The document discusses meiosis, gametogenesis, and genetic variation, explaining the significance of homologous chromosomes, haploid and diploid terms, and the processes that generate genetic diversity through crossing over and independent assortment. It also covers Mendelian genetics, including monohybrid and dihybrid crosses, along with the principles of dominance and recessiveness in alleles. Additionally, it addresses mutations, gene control in prokaryotes, and the specifics of operons, shedding light on the complexities of inheritance and gene expression.

1. Inherited Change

3. OBJECTIVE
• explain what is meant by homologous pairs of chromosomes
• explain the meanings of the terms haploid and diploid and the need
for a reduction division (meiosis) prior to fertilisation in sexual
reproduction
• Outline the role of meiosis in gametogenesis in humans and in the
formation of pollen grains and embryo sacs in flowering plants
• describe, with the aid of photomicrographs and diagrams, the
behaviour of chromosomes in plant and animal cells during meiosis,
and the associated behaviour of the nuclear envelope, cell surface
membrane and the spindle
• explain how crossing over and random assortment of homologous
chromosomes during meiosis and random fusion of gametes at
fertilisation lead to genetic variation including the expression of rare,
recessive alleles

8. • Diploid organisms contain pairs of homologous
chromosomes.
• Homologous chromosomes:
– a pair of chromosomes in a diploid cell
– same structure, genes, loci
– pair together to form bivalent during the first
division of meiosis

9. • Haploid: possesses 1 complete set of
chromosomes: n
• Diploid: possesses 2 complete sets of
chromosomes: 2n
• Meiosis: reduction division
• introduces genetic variation mutation

10. Life cycle of an animal

11. GAMETOGENESIS

12. Meiosis
• Used in sexual reproduction to allow for
variation.
• Homologous chromosome – chromosomes that
have the same genes but not the same alleles.
One from mum and one from dad.
• Bivalent – when the homologous chromosomes
have replicated they join together by a chiasma
and this forms a bivalent made of 4 chromatids.
• Crossing over – the non-sister chromatids that
lie next to each other in the bivalent may join
temporarily and then break off swopping some
of their genetic information.

15. Variation
• Due to:
– Crossing over of non-sister chromatids in the
bivalent
– Independent Assortment of bivalents and
sister chromotids during metaphase I and II
– Random fertilisation

17. Gametogenesis

22. OBJECTIVE
• explain what is meant by homologous pairs of chromosomes
• explain the meanings of the terms haploid and diploid and the need
for a reduction division (meiosis) prior to fertilisation in sexual
reproduction
• Outline the role of meiosis in gametogenesis in humans and in the
formation of pollen grains and embryo sacs in flowering plants
• describe, with the aid of photomicrographs and diagrams, the
behaviour of chromosomes in plant and animal cells during meiosis,
and the associated behaviour of the nuclear envelope, cell surface
membrane and the spindle
• explain how crossing over and random assortment of homologous
chromosomes during meiosis and random fusion of gametes at
fertilisation lead to genetic variation including the expression of rare,
recessive alleles

23. How meiosis causes variation
Crossing over:
prophase of meoisis I
chromatids of 2 homologous chromosomes break and rejoin --> part of
one chromatid swaps places with the same part of the
other
chiasmata: point where crossing over occurs
Recombinant organisms result from crossing over and so 'recombines' the
characteristics of the parent organisms. The cross over value is the
percentage of offspring that belong to the recombinant class.

24. Independent assortment:
random alignment of bivalents on the equator during meiosis I
different alleles of genes on different chromosomes may end up in any
combination in gametes

25. A B
C
D

26. • Studies of human genetic conditions have
revealed the links between genes,
enzymes and the phenotypes

28. Definitions
• gene: length of DNA that codes for a particular
protein/polypeptide
• locus: position at which a particular gene is found
on a particular chromosome; same gene on same
locus
•
• allele: particular variety of a gene
•
• dominant: the allele whose effect on the phenotype
of a heterozygote is identical to its effect on a
homozygote
• recessive: the allele that is only expressed when
no dominant allele is present

29. • codominant: alleles that both have an effect on the
phenotype of a heterozygous organism
• linkage: the presence of 2 genes on the same
chromosome so that they tend to be inherited
together and do not assort independently
•
• test cross: a genetic cross in which an organsim
showing a characteristic caused by a dominant
allele is corssed with an organism that is
homozygous recessive --> phenotype of offspring
is a guide to whether the 1st organism is
homozygous or heterozygous

30. • F1: generation of offspring produced from
homozygous dominant x homozygous
recessive genotype
• F2: generation of offspring produced from
cross between 2 F1 organisms
• phenotype: organisms' characteristics;
often resulting from an interaction between
its genotype and the environment
• genotype: alleles possessed by an
organism

31. Monohybrid crosses
• Genetic crosses that involve just one
gene.
• Eg. Eye colour

32. Let B represent brown eyes. (parent is dominant homozygous)
Let b represent blue eyes (parent is recessive homozygous)
Parent phenotypes?
Parent genotypes?
Gametes?
F1 genotype?
F1 phenotype?
Ratio?

