Genetic information, variation and relationships between organisms

• Compare and contrast DNA in eukaryotic cells with DNA in prokaryotic cells : Similarities


  ● Nucleotide structure is identical - deoxyribose attached to phosphate and a base

  ● Adjacent nucleotides joined by phosphodiester bonds, complementary bases joined by hydrogen bonds

  ● DNA in mitochondria / chloroplasts have similar structure to DNA in prokaryotes

  ○ Short, circular, not associated with proteins
• Differences:


  ● Eukaryotic DNA is longer

  ● Eukaryotic DNA is linear, prokaryotic DNA is circular

  ● Eukaryotic DNA is associated with histone proteins, prokaryotic DNA is not

  ● Eukaryotic DNA contain introns, prokaryotic DNA does not
• What is a chromosome?


  ● Long, linear DNA + its associated histone proteins

  ● In the nucleus of eukaryotic cells
• What is a gene?


  A sequence of DNA (nucleotide) bases that codes for:

  ● The amino acid sequence of a polypeptide

  ● Or a functional RNA (eg. ribosomal RNA or tRNA)
• What is a locus?


  Fixed position a gene occupies on a particular DNA molecule.
• Describe the nature of the genetic code - Triplet code


  A sequence of 3 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 so each triplet is read as a discrete unit
• Degenerate


  An amino acid can be coded for by more than one base triplet
• What are 'non-coding base sequences' and where are they found?

  Non-coding base sequence - DNA that does not code for amino acid sequences / polypeptides:

  1. Between genes - eg. non-coding multiple repeats

  2. Within genes - introns In eukaryotes, much of the nuclear DNA does not code for polypeptides.
• What are introns and exons?


  Exon - Base sequence of a gene coding for amino acid sequences (in a polypeptide)

  Intron - Base sequence of a gene that doesn't code for amino acids, in eukaryotic cells
• 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)
• Describe the two stages of protein synthesis - 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
• Compare and contrast the structure of tRNA and mRNA - Comparison (similarities)


  ● Both single polynucleotide strand
• Contrast (differences)


  ● tRNA is folded into a 'clover leaf shape', whereas mRNA is linear / straight

  ● tRNA has hydrogen bonds between paired bases, mRNA doesn't

  ● tRNA is a shorter, fixed length, whereas mRNA is a longer, variable length (more nucleotides)

  ● tRNA has an anticodon, mRNA has codons

  ● tRNA has an amino acid binding site, mRNA doesn't
• Describe how mRNA is formed by transcription in eukaryotic cells


  1. Hydrogen bonds between DNA bases break

  2. Only one DNA strand acts as a template

  3. Free RNA nucleotides align next to their complementary bases on the template strand

  ● In RNA, uracil is used in place of thymine (pairing with adenine in DNA)

  4. RNA polymerase joins adjacent RNA nucleotides

  5. This forms phosphodiester bonds via condensation reactions

  6. Pre-mRNA is formed and this is spliced to remove introns, forming (mature) mRNA
• Describe how production of messenger RNA (mRNA) in a eukaryotic cell is different from the production of mRNA in a prokaryotic cell


  ● Pre-mRNA produced in eukaryotic cells whereas mRNA is produced directly in prokaryotic cells

  ● Because genes in prokaryotic cells don't contain introns so no splicing in prokaryotic cells
• Describe how translation leads to the production of a polypeptide


  1. mRNA attaches to a ribosome and the ribosome moves to a start codon (AUG)

  2. tRNA brings a specific amino acid

  3. tRNA anticodon binds to complementary mRNA codon

  4. Ribosome moves along to next codon and another tRNA binds so 2 amino acids can be joined by a condensation reaction forming a peptide bond

  ● Using energy from hydrolysis of ATP

  5. tRNA released after amino acid joined polypeptide

  6. Ribosome moves along mRNA to form the polypeptide, until a stop codon is reached
• Describe the role of ATP, tRNA and ribosomes in translation - ATP


  ● Hydrolysis of ATP to ADP + Pi releases energy

  ● So amino acids join to tRNAs and peptide bonds form between amino acids
• tRNA


  ● Attaches to / transports a specific amino acid, in relation to its anticodon

  ● tRNA anticodon complementary base pairs to mRNA codon, forming hydrogen bonds

  ● 2 tRNAs bring amino acids together so peptide bond can form
• Ribosomes


  ● mRNA binds to ribosome, with space for 2 codons

  ● Allows tRNA with anticodons to bind

  ● Catalyses formation of peptide bond between amino acids (held by tRNA molecules)

  ● Moves along (mRNA to the next codon) / translocation
• What is a gene mutation?


  ● A change in the base sequence of DNA (on chromosomes)

  ● Can arise spontaneously during DNA replication (interphase) Examples - base deletion or substitution
• What is a mutagenic agent?


