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Aqa a level biology > Topic 4: Genetic information, variation and relationships between organisms > Flashcards

Topic 4: Genetic information, variation and relationships between organisms Flashcards

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1
Q

why is the genetic code described as universal

A

each triplet codes for the same amino acid in all organisms

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2
Q

prokaryotic DNA

A

-short

-circular

-not associated with proteins

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3
Q

eukaryotic DNA

A

-long

-linear

-associated with proteins called histones

-forms chromosomes

-mitochondria and chloroplasts contains prokaryotic DNA

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4
Q

what do genes code for

A

A gene is a sequence of DNA bases that code for..:

-amino acid sequence of a polypeptide

-a functional RNA

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5
Q

what is a gene

A

sequence of DNA bases that codes for a specific sequence of amino acids in a polypeptide

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6
Q

triplet

A

A sequence of three DNA bases is called a triplet

A triplet codes for a specific amino acid. The genetic code is universal, non-overlapping and degenerate

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7
Q

degenerate

A

most amino acids are coded for by more than one triplet

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8
Q

non overlapping

A

each base in an exon is only read once

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9
Q

universal

A

each triplet codes for the same amino acid in all organisms

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10
Q

what is the fixed position which a gene occupies on a DNA molecule

A

locus/loci

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11
Q

gene coding

A

A gene includes coding exons and non-coding introns sections of DNA

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12
Q

How many bases code for a polypeptide of 24 amino acids

A

24 x 3 = 72

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13
Q

Explain how a change in sequence of DNA bases could result in a non-functional enzyme

A

-change in sequence of amino acids
-change in bonding of tertiary structure so shape of active site is altered
-no enzyme substrate complexes can be formed

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14
Q

Give two differences between the structure of mRNA and the structure of tRNA

A

-tRNA is clover shaped whereas mRNA is straight chained
-tRNA is only 80 bases (short) whereas mRNA is longer
-mRNA has no paired bases whereas tRNA does

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15
Q

describe how mRNA is produced from an exposed template strand of DNA

A

-free RNA nucleotides form complimentary base pairs
-phosphodiester bonds form by action of RNA polymerase

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16
Q

Describe how mRNA is formed by transcription in eukaryotes

A

-hydrogen bonds break between complimentary base pairs of DNA molecule by DNA helicase
-one DNA strand acts as a template
-free RNA nucleotides align by complimentary base pairings
-uracil replaced thymine
-RNA polymerase joins adjacent nucleotides by phosphodiester bonds
-pMRNA is spliced

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17
Q

what is a codon

A

sequence of 3 bases on mRNA that codes for a single amino acid

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18
Q

anticodon

A

a sequence of three nucleotides forming a unit of genetic code in a transfer RNA molecule, corresponding to a complementary codon in messenger RNA.

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19
Q

true or false - in eukaryotic cells is only where pre-mRNA is formed

A

true

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20
Q

introns

A

Introns –> sections of DNA that do not code for a polypeptide chain (protein)

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21
Q

exons

A

sections of DNA that code for a polypeptide chain (protein)

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22
Q

genome

A

-genome –> complete set of genes in an organism

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23
Q

proteome

A

-proteome –> full range of proteins produced by the genome

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24
Q

RNA

A

RNA –> single stranded, short, uracil replaces thymine, RNA polymerase, ribose pentose sugar

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25
Q

mRNA

A

mRNA –> contains genetic information from inside of the nucleus. DNA moves from cytoplasm to ribosomes for protein synthesis

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26
Q

tRNA

A

tRNA –> brings amino acids to mRNA to form proteins –> clover shape, short, only 80 bases long, amino acid attachment site, complimentary base pairs, anticodon

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27
Q

define transcription

A

-copy of DNA is made. This copy is known as messenger RNA which occurs in the nucleus of the cell

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28
Q

transcription

A

1) in nucleus, DNA helicase hydrolyses hydrogen bonds between complimentary base pairs to unwind double helix structure at target gene

2) free RNA nucleotides bind to complimentary base pairs on exposed template DNA strand

3) uracil replaces thymine

4) RNA polymerase catalyses formation of phosphodiester bonds between ribose sugar and phosphate groups of adjoining nucleotides to form sugar phosphate backbone

5) messenger RNA is formed and moves out of nucleus into cytoplasm. Pre-mRNA strand is spliced before leaving nucleus

