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AQA A-Level Biology > [3.4] Genetic Information, Variation & Relationships Between Organisms > Flashcards

[3.4] Genetic Information, Variation & Relationships Between Organisms Flashcards

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1
Q

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

A

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 contains introns, prokaryotic DNA does not.
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2
Q

What is a chromosome?

A
  • Long, linear DNA + its associated with histone proteins.
  • In the nucleus of eukaryotic cells.
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3
Q

What is a gene?

A

A sequence of DNA (nucleotide) bases that codes for:
- The amino acid sequence of a polypeptide.
- Or a functional RNA (e.g. ribosomal RNA or tRNA).

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4
Q

What is a locus?

A

Fixed position a gene occupies on a particular DNA molecule.

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5
Q

Describe the nature of the genetic code.

A
  1. Triplet code.
    • A sequence of 3 DNA bases, called a triplet, codes for a specific amino acid.
  2. Universal.
    • The same base triplets code for the same amino acids in all organisms.
  3. Non-overlapping.
    • Each base is part of only one triplet so each triplet is read as a discrete unit.
  4. Degenerate.
    • An amino acid can be coded for by more than one base triplet.
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6
Q

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

A
  • Non-coding base sequence - DNA that does not code for amino acid sequences/polypeptides. It is found:
    • Between genes - e.g. non-coding multiple repeats.
    • Within genes - introns.

(In eukaryotes, much of the nuclear DNA do not code for polypeptides)

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7
Q

What are introns and exons?

A

EXON

  • Base sequence of 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.
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8
Q

Define ‘genome’ and ‘proteome’.

A

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).
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9
Q

Describe the two stages of protein synthesis.

A

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..
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10
Q

Compare and contrast the structure of tRNA and mRNA.

A

COMPARISON

  • Both single polynucleotide strand.

CONTRAST

  • 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 had an anticodon, mRNA has codons.
  • tRNA has an amino acid binding site, mRNA doesn’t.
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11
Q

Describe how mRNA is formed by transcription in eukaryotic cells.

A
  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.
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12
Q

Describe how the production of messenger RNA (mRNA) in a eukaryotic cell is different from the production of mRNA in a prokaryotic cell.

A
  • 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.
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13
Q

Describe how translation leads to the production of a polypeptide.

A
  1. mRNA attaches to a ribosome and the ribosome moves to a start codon.
  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.
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14
Q

Describe the role of ATP, tRNA and ribosomes in translation.

A

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 aids (held by tRNA molecules).
  • Moves along mRNA to the next codon.
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15
Q

Describe how the base sequence of nucleic acids can be related to the amino acid sequence of polypeptides when provided with suitable data

A
  • You may be provided with a genetic code to identify which triplets/codons produce which amino acids.
  • tRNA anticodons are complementary to mRNA codons.
    • e.g. mRNA codon = ACG
      -> tRNA anticodon = UGC
  • Sequence of codons on mRNA are complementary to sequence of triplets on DNA template strand.
    • e.g. mRNA base sequence = ACG UAG AAC
      -> DNA base sequence = TGC ATC TTG
  • In RNA, uracil replaces thymine.
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16
Q

What is a gene mutation?

A
  • A change in the base sequence of DNA on chromosomes.
  • Can arise spontaneously during DNA replication (in interphase).
  • Examples include base deletion and substitution.
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17
Q

What is a mutagenic agent?

A

A factor that increases rate of gene mutation e.g. ultraviolet (UV) light or alpha particles.

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18
Q

Explain how a mutation can lead to the production of a non-functional protein or enzyme.

A
  1. Changes 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 of protein.
  5. Enzymes active site changes shape so substrate can’t bind, enzyme-substrate complex can’t form.
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19
Q

Explain the possible effects of a substitution mutation.

A
  1. Base/nucleotide in DNA replaced by a different base/nucleotide.
  2. This changes one triplet so changes 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.
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20
Q

Explain the possible effects of a deletion mutation.