  A factor that increases rate of gene mutation, eg. ultraviolet (UV) light or alpha particles.
• Explain how a mutation can lead to the production of a non-functional protein or enzyme


  1. Changes sequence of base triplets in DNA (in a gene) so changes sequence of codons on mRNA

  2. So changes sequence of amino acids in the polypeptide

  3. So changes position of hydrogen / ionic / disulphide bonds (between amino acids)

  4. So changes protein tertiary structure (shape) of protein

  5. Enzymes - active site changes shape so substrate can't bind, enzyme-substrate complex can't form
• Explain the possible effects of a substitution mutation


  1. Base / nucleotide in DNA replaced by a different base / nucleotide

  2. This changes one triplet so changes one mRNA codon

  3. So one amino acid in polypeptide changes

  ● Tertiary structure may change if position of hydrogen / ionic / disulphide bonds change OR amino acid doesn't change

  ● Due to degenerate nature of genetic code (triplet could code for same amino acid) OR if mutation is in an intron
• Explain the possible effects of a deletion mutation

  1. One nucleotide / base removed from DNA sequence

  2. Changes sequence of DNA triplets from point of mutation (frameshift)

  3. Changes sequence of mRNA codons after point of mutation

  4. Changes sequence of amino acids in primary structure of polypeptide

  5. Changes position of hydrogen / ionic / disulphide bonds in tertiary structure of protein

  6. Changes tertiary structure / shape of protein
• Describe features of homologous chromosomes


  Same length, same genes at same loci, but may have different alleles.
• Describe the difference between diploid and haploid cells


  ● Diploid - has 2 complete sets of chromosomes, represented as 2n

  ● Haploid - has a single set of unpaired chromosomes, represented as n
• Describe how a cell divides by meiosis


  In interphase, DNA replicates → 2 copies of each chromosome (sister chromatids), joined by a centromere. 1. Meiosis I (first nuclear division) separates homologous chromosomes ● Chromosomes arrange into homologous pairs ● Crossing over between homologous chromosomes ● Independent segregation of homologous chromosomes 2. Meiosis II (second nuclear division) separates chromatids ● Outcome = 4 genetically varied daughter cells ● Daughter cells are normally haploid (if diploid parent cell)
• Explain why the number of chromosomes is halved during meiosis


  Homologous chromosomes are separated during meiosis I (first division).
• Explain how crossing over creates genetic variation


  ● Homologous pairs of chromosomes associate / form a bivalent

  ● Chiasmata form (point of contact between (non-sister) chromatids)

  ● Alleles / (equal) lengths of (non-sister) chromatids exchanged between chromosomes

  ● Creating new combinations of (maternal & paternal) alleles on chromosomes
• Explain how independent segregation creates genetic variation


  ● Homologous pairs randomly align at equator → so random which chromosome from each pair goes into each daughter cell

  ● Creating different combinations of maternal & paternal chromosomes / alleles in daughter cells
• Other than mutation and meiosis, explain how genetic variation within a species is increased


  ● Random fertilisation / fusion of gametes

  ● Creating new allele combinations / new maternal and paternal chromosome combinations
• Explain the different outcomes of mitosis and meiosis


  1. Mitosis produces 2 daughter cells, whereas meiosis produces 4 daughter cells

  ● As 1 division in mitosis, whereas 2 divisions in meiosis

  2. Mitosis maintains the chromosome number (eg. diploid → diploid or haploid → haploid) whereas meiosis halves the chromosome number (eg. diploid → haploid)

  ● As homologous chromosomes separate in meiosis but not mitosis

  3. Mitosis produces genetically identical daughter cells, whereas meiosis produces genetically varied daughter cells

  ● As crossing over and independent segregation happen in meiosis but not mitosis
• Explain the importance of meiosis


  ● Two divisions creates haploid gametes (halves number of chromosomes)

  ● So diploid number is restored at fertilisation → chromosome number maintained between generations

  ● Independent segregation and crossing over creates genetic variation
• How can you recognise where meiosis and mitosis occur in a life cycle?


  ● Mitosis occurs between stages where chromosome number is maintained (eg. diploid (2n) → diploid (2n) OR haploid (n) → haploid (n))

  ● Meiosis occurs between stages where chromosome number halves (eg. diploid (2n) → haploid (n))
• Describe how mutations in the number of chromosomes arise


  ● Spontaneously by chromosome non-disjunction during meiosis

  ● Homologous chromosomes (meiosis I) or sister chromatids (meiosis II) fail to separate during meiosis

  ● So some gametes have an extra copy (n+1) of a particular chromosome and others have none (n-1)
• Suggest how the number of possible combinations of chromosomes in daughter cells following meiosis can be calculated

  2 n where n = number of pairs of homologous chromosomes (half the diploid number)