6) DNA recoils as hydrogens bonds reform between template and coding strands

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29
Q

splicing of pre-mRNA

A

-removal of introns from mRNA to leave only exons

-functional exons joined together by splicing

-because most prokaryotic cells don’t have introns they don’t require splicing

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30
Q

define translation

A

-process where mRNaA is read and translated into a protein

-translation occurs on the ribosomes

-translation involves another type of RNA molecules known as tRNA

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31
Q

similarities between mRNA and tRNA

A

-both contain uracil

-both single stranded

-both contain codons (triplet bases)

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32
Q

differences between mRNA and tRNA

A

-mRNA is straight whereas tRNA is clover shaped

-mRNA is less stable than tRNA

-mRNA is larger than tRNA

-tRNA has an anticodon whereas mRNA doesn’t

-tRNA has an amino acid binding site whereas mRNA doesn’t

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33
Q

translation

A

1) mRNA leaves nucleus via a nuclear pore

2) mRNA attaches to a ribosome at start of codon

3) tRNA with an anticodon complimentary to start codon arrives at ribosome. TRNA molecule has specific amino acid attached to it

4) another tRNA molecule binds to second codon on mRNA. This brings another amino acid with it

5) Peptide bond forms between two amino acids

6) First tRNA molecule then leaves and process continues until stop codon is released. This is complimentary to the anticodon on a tRNA which does not have an associated amino acid. This signal the end of a polypeptide chain

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34
Q

define the term exon

A

sequence of DNA that codes for a polypeptide chain

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35
Q

explain how the proteome of a genetically modifed cell differs to one that is not genetically modified

A

-expression of a gene from different species
-new protein is formed

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36
Q

explain how can increase in rate of transcription of PIP1b gene will affect permeability of tobacco plant

A

-more aquaporin channels
-increase permability

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37
Q

describe the role of a ribosome in the production of a polypeptide

A

-mRNA attaches to ribosome at start codon
-2 codons
-allows tRNA iwth anticodons to bind
-specific sequence of amino acids with peptide bonds formed

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38
Q

true or false - pre-mrna has introns and mRNA doesnt

A

true

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39
Q

starting with mRNA in the cytoplasm, describe how translation leads to the production of a polypeptide

A

-mRNA attaches to ribosome
-codon on mRNA binds to anticodon on tRNA
-each tRNA brings a specific amino acid
-sequence of codons determines order of amino acids
-formation of peptide bonds

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40
Q

define species

A

organisms with similar features that are capable of breeding to produce living fertile offspring

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41
Q

naming organism

A

binomial systems
genus + species

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42
Q

signs of a recent common ancestor

A

-physical features, courtship behaviour, genome sequencing, amino acid sequencing, immune response

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43
Q

courtship behaviour (essential for successful mating)

A

-attracts members of the same species
-attracts member of the opposite sex
-indicated readiness to mate
-simulates making and release of gametes
-encouraged bond to raise young

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44
Q

start vs stop codon

A

start -> first 3 bases on DNA or mRNA
stop –> final 3 bases that do not code for an amino acid (mark end of polypeptide chain and cause ribosomes to detach)

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45
Q

how many amino acids does the genetic code, code for

A

20

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46
Q

what is a mutation

A

change in base sequence on a chromosome which results in the formation of a new allele

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47
Q

allele

A

different version of a gene

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48
Q

explain how a gene mutation could result in a new protein

A

-change in sequence of bases as a result of subsitution
-change in primary amino acid sequence
-change in tertiary structure of a protein

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49
Q

Explain why mutation 1 leads to the production of non-functional protein

A

-deletion of a base alters triplet/codon on mRNA molecule
-this change results in change in primary sequence of amino acids
-this change alters the hydrogen bonding in tertiary structure

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50
Q

define gene mutation and explain how a gene mutation can have no effect and a positive effect

A

-gene mutation is a change in base sequence of a chromosome molecule which results in the formation of a new allele

no effect –> when mutation occurs in an intron as it does not code for a polypeptide or if mutation occurs in recessive allele

positive effect –> change in properities of protein which leads to increased survival and reproductive success

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51
Q

define a mutagenic agent and give an example

A

-environmental factors which increase rate of mutation

e.g high energy radiation

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52
Q

what is a substitution mutation

A

replacement of a base by a different base in DNA

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53
Q

mutations

A

-a mutation is a change in the base sequence on a chromosome which results in the formation of a new allele

-mutations are random events which occur in S phase of cell cycle

-Some environmental factors can increase rate of mutation e,g toxins/radiation

-mutations don’t always change the protein formed e.g when it occurs in introns or bc the genetic code is degenerate bc same amino acid is being coded for

-some mutations result in a change in protein structure due to change in amino acid sequence which affects bonding in teritary structure

-mutations don’t always have negative effects e.g can increase an organisms chance of survival

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54
Q

causes of mutations

A

-occur spontaneously during interphase as DNA is replicated and it is the longest stage of the cell cycle

-high energy radiation or ionising energy can disrupt DNA e.g x-rays or UV light

-chemicals e.g nitrogen dioxide can disrupt DNA or interfere with transcription. Some chemicals in cigarette smoke can inactivate a tumor suppressor gene leading to cancer

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55
Q

natural mechanisms

A

-Natural mechanisms exist within cells to help identify and repair damaged to DNA but can become ineffective if rate of mutation increases above the normal rate

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56
Q

environmental factors which increase mutation rate

A

–> toxic chemicals e.g peroxides or bromine compounds

—> ionising radiation e.g gamma rays

–> high energy radiation e.g UV

–> some viruses – transfer of viral DNA into host DNA

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57
Q

what are the 2 types of mutation

A

-substiution
-deletion

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58
Q

substitution mutation

A

-wrong base is included in base sequence. May result in different amino acid being included in polypeptide chain

-however, if substitution results in a triplet that still codes for the same amino acid it may not change the sequence of amino acids at all bc the genetic code is degenerate

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59
Q

deletion mutation

A

-base is lost from the base sequence

-as a result the whole base sequence following the deleted base moves back one place (frame shift) which often has a significant effect on the encoded protein bc it can alter sequence of all codons following base deletion

-change of primary structure = fault enzymes/altered teritary structure

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60
Q

Describe how a gene is code for polypeptide

A

-nucleotide sequence in triplet determines primary sequence of amino acids

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61
Q

define non-coding base sequences and describe where it is positioned in the genome

A

-DNA that doesnt code for a polypeptide
-positioned between genes

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62
Q

similarites of DNA in prokaryotes and eukaryotes

A

-nucelotide structure is identical
-joined by phosphodiester bonds

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63
Q

suggest one way the structure of the chromosome could differ along its length to result in the stain binding to some more areas

A

-difference in base sequence

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64
Q

what is a homogolous pair of chromosomes

A

two chromosomes that carry the same genes

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65
Q

describe and explain the appearance of one of the chromosomes in cell X

A

-chromosomes are constructed from 2 chromatids
-due to cells undergoing replication these are held together by the centromere

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66
Q

describe what has happened during division 1 of meiosis

A

homologous pair of chromosomes have been separated into each daughter cell

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67
Q

identify one event that occurred during division 2 but not during division 1

A

separation of chromatids

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68
Q

name 2 ways in which meiosis produces genetic variation

A

-independent segregation of homologous chromosomes
-crossing over

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69
Q

name how two amino acids can differ from one another

A

have different R groups

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70
Q

why is the genetic code described as being universal

A

in all organisms the same triplet codes for the same amino acid

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71
Q

explain why homologous chromosomes carry the same genes but are not genetically identical

A

carry different alleles

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72
Q

give one way which meiosis allows the production of genetically different cells

A

crossing over (alleles exchanged between chromosomes)

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73
Q

meiosis

A

-In meiosis the daughter cells are haploid and genetically different (4 daughter cells)

-involves 2 nuclear divisions

-2 stages in meiosis

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74
Q

diploid vs haploid

A

Diploid = 2 cells = full sets of chromosomes

Haploid = 4 cells = half sets of chromosomes

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75
Q

define a tetrad

A

collection of 2 sets of homologous pairs (crossover)

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76
Q

prophase 1

A

Prophase 1 (variation):

-chromosomes condense and become visible

-crossing over = recombination of genetic material from chromatids

-provides genetic variation

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77
Q

metaphase 1

A

Metaphase 1 (random)

-chromosomes line up in homologous pairs at the equator

-chromosomes separate on the spindle fibers

-centromere with kinetochore

-microtubule attached to kinetochore

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78
Q

anaphase 1

A

Anaphase 1 (genetic crossover):

-independent segregation of chromosomes

-spindle fibres shorten

-pairs of homologous chromosomes split up

79
Q

homologous pairs

A

two chromosomes with the same genes at the same loci

80
Q

variation

A

Meiosis creates genetic variation through crossing over and random arrangement

81
Q

first vs second division

A

First division:

-homologous chromosomes pair up and their chromatids wrap around eachother

-crossing over

-homologous pairs separate with one chromosome from each pair going into one of 2 daugther cells

Second division:

-chromatids move apart

-4 haploid daughter cells formed

82
Q

name 2 biological molecules that can be codes for by a gene

A

-tRNA
-rRNA

83
Q

kinetochore

A

a protein structure that is important for linking the chromosomes to the mitotic spindle and is assembled on the centromere.

84
Q

explain how natural selection would produce a copper tolerant population in the mine waste

A

-a random mutation occured causing the grass to have higher copper tolerance
-natural selection will then occur as the grass with the copper tolerance is better adapted to the environment and is more likely to survive
-This means that the newly mutated grass is more likely to reproduce and pass on their advantageous alleles to their offpsring creating a new population of copper tolerant grass
-thusincreasing the frequency of copper tolerant alleles

85
Q

define species

A

Species –> group of organisms that are able to interbreed to produce fertile offspring

86
Q

define population

A

Population –> a group of organisms of the same species living in the same area at the same time

87
Q

define genetic diversity

A

Genetic diversity –> total number of different alleles of genes in a population

The greater the genetic diversity the greater the likelihood that the same organisms will survive an environmental change

DNA results in genetic diversity

88
Q

mutation = competitive

A

Most mutations are harmful but can lead to a competitive advantage in a population. This mutation increases the possessors evolutionary success

89
Q

steps for natural selection

A

1) gene pool (wide variety of alleles)

2) Random mutation (new alleles)

3) Better adapted and better competitors

4) more likely to reproduce

5) Pass on successful advantageous alleles

6) survive and reproduce

7) Over time increased frequency of alleles

90
Q

directional selection

A

-Normal distribution which shows where most common phenotype is

-shift to one extreme so one of the extreme phentoypes becomes most common

91
Q

stabilising selection

A

-Shift to the middle = less extremes because extremes give a rise to disadvantages
-favour average

92
Q

selection definition

A

Differences between the reproductive success of individuals affects allele frequency in populations

Selection –> process by which organisms that are better adapted to their environment tend to survive and breed

93
Q

suggest one type of guppy behaviour that could be affected by the presence of brightly coloured stones

A

-mating habits change

94
Q

describe a further investigation and null hypothesis

A

-include a control group with no coloured stones to compare
-there will be no difference….any difference will be due to chance

95
Q

name the type of selection and explain why the frequency of alleles was affected

A

-directional selection
-extreme values are favoured
-random mutation becomes advantageous

96
Q

the enzyme coded for one allele results in faster ctalysis than the other explain why

A

-different primary amino acid sequence
-difference tertiary structure so shape of active site changes
-formation of E-S complexes more likely

97
Q

suggest why the scientist took his sample from the population at random

A

-so results are more likely to be reliable

98
Q

stages of meiosis

A

The stages of meiosis can be split into two nuclear divisions, they are summarised below:
Meiosis 1 - homologous chromosomes pair up whereby crossing over at the chiasmata may take
place. The cell then divides whereby each daughter cell contains one chromosome from each
homologous pair.
Meiosis 2 - the chromatids of each chromosome are separated producing 4 haploid daughter
cells.

99
Q

independent segregation of chromosomes

A

Independent assortment of chromosomes – there are various combinations of
chromosome arrangement. During meiosis 1 homologous chromosomes line up in pairs,
the arrangement of these pairs is random, meaning that the division into the daughter
cells is also random.

100
Q

crossing over of chromatids

A

Crossing over of chromatids - When pairs of chromosomes line up they can exchange
some of their genetic material. Crossing over occurs when one chromosome may swap
places with the same part of its homologous pair leading to a different combination of
alleles on the gene.

101
Q

directional vs stabilising selection

A

directional selection occurs when the environmental conditions change and the phenotypes best sutited to the new condition is more likely to survive whereas stabilising selection is when phenotypes with successful characteristics are preserved and genetic diversity is reduced

102
Q

xerophytic adaptations

A

-spines for leaves
-extensive root network
-sunken stomata
-low stomatal density
-rolled leaves
-fleshy stem

103
Q

types of adaptations

A

-Adaptations can be behavioural, physiological or anatomical

104
Q

behavioral adaptation

A

Behavioural = changes in an organisms behaviour e.g closing of stomata when little water avaliable

105
Q

physiological adaptation

A

Physiological = changes in the mechanisms within and an organism e.g plant mechanism to close the stomata or rolled leaves

106
Q

anatomical adaptation

A

Anatomical = changes in the physical structure of an organism e.g spines for leaves

107
Q

importance of courtship behaviour and what does it ensure

A

-courtship behaviour ensures successful production of fertile offspring

Importance of courtship behaviour:

-recognise/identify/attract organisms of the same species

-indicates sexual maturity and fertility

-formation of a paid bond

-stimulate mating and production of gametes

108
Q

what is the importance of meiosis

A

halves chromosome number

109
Q

the genetic does is described as being degenerate, What do this mean

A

one amino acid can be coded for by more than one triplet

110
Q

what is a codon

A

triplet that codes for an amino acid

111
Q

what is the role of RNA polymerase during transcription

A

join nucleotides together to form mRNA

112
Q

mRNA can be converted to cDNA - what is the enzyme

A

reverse transcriptase

113
Q

two molecules from which a ribosome is made

A

RNA and proteins

114
Q

steps for meiosis

A

1) production of sister chromosomes/chromatids (replication)

2) chromosomes are found in homologous pairs and are arranged at equator of the cell

3) homologous chromosomes are separated and pulled to opposite parts of the cell (random)

4) meiosis 1 (2 genetically different cells)

5) sister chromatids separated again

6) Meiosis 2 = 4 haploid genetically different daughter cells

115
Q

suggest why the frequency of ADf allele changed

A

-ADF allele has a selective advantage
-more likely to reproduce and pass on allele
-allele frequency increases

116
Q

suggest and explain one reason why bacteria resistant to tetracycline was more common

A

-tetracyline used in higher doses
-resistance bacteria more likely to survive, reproduce and pass on allele for resistance

117
Q

non-disjunction mutation

A

in meiosis, chromosomes not separated so dont form homologous pairs

118
Q

apart from mutation select another way in which genetic variation within a species is increased

A

-random fertilisation
-produces new allele combination

119
Q

Explain how the chromosome pair is halved during meiosis

A

one of each homologous pair goes to opposite poles of the cell

120
Q

Describe the process of crossing over and how it increases genetic diversity

A

-homologus pairs of chromosomes associate
-alleles are exchanged on the chiasma producing new combinations of alleles

121
Q

Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation

A

-tRNA brings a specific amino acid
-anticodon on tRNA binds to codon on mRNA
-amino acids join by condensation reaction to form peptide bond

122
Q

describe how a gene is code for the production of the polypeptide

A

-base sequence in triplet determines sequence of amino acid

123
Q

calculate no. of chromosomes - meiosis

A

2 to the power of n

n = number of homologous chromosomes (same genes, different alleles)

124
Q

crossing over

A

-chromatids twisted around eachother
-this puts tension on chromatids causing it to break off
-when it recombines crossing over happens

125
Q

describe and explain 3 ways in which the scientist would ensure he used aseptic techniques to move each cube of agar onto a new agar plate

A

-flame instruments to sterilise and kill any contaminating microbes
-lift lid gently to prevent microbes getting in
-disinfect working surface to prevent contamination

126
Q

give two differences between mitosis and meiosis

A

-mitosis produces 2 genetically identical diploid daughter cells whereas meiosis produces 4 genetically different haploid daughter cells
-mitosis has 1 cell division whereas meiosis has 2

127
Q

describe how the process of meiosis results in the formation of haploid cells (4 marks)

A

-DNA replication
-2 divisions
-separation of homologous chromosomes in first division
-formation of 4 genetically different haploid daughter cells

128
Q

suggest two reasons why it was important that they were of similar age

A

-to ensure they attract a mate
-so they have similar sexual maturity

129
Q

explain the principles which biologists use to classify organisms into groups

A

-large groups divided into smaller groups based on evolutionary relationships
-members of each group have common features (homologous characteristics)

130
Q

Give one feature of a hierarchy that is shown in the diagram

A

no overlap

131
Q

what is meant by a hierarchy

A

-large groups divided into smaller groups (groups within groups) and no overlap

132
Q

explain the role of independent segregation in meiosis

A

-provides genetic variation as it allows different combinations of maternal and paternal alleles

133
Q

who developed the first classification system

A

Carl Linneaus

134
Q

order of classification

A

-Domain –> Kingdom –> Phylum –> Class –> order –> family –> genus –> species

135
Q

classification vs taxonomy

A

-Classification = group of organisms

-Taxonomy –> theory and practice of biological classification

136
Q

two types of classification

A

phylogenetic and artificial

137
Q

artificial classification

A

-Artificial classification = divides organisms according to features and differences

138
Q

phylogenetic classification

A

-Phylogenetic classification = based on evolutionary relationships between organisms and their ancestors with no overlap.

Uses homologous characteristics = characteristics that have similar evolutionary origin

139
Q

3 domains

A

3 domains –> bacteria, archaea, eukarya (classification based on DNA)

140
Q

describe how organisms are grouped in a phylogenetic classification system

A

-heirachy of groups with no overlap
-groups according to evolutionary relationships

141
Q

Explain how these sequences could provide evidence for different species

A

-different species would have different sequences of amino acids
-amino acid sequence is the result of DNA

142
Q

state 3 comparisons of genetic diversity that the scientists could use

A

-base sequence of DNA
-base sequence of mRNA
-amino acid sequence of proteins

143
Q

in this investigation what is meant by genetic diversity

A

the number of different alleles of each gene

144
Q

suggest a reason why the diversity index for the lacewings is different between the two crops

A

more prey found on strawberries

145
Q

bacteria

A

-single celled prokaryotes
-no membrane bound organelles such as nuclei
-cell walls made of murein
-circular DNA

146
Q

archaea (differ to bacteria)

A

-genes and protein synthesis more similar to eukaryotes
-membranes contain ether linkages
-no murein in cell walls
-complex form of RNA polymerase

147
Q

Eukarya

A

-membrane bound organelles
-membranes have ester linkages
-cell wall
-larger ribosomes

148
Q

Eukaraya domain divided into 4 kingdoms

A

-Protoctistica
-Fungi
-plantae
-animalia

149
Q

biodiversity

A

term used to describe variety in the living world

150
Q

species diversity

A

can be measured by looking at the species richness and evenness

151
Q

species richness and evenness

A

Species richness = total number of different species

Species evenness –> number of individuals of each species number of individuals of each

152
Q

genetic diversity

A

Genetic diversity –> number of alleles within a species (variety of genes possessed)

153
Q

ecosystem diversity

A

Ecosystem diversity –> range of different habitats

154
Q

how to measure genetic diversity

A

-amino acid sequence of proteins encoded by DNA and mRNA

-the base sequence of DNA

-base sequence of mRNA

-frequency of measurable or observable characteristics

155
Q

index of diversity

A

An index of diversity is a measurement that describes the relationship between the number of species present and how each species contributes to the total number of organisms that are present in that community

156
Q

formula for index of diversity

A

n = total no. of organisms for a single species in the community

N = total no. of organisms in the community

Σ = sum of

The larger the number obtained, the higher the level of diversity

157
Q

how to calculate index of diversity

A

To calculate:

Step 1: Calculate N(N-1) to find value A

Step 2: Calculate n(n-1) for each species

Step 3: Add these numbers together to find value B

Step 4: Divide value A by value B

158
Q

give one advantage of calculating the index of diversity rather than just recording the number of species present

A

almost all of the sample could be the same species

159
Q

using SD date from the table describe the differences in prokaryotic diversity found in the soil with these 2 farming methods

Define species richness and index of diversity in ur answe

A

-species richness = number of species in a community

index of diversity = relationship between number of species in a community and number of individuals in each species

-no SD overlap for species richness so it is significant

-index of diversity is not signifcant due to SD overlap

160
Q

limitations of DNA sequencing

A

-non-coding DNA
-amino acid code is degenerate

161
Q

suggest why in recent years has our knowledge about prokaryotic diversity in the soil increased

A

-DNA sequencing is now used
-observe more prokaryotes than before

162
Q

Using a phylogenetic classification, all of these species have names that start with Apodemus. What information does this give

A

-they have the same genus
-they all share a common ancestor

163
Q

common ancestor =

A

phylogenetic classification

164
Q

describe how breeding experiments could determine whether the two populations are from the same species

A

-breed 2 mice from different populations together
-if produce fertile offspring = same species

165
Q

why do humans and grasshoppers have similar percentages of each base in their DNA but are very different organisms

A

-different genes
-different triplet sequence
-different primary sequence of amino acids

166
Q

why is DNA of virus different to that of other organisms

A

-no base pairing
-DNA single stranded

167
Q

describe the primary structure of all proteins

A

-sequence of amino acids joined by peptide bonds

168
Q

universal, non-overlapping, degenerate

A

universal = same triplet codes for the same amino acid

non-overlapping = each base is only part of one triplet

degenerate = more than one triplet codes for an amino acid

169
Q

outline the similarties and differences between structure of chloroplasts and mitochondria

A

Similarities:
-ribosomes, circular DNA, double membrane

Differences:
-stroma vs matrix
-pigment vs no pigment
-starch grain vs no starch grain

170
Q

explain how a gene codes for a protein

A

-base sequence in gene determines sequence of amino acids
-triplet codes for amino acid

171
Q

define homologous chromosomes

A

-pairs of chromosomes with the same genes at the same loci

172
Q

the method the scientists used resulted in them getting different scores for the same band. explain why

A

band width is not the same on both sides of the tail

173
Q

explain how scientists could use this information to show that some variation in tail banding was genetic

A

-same family = look similar bc genetically similar bc from same parent
-random = more different

174
Q

describe the cohesion tension theory

A

-water lost from leaf due to transpiration > evaporation through the stomata
-lowered water potention
-water pulled up xylem creating tension
-sticky H bonds = cohesion
-adhesion of water molecules to wall of the xylem

175
Q

explain the importance of meioisis

A

-meiosis halves number of chromosomes (haploid cells)
-during fertilisation diploid number is resotred
-chromosome number remains constant
-introduces genetic variation

176
Q

what happens to chromosomes during meiosis

A

-chromosomes condense
-associate in homologous pairs
-crossing over
-join to spindle fibres at equator of cell. Joined via centromere
-homologous chromosomes move to opposite poles
-chromatids seperated in 2nd division

177
Q

meiosis and genetic variation

A

How it happens:
-crossing over
-independent segregation of chromosomes in meiosis I
-independent segregation of chromatids in meiosis II

Advantages:
-better adapted = increased survival
-reproduce and pass on genes

178
Q

give one advantage of calculating the index of diversity rather than just recodring the number of species present

A

-takes account the number of individuals/population size

179
Q

explain how differences in the primary structure of haemolgobin molecules can provide evidence of phylogenetic relationships between species

A

-mutations change base sequence
-causing change in amino acid sequence
-mutations build up over time
-more mutations = more differences in species that are closely related
-distantly related species have easier common ancestor

180
Q

role of ribosomes and tRNA in translation

A

ribosomes = assemble amino acids to proteins by faciliating tRNA binding

tRNA = contains anticodon which is specific to codon on mRNA molecule

181
Q

term used to describe ionising radiation

A

mutagenic agent

182
Q

how does DNA replicate to produce sister chromatids

A

semi-conservative replication

183
Q

why is standard deviation more useful than range

A

-shows spread of data around the mean
-reduces effect of anomalies
-indicate signifiance of results

184
Q

dm3 =

A

1 L

185
Q

structure of evaluate question

A

there is a positive correlation between
However data overlaps so correlation doesnt prove causation
no correlation coefficient to know signifiance

186
Q

10^-7 is greater than 10^-8 true or false

A

true

187
Q

minimum percentage decrease

A

distorted value - top end value given / distorted value

188
Q

how to reduce uncertainity

A

-increase resolution of instrument used (mm instead of cm)
-take repeated readings
-incrase size

189
Q

reading vs measurement

A

reading = single judgement e.g thermometer = +- 0.5

measurement = two judgements = ruler or stopwatch = +- 1.0 of the smallest division

190
Q

why might his conclusion not be valid

A

only done on 1 type of cancer-> should be done on more types

dont know how much ECGC is in green tea

not all cancers are affected the same way

191
Q

antibiotic resistance bacteria

A

-some bacteria have an allele for antibiotic resistance
-resistant bacteria survivie and reproduce
-high frequency of resistant allele in bacterial population

192
Q

greater inhibition zone =

A

more effective antibiotic (more bacteria killed)

193
Q

index of diversity

A

adundance of species + how many species

Aqa a level biology (8 decks)

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