A
  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.
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21
Q

Describe features of homologous chromosomes.

A

Same length, same genes at same loci, but may have different alleles.

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22
Q

Describe the difference between diploid and haploid cells.

A
  • Diploid - has 2 complete sets of chromosomes, represented as 2n.
  • Haploid - has a single set of unpaired chromosomes, represented as n.
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23
Q

Describe how a cell divides by meiosis.

A

In interphase, DNA replicates to form 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 that are haploid.
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24
Q

Draw a diagram to show the chromosome content of cells during meiosis.

A
  • You should be able to complete diagrams showing the chromosome contents of cells after the first and second meiotic division when given the chromosome content of the parent cell.
  • In the example shown below:
    • The parent cell has 4 chromosomes = 2 homologous pairs.
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25
Q

Explain why the number of chromosomes is halved during meiosis.

A

Homologous chromosomes are separated during meiosis I (frist division).

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26
Q

Explain how crossing over creates genetic variation.

A
  1. Homologous pairs of chromosomes associate/form a bivalent.
  2. Chiasmata form.
    • Point of contact between non-sister chromatids.
  3. Alleles exchanges between chromosomes.
  4. Creating new combinations of maternal and paternal alleles on chromosomes.
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27
Q

Explain how independent segregation creates genetic variation.

A
  1. Homologous pairs randomly align at equator.
    • So random which chromosome from each pair goes into each daughter cell.
  2. Creating different combinations of maternal and paternal chromosomes/alleles in daughter cells.
28
Q

Other than mutation and meiosis, explain how genetic variation within a species is increased.

A
  1. Random fertilisation/fusion of gametes.
    • Creating new allele combinations/new maternal and paternal chromosome combinations.
29
Q

Explain the different outcomes of mitosis and meiosis.

A
  1. Mitosis produces 2 daughter cells, whereas meiosis produces 4 daughter cells.
    • As 1 divison in mitosis, where as 2 divisions in meiosis.
  2. Mitosis maintains the chromosome number (e.g. diploid -> diploid or haploid -> haploid) whereas mitosis halves the chromosome number (e.g. 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 independant segregation happen in meiosis but not mitosis.
30
Q

Explain the importance of meiosis.

A
  • Two divison creates haploid gametes (halves number of chromosomes).
  • So diploid number is restored at fertilisation.
    • Chromosome number maintained between generations.
  • Independant segregation and crossing over creates genetic variation.
31
Q

How can you recognise where meiosis and mitosis occur in a life cycle?

A
  • Mitosis occurs between stages where chromosome number is maintained (e.g. diploid (2n) -> diploid (2n) OR haploid (n) -> haploid (n)).
  • Meiosis occurs between stages where chromosome number halves (e.g. diploid (2n) -> haploid (n)).
32
Q

Describe how mutations in the number of chromosomes arise.

A
  • Spontaneously by chromosomes 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).
33
Q

Suggest how the number of possible combinations of chromosomes in daughter cells following meiosis without crossing over can be calculated.

A

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

34
Q

Suggest how the number of possible combinations of chromosomes following random fertilisation of two gametes can be calculated.

A

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

35
Q

What is genetic diversity?

A

Number of different alleles of genes in a population.

36
Q

What are alleles and how do they arise?

A
  • Variations of a particular gene due to ifferent DNA base sequence.
  • Arises by mutation.
37
Q

What is a population?

A

A group of interbreeding individuals of the same species.

38
Q

Explain the importance of genetic diversity.

A
  1. Enables natural selection to occur.
  2. As in certain environments, a new allele of a gene might benefit its possessor.
  3. Benefit due to positive changes in polypeptide’s properties which are coded for by that new allele.
  4. Giving possessor a selective advantage.
    • Such as increased chances of survival or reproductive success.
39
Q

What is evolution?

A
  • Change in allele frequency over many generations in a population.
  • Occurring through the process of natural selection.