33. Let B represent brown eyes. (parent is dominant homozygous)
Let b represent blue eyes (parent is recessive homozygous)
Parent phenotypes Brown eyes x Blue eyes
Parent genotype BB x bb
Gametes?
F1 genotype Bb
F1 phenotype Brown eye
Ratio 100%
b b
B Bb Bb
B Bb Bb
B b

36. Dihybrid Crosses
• Mendel
• Looked at pea plants
• Colour of seeds and round or wrinkled
• Colour of flowers and tall or short
• Dihybrid means showing two
characteristics in one genetic cross.

37. • Round seeds are dominant over wrinkled
seeds
• Yellow seeds are dominant over green
seeds
• Let the dominent allele for shape be?
• Let the recessive allele for shape be?
• Let the dominant allele for colour be?
• Let the recessive allele for colour be?
R
r
Y
y

38. • Purple stem is dominant over green stem
• cut leaves is dominant over potatoleaves
• Let the dominent allele for shape be?
• Let the recessive allele for shape be?
• Let the dominant allele for colour be?
• Let the recessive allele for colour be?

39. Both parents are
heterozygous

40. • Parent Phenotypes
• Parent Genotype
• Gametes?
(there are four
Possible)

41. • Parent Phenotypes?
• Parent Genotype (pure breeding)?
• Gametes?
(there are four
Possible)

42. • F1 genotype (see punnett square)
• F1 phenotype

43. Monohybrid phenotype outcomes
• 4 : 0 homozygous dominant crossed with
homozygous recessive
or homozygous dominant crossed with
heterozygous
• 3 : 1 heterozygous crossed with heterozygous
• 1 : 1 homozygous recessive crossed with
heterozygous
• Codominance – heterozygous cross
heterozygous 1 : 2 : 1

44. Codominance
• If both alleles show in the phenotype in the
heterozygous condition this is called
codominance.
• Eg. Red and white alleles make pink.
• Still use one letter though.
• RR is red, rr is white, Rr is pink
• Main example is blood groups

45. • PRACTICE

46. Sex Linkage
• When genes are located on the either the X
chromosome or Y chromosome they are said to
be sex linked.
• Eg. Ability to see particular colours and blood
clotting.
• These are both found on the X chromosome.
• Therefore a recessive allele will be more likely to
show in a male than in a female. As there is no
other X chromosome to mask it.

47. Haemophilia
• When blood does not clot normally.
• Sex-linked character caused by a recessive
allele carried on the X chromosome.
• If the male has the recessive allele then he has
the disease.
• The female only gets the disease if she inherits
two recessive alleles.
• If the female is heterozygous she is known as a
carrier.

48. Let Xh represent haemophiliac
Let XH represent normal clotting
• Parent Phenotype
• Parent Genotype
• Gametes

49. Test cross
• If we don’t know the genotype of an
individual but we can find out by doing a
test cross.
• This is a cross with a homozygous
recessive individual.

50. Let T represent tall
Let t represent dwarf
• Parent phenotype Tall Dwarf
• Parent genotype TT tt
• Gametes

51. Chi – squared Test
• To see if there is a difference between
numbers observed in an experiment and
the numbers given in an hypothesis.
• And if that difference is real or if it is due to
chance or sampling error.
• Chi-squared =

57. • probability of 0.05 means that we would expect these
differences to occur in five out of every 100 experiments, or one
in 20, just by chance.
• A probability of 0.01 means that we would expect these
differences to occur in one out of every 100 experiments, just
by chance.
• In biological experiments, we usually take a probability of 0.05
as being the critical one.
• If our χ2 value represents a probability of 0.05 or larger, then
we can be fairly certain that the differences between our
observed and expected results are due to chance – the
differences between them are not significant.
• However, if the probability is smaller than 0.05, then it is likely
that the difference is significant, and we must reconsider our
assumptions about what was going on in the cross

59. • To work out the number of degrees of
freedom, simply calculate the number of
classes of data minus 1.
• Here we have four classes of data (the
four possible sets of phenotypes), so the
degrees of freedom are: 4 − 1 = 3.

60. • find the χ2 value that represents a
probability of 0.05.
• You can see that this is 7.82. Our
calculated value of χ2 was 0.79. So our
value is a much, much smaller value than
the one we have read from the table.

61. • Null hypothesis – where we assume there
is no difference between the expected and
the observed.
• Which means our hypothesis (genetic
diagram) will be backed up by actual
results if we did the experiment.
• Eg. If we said there should be a 3:1 ratio
then the actual result will be 9 white
rabbits and 3 pink ones.

62. • To calculate the chi-squared value
• first determine the number expected in
each category.
• If the ratio is 3:1 and the total number of
observed individuals is 880, then
the expected numerical values should be
660 white and 220 pink.

63. • Say that the actual cross between two
rabbits yields a population of 880 rabbitss,
639 with white fur and 241 with pink fur.
• Now plug it into the equation:

64. • The chi-squared value will be changed into
a probability value using a chi-squared
table.
• The table contains degrees of freedom.
• Degrees of freedom ≡ mathematical term
relating to the number of free variables in
the system. It is always (n – 1).
• n is the number of categories or
phenotypes in a given example. In our
example 2 (white and pink) therefore
degrees of freedom would be 2 – 1 = 1

66. • Use the chi-square distribution table to
determine significance of the value.
a) Determine degrees of freedom and
locate the value in the appropriate column.
b) Locate the value closest to your
calculated chi-squared value on that
degrees of freedom row.
• Move up the column to determine the p
value.

67. • Probability value means the probability that our
observed will match up to our expected every
time we do the practical.
• State your conclusion in terms of your
hypothesis.
• If the p value is p > 0.05, accept your
hypothesis.
– 'The deviation is small enough that chance alone
accounts for it. A p value of 0.6, for example, means
that there is a 60% probability that any deviation from
expected is due to chance only. This is within the
range of acceptable deviation.
• If the p value is p < 0.05, reject your hypothesis,
– conclude that some factor other than chance is
operating for the deviation to be so great. For
example, a p value of 0.01 means that there is only a
1% chance that this deviation is due to chance alone.
Therefore, other factors must be involved.

68. • So bigger than 0.05 backs the null
hypothesis.
• Smaller than 0.05 means there is a
significant difference between the
hypothesis and the actual results so chuck
your hypothesis out and start again.
• Eg. Pg 233 and 235

73. Variation
• Phenotype results from the interaction between
the genotype and the environment.
• The genotype determines the potential of an
individual but the environment determines to
what extent that potential is fulfilled.
• Divided into
– Discontinuous – usually genetic only with an either or
outcome. Distinct groups.
– Continuous – environmental effects come in to play.
Groups merge in to one another.

74. Mutation and phenotype
• Gene mutation is a change in the base
sequence of DNA
• Alters primary structure of proteins
– Substitution - one base is swopped for
another
– Deletion – one base is removed
– Insertion – one base is added
– Duplication – one or more bases are repeated
– Inversion – a sequence of bases is reversed

75. examples
• Sickle-cell anaemia – base substitution
• Normal haemoglobin is:
– Val – His – Leu – Thr – Pro – Glu – Glu – Lys
• Sickle-cell is:
– Val – His - Leu – Thr – Pro – Val – Glu – Lys
Valine is non-polar and hydrophobic
In low oxygen concentrations haemoglobin becomes
less soluble and crystallises
Sticky fibres then distort the red blood cell shape.

76. • Cystic fibrosis
– Recessive mutation on chromosome 7
• Normal gene codes for a chloride channel
protein called CFTR
• In cystic fibrosis sufferers channel defective.
• Result –
– as chloride ions do not flow out
– sodium ions rush in to balance charge and this
prevents water leaving the cell
– Mucus becomes thick and sticky.
– Affects lungs, pancreas, liver
– Mucus clogs ducts and passages

77. • Down’s syndrome
– Chromosome mutation on 21
• Sufferers have 47 chromosomes instead
of 46 as there is an extra copy of 21.
• Called trisomy
• Symptoms
– Mental retardation
– Short stature
– heart defects
– Coarse hair

79. Albinism
• A mutation in the gene for the enzyme
tyrosinase results in either the absence of
tyrosinase or the presence of inactive
tyrosinase in the cells responsible for melanin
production.
Tyrosine cannot be converted into DOPA and
dopaquinone.
(tyrosinase)
• tyrosine → DOPA → dopaquinone → melanin

80. Huntington’s disease
• Dominant allele
• HD is a neurological disorder resulting in
involuntary movements (chorea) and
progressive mental deterioration.
• Brain cells are lost and the ventricles of
the brain become larger.
• The age of onset is variable, but occurs
most commonly in middle age.

81. • The mutation is an unstable segment in a gene
on chromosome 4 coding for a protein,
huntingtin.
• In people who do not have HD, the segment is
made up of a small number of repeats of the
triplet of bases CAG.
• People with HD have a larger number of repeats
of the CAG triplet. This is called a ‘stutter’.
• There is a rough inverse correