
Example: Human birth weight.
Mechanism: Favors individuals with average traits in a stable environment.
Outcome: Reduction in trait variation around the mean.

Question 15

What is a species, and how are species classified?

Answer 15

Species: A group of organisms that can interbreed to produce fertile offspring.
Taxonomy: Classification of organisms into a hierarchy based on evolutionary relationships, including domain, kingdom, phylum, class, order, family, genus, and species.

Question 16

Describe the phylogenetic classification system and how phylogenetic trees are interpreted.

Answer 16

Phylogenetic Classification: Groups species based on common ancestors and evolutionary relationships, forming a hierarchy with no overlap.
Phylogenetic Trees:
Branch Point: Represents a common ancestor.
Branches: Show evolutionary paths; closely related species share a recent common ancestor.

Question 17

What is biodiversity, and how is it measured using the index of diversity?

Answer 17

Biodiversity: The variety of living organisms, encompassing species, genetic, and ecosystem diversity.
Index of Diversity: Measures the number of species and the number of individuals in each species within a community. The formula:
​where
𝑁
N is the total number of organisms of all species, and
𝑛
n is the number of organisms of each species.

Question 17

Explain how certain farming techniques reduce biodiversity.

Answer 17

Removal of woodland and hedgerows reduces plant variety and habitats.
Monoculture decreases the variety of food sources.
Use of herbicides and pesticides reduces species diversity.
Conservation Balance: Implementing biodiversity-increasing methods like reintroducing hedgerows and reducing pesticide use, balanced with maintaining agricultural productivity.

Question 18

How can genetic diversity be measured, and what considerations should be made in quantitative investigations?

Answer 18

Measurement Methods:
Compare measurable/observable characteristics, DNA base sequences, mRNA sequences, and amino acid sequences.
Key Considerations:
Use random sampling, ensure a large sample size, ethical considerations, calculate mean values and standard deviations, and perform statistical tests to determine the significance of differences between populations.