(Adaptation and selection are major factors in evolution and contribute to the diversity of living organisms)

40
Q

Explain the principles of natural selection in the evolution of populations.

A
  1. Mutation
  • Random gene mutations can result in new alleles of a gene.
  1. Advantage
  • In certain environments, the new allele might benefit its possessor [explain why].
    • Organism has a selective advantage.
  1. Reproductive
  • Possessors are more likely to survive and have increased reproductive success.
  1. Inheritance
  • Advantageous allele is inherited by members of the next generation (offspring).
  1. Allele frequency
  • Over many generations, allele increases in frequency in the population.

(Natural selection results in species that are better adapted to their environment)

41
Q

Describe 3 types of adaptations.

A

ANATOMICAL

  • Structural/physical features that increase chance of survival.

PHYSIOLOGICAL

  • Processes/chemical reactions that increase chance of survival.

BEHAVIOURAL

  • Ways in which the organism acts that increase chance of survival.
42
Q

Explain directional selection using an example.

A

Example

  • Antibiotic resistance in bacteria.

Key feature - who has a selective advantage?

  • Organisms with an extreme variation of a trait.
    • e.g. Bacteria with high level of resistance to a particular antibiotic.

Change in environment?

  • Yes, usually.
    • e.g. antibiotic introduced.

Effect on population over many generations

  • Increased frequency of organisms with/alleles for extreme trait.
  • Normal distribution curve shifts towards extreme trait.
43
Q

Explain stabilising selection using an example.

A

Example

  • Human birth weight.

Key feature - who has a selective advantage?

  • Organisms with an average/modal variation of a trait.
    • e.g. babies with an average weight.

Change in environment?

  • No, usually stable.

Effect on population over many generations

  • Increased frequency of organisms with/alles for average trait.
  • Normal destruction curve similar, less variation around mean.
44
Q

What is a species?

A

A group of organisms that can interbreed to produce fertile offspring.

45
Q

Suggest why 2 different species are unable to produce fertile offspring.

A
  • Different species have different chromosome numbers -> offspring may have odd chromosome number.
  • So homologous pairs cannot form -> meiosis cannot occur to produce gametes.
46
Q

Explain why courtship behaviour is a necessary precursor to successful mating.

A
  1. Allows recognition of members of same species so fertile offspring produced.
  2. Allows recognition/attraction of opposite sex.
  3. Stimulates/synchronises mating/production/release of gametes.
  4. Indicates sexual maturity/fertility.
  5. Establishes a pair bond to raise young.
47
Q

Describe a phylogenetic classification system.

A
  • Species arranged into groups, called taxa, based on their evolutionary origins and relationships.
  • Uses a hierarchy:
    • Smaller groups are placed within larger groups.
    • No overlap between groups.
48
Q

Name the taxa in the hierarchy of classification.

A

Domain (largest/broadest) -> Kingdom -> Phylum -> Class -> Order -> Family -> Genus -> Species (smallest)

49
Q

How is each species universally identified?

A
  • A binomial consisting of the name of its genus and species.
    • e.g. Homo sapiens.
50
Q

Suggest an advantage to binomial naming.

A

Universal so no confusion as many organisms have more than one common name.

51
Q

How can phylogenic trees be interpreted?

A
  • Branch point = common ancestor.
  • Branch = evolutionary path.
  • If two species have a more recent common ancestor, they are more closely related.
    • e.g. C & D in the diagram below.
52
Q

Describe two advances that have helped to clarify evolutionary relationships between organisms.

A
  1. Advances in genome sequencing allow for comparison of DNA base sequences.
    • More differences in DNA sequences, more distantly related/earlier common ancestor.
    • As mutations build up over time.
  2. Advances in immunology allow for comparison of protein tertiary structure.
    • Higher amount of protein from one species binds to antibody against the same protein from another species, more closely related/more recent common ancestor.
    • As indicates a similar amino acid sequence and tertiary structure.
    • So less time for mutations to build up.
53
Q

What is biodiversity?

A
  • Variety of living organisms (species, genetic & ecosystem diversity).
  • Can relate to a range of habitats, from a small local habitat to the Earth.
54
Q

What is a community?

A

All populations of different species that live in an area.

55
Q

What is species richness?

A

A measure of the number of different species in a community.

56
Q

What does an index of diversity do?

A

Describes the relationship between:
1. The number of species in a community (species richness).
2. The number of individuals in each species (population size).

57
Q

Suggest why index of diversity is more useful than species richness.

A
  • Also takes into account number of individuals in each species
  • So takes into account that some species may be present in small or high numbers.
58
Q

What is the formula for index of diversity?

A
59
Q

List the steps involved in calculating an index of diversity.

A
  1. Calculate the total number of organims (N), if not given.
  2. Multiply N by (N-1).
  3. For each species, multiply the number of organisms (n) by (n-1).
  4. Add up all the values of n(n-1) to get ∑n(n-1).
  5. Divide N(N-1) by ∑n(n-1).
60
Q

Describe how index of diversity values can be interpreted.

A

HIGH

  • Many species present (high species richness) and species evenly represented.

LOW

  • Habitat dominated by one/a few species.
61
Q

Explain how some farming techniques reduce biodiversity.

A

FARMING TECHNIQUES AND HOW THEY REDUCE BIODIVERSITY

  1. Removal of woodland and hedgerows.
  2. Monoculture (growing one type of crop).
  3. Use of herbicides to kill weeds.
    • Reduces variety of plants
    • So fewer habitats and niches.
    • And less variety of food sources.
  4. Use of pesticides to kill pests.
    • Predator population of pests decreases.
62
Q

Explain the balance between conservation and farming.

A
  • Conservation required to increase biodiversity.
  • But when implemented on farms, yields can be reduce, reducing profit/income for farmers.
    • e.g. by reducing land area for crop growth, increasing competition, increasing pest population.
  • To offset loss, financial incentives/grants are offered.
63
Q

Give examples of how biodiversity can be increased in areas of agriculture.

A
  • Reintroduction of field margins and hedgerows.
  • Reduce use of pesticides.
  • Growing different crops in the same area (intercropping).
  • Using crop rotation of nitrogen-fixing crops instead of fertilisers.
64
Q

How can genetic diversity within or between species be measured?

A
  • Comparing frequency of measurable or observable characterisitcs.
  • Comparing base sequence of DNA.
  • Comparing base sequence of mRNA.
  • Comparing amino acid sequenceof a specific potein encoded by DNA or mRNA.
65
Q

Explain how comparing DNA, mRNA and amino acid sequences can indicate relationships between organisms within a species and between species.

A
  1. More differences in sequence, more distantly related/earlier common ancestor.
  2. As mutations build up overtime.
  3. More mutations cause more changes in amino acid sequences.
66
Q

Explain the change in methods of investigating genetic diversity over time.

A
  1. Early estimates made by inferring DNA differences from measurable or observable characteristics.
    • Many coded for by more than one gene.
      • Difficult to distinguish one from another.
    • Many influenced by environment.
      • Differences due to environment, not genes.
  2. Gene technologies allowed this to be replaced by direct investigation of DNA sequences.
67
Q

Explain the key considerations in quantitative investigations of variation within a species.

A
  1. Collect data from random samples (use random number generator) to remove bias.
  2. Use large sample size so representative of whole population.
  3. Ethical sampling so to not harm organism/allow release unchanged.
  4. Calculate a mean value of collected data and standard deviation of that mean.
  5. Interpret mean values and their standard deviations.
    • Standard deviations shows spread of values about the mean.
      • Higher standard deviation = higher variation.
    • If standard deviations overlap, causing values of two sets of data to be shared, any difference between the two may be due to chance/not significant.
  6. Use [named] statistical test to analyse whether there is a significant difference between populations.

AQA A-Level Biology (5 decks)

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