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Y12 Biology > 4. Genetic Variation & Biodiversity > Flashcards

4. Genetic Variation & Biodiversity 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 and not associated with histone 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 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 (chromosome)

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5
Q

Describe the nature of the genetic code

A

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
Degenerate —> an amino acid can be coded for by more than 1 base triplet
Non-overlapping —> each base is part of only one triplet so each triplet is read as a discrete unit

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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:
1. Between genes - e.g non-coding multiple repeats
2. Within genes - introns

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7
Q

What are introns and exons?

A

Introns: base sequence of a gene that doesn’t code for amino acids, in eukaryotic cells
Exons: base sequence of a gene coding for amino acid sequences (in a polypeptide)

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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

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9
Q

Describe the 2 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

Similarities:
- Both single polynucleotide strand
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 nucelotides)
- tRNA has 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 floating 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 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 (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 energ from the 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 in translation

A
  • Hydrolysis of ATP to ADP + Pi releases energy
  • So amino acids join to tRNA’s and peptide bonds form between amino acids
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15
Q

Describe the role of tRNA in translation

A
  • Attaches to/transports a specific amino acid, in relation to its anticodon
  • tRNA anticodon complementary base pairs to mRNA codon, forming hydrogen bonds
  • 2 tRNA’s bring amino acids together so a peptide bond can form
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16
Q

Describe the role of ribosomes in translation

A
  • 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 next codon)/translocation
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17
Q

What is a gene mutation?

A
  • A change in the base sequence of DNA (on chromosomes)
  • Can arise spontaneously during DNA replication (interphase)
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18
Q

What is a mutagenic agent?

A

A factor that increases rate of gene mutation, e.g UV light

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19
Q

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

A
  1. changes sequence of base triplets in DNA so changes sequence of codons on mRNA
  2. so changes sequence of amino acids in the polypeptide
  3. so changes position of hydrogen/ionic/disulfide bonds between amino acids
  4. so changes tertiary structure of protein
  5. Enzymes - active site changes shape so substrate cannot bind and e/s complexes can’t form
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20
Q

Explain the possible effects of a substitution mutation

A
  1. Base/nucleoide 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/disulfide bonds change
    OR amino acid doesn’t change —> due to degenerate nature of genetic code (triplet could code for the same amino acid) OR if mutation is in an intron (non-coding parts of DNA)
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21
Q

Explain the possible effects of a deletion mutation

A
  1. one nucleotide/base is 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/disulfide bonds in tertiary structure of protein
  6. changes tertiary structure/shape of protein
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22
Q

Describe features of homologous chromosomes

A

Same length, same genes at same loci, but may have different alleles
Maternal and paternal chromatids joined at the centromere

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23
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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24
Q

Describe how a cell divides by meiosis

A

In interphase, DNA replicates —> 2 copies of each chromosome (sister chromatids), joined by a centromere
1. Meiosis I (first division) seperated homologous chromosomes
> chromosomes arrange into homologous pairs
> crossing over between homologous chromosomes
> independent segregation of homologous chromosomes
2. Meiosis II (second division) seperated chromatids

Outcome = 4 genetically varied daughter cells
Daughter cells are normally haploid if parent cell is diploid

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25
Q

Draw a diagram to show the chromosome content of cells during meiosis (parent cell has 4 chromosomes)

A

Parent cell has 4 chromosomes
Splits into 2 cells each with 2 chromosomes
Splits into 4 cells, each with 2 chromatids

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26
Q

Explain why the number of chromosomes is halved during meiosis

A

Homologous chromosomes are seperated during meiosis I

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27
Q

Explain how crossing over creates genetic variation

A
  • Homologous pairs of chromosomes associate/form a bivalent
  • Chiasmata form (point of contact between 2 non-sister chromatids)
  • Alleles are exchanged between chromosomes
  • Creating new combinations of alleles on chromosomes
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28
Q

Explain how independent segregation creates genetic variation

A
  • Homolgous pairs randomly align at equator —> so random which chromosome from each pair goes into each daughter cell
  • Creating different combinations of alleles in daughter cells (maternal & paternal chromosomes)
29
Q

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

A
  • Random fertilisation
  • Creating new allele combinations
30
Q

Explain the different outcomes of mitosis and meiosis

A
  1. Mitosis produces 2 daughter cells, whereas meiosis produces 4 daughter cells
    > as 1 division in mitosis but 2 divisions in meiosis
  2. Mitosis maintains the chromosome number (D>D) or (H>H), whereas meiosis halves the chromosome number (D>H)
    > as homologous chromosomes seperate 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
31
Q

Explain the importance of meiosis

A
  • 2 divisions create haploid gametes
  • So diploid number is restored at fertilisation —> chromosome number maintained between generations
  • Independent segregation and crossing over creates genetic variation
32
Q

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

A
  • Mitosis occurs between stages where chromosome number is maintained
  • Meiosis occurs between stages where chromosome number halves (diploid (2n) —> haploid (n) )
33
Q

Describe how mutations in the number of chromosomes arise

A
  • Spontaneously by chromosome non-disjunction during meiosis
  • Homologous chromosomes or sister chromatids fail to seperate during meiosis
  • So some gametes have an extra copy of a particular chromosome and others have none
34
Q

Suggest how the number of possible combination of chromosomes in daughter cells following meiosis can be calculated

A

2 to the power of n, where n = number of pairs of homologous chromosomes

35
Q

Suggest how the number of possible combinations of chromosomes following random fertilisation of 2 gametes can be calculated

A

2 to the power of n, all squared, where n = the number of pairs of homologous chromosomes

36
Q

What is genetic diversity?

A

Number of different alleles of genes in a population

37
Q

What are alleles and how do they arise?

A
  • Variations of a particular gene (same locus) —> different DNA base sequence
  • Arise by mutation
38
Q

What is a population?

A

A group of interbreeding individuals of the same species

39
Q

Explain the importance of genetic diversity

A
  • Enables natural selection to occur
  • As in certain environments, a new allele of a gene might benefit its possessor
  • By resulting in a change in the polypeptide coded for that positively changes it’s properties
  • Giving possessor a selective advantage (increased chances of survival and reproductive success)
40
Q

What is evolution?

A
  • Change in the allele frequency over many generations in a population
  • Occuring through the process of natural selection
41
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
  2. Advantage —> in certain environments, the new allele might benefit its possessor (selective advantage)
  3. Reproductive success —> possessors are more likely to survive and have reproductive success
  4. Inheritance —> advantageous allele is inherited by members of the next generation (offspring)
  5. Allele frequency —> over any generations, allele increases in frequency in the population
42
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 an organisms acts that increase chance of survival
43
Q

Explain 2 types of selection, with examples

A

DIRECTIONAL SELECTION
e.g antibiotic resistance in bacteria
- Organisms with an extreme variation of a trait
- Due to change in environment (antibiotic introduced)
- Increased frequency of organisms with alleles for extreme trait
- Normal distribution curve shifts towards extreme trait
STABILISING SELECTION
e.g human birth weight
- Organisms iwith an average/modal variation of a trait
- Usually stable, no change in environment
- Increased frequency of organisms with alleles for average trait
- Normal distribution curve similar, less variation around the mean, taller and more central to 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
  • Allows recognition of members of same species —> so fertile offspring produced
  • Allows recognition/attraction of opposite sex
  • Stimulates/synchronises mating/production/release of gametes
  • Indicates sexual maturity/fertility
  • 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 (common ancestors) 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
Kingdom
Phylum
Class
Order
Family
Genus
Species

49
Q

How is each species universally identified?

A

A binomial consisting of the name of its genus and species, genus is capitalised and all written in italics

50
Q

Suggest and advantage of binomial naming

A

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

51
Q

How can phylogenetic trees be interpreted?

A
  • Branch point = common ancestor
  • Branch = evolutionary path
  • If 2 species have a more recent common ancestor, they are more closely related
52
Q

Describe 2 advances that have helped to clarify evolutionary relationships between organisms

A
  1. Advances in genome sequencing —> allows comparison of DNA base sequences
    - more differences in sequences = more distantly related/earlier common ancestor
    - as mutation build up over time
  2. Advances in immunology —> allowing comparisons of protein tertiary structure (e.g albumin)
    - higher amount of protein from 1 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
  • 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 the 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

d = N(N-1)/sum of n(n-1)

N = total number of organisms of all species
n = total number of organisms of each species (sum of all added together)

59
Q

List the steps involved in calculating an index of diversity

A
  1. Calculate the total number of organisms (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 the sum of them all (denominator)
  5. Divide N(N-1) by sum of 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
  • Removal of woodland and hedgerows
  • Monoculture (growing 1 type of crop)
  • Use of herbicides to kill weeds
    > reduces variety of plants
    > so fewer habitats and niches
    > and less variety of food sources
  • Use of pesticides to kill pests
    > predator population of pest decreases
62
Q

What is meant by the term ‘niche’?

A

The role an organism plays in a community

63
Q

Explain the balance between conservation and farming

A
  • Conservation required to increase biodiversity
  • But when implemented on farms, yields can be reduced, 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
64
Q

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

A
  • Reintroduction of field margins and hedgerows (next to monocultures)
  • Reduce use of pesticides
  • Growing different crops in the same area (intercropping)
  • Using crop rotation of nitrogen fixing crops instead of fertilisers
65
Q

How can genetic diversity within of between species be measured?

A
  • Comparing frequency of measurable or observable characteristics
  • Comparing base sequence of DNA
  • Comparing base sequence of mRNA
  • Comparing amino acid sequence of a specific protein encoded by DNA and mRNA
66
Q

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

A
  • More differences in sequences —> more distantly related/earlier common ancestor
  • As mutations build up over time
  • More mutations cause more changes in amino acid sequences
67
Q

Explain the changes in methods of investigating genetic diversity over time

A
  • Early estimates made by inferring DNA differences from measurable or observable characteristics
    > many coded for by more than 1 gene —> difficult to distinguish one from another
    > many influenced by environment —> differences due to environment not genes
  • Gene technologies allowed this to be replaced but direct investigation of DNA sequences
68
Q

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

A
  • Collect data from random samples (use a random number generator) —> removes bias
  • Use large sample sizes (or sample until stable running mean) —> representative of whole population
  • Ethical sampling —> must not harm organism/allow release unchanged
  • Calculate a mean value of collected data and SD of that mean
    > SD shows spread of values about that mean —> higher SD = higher variation
    > if SDs overlap, causing values of 2 sets of data to be shared, any difference between the 2 may be due to chance/not significant
  • Use a statistical test —> analyse whether there is a significant difference between populations

Y12 Biology (4 decks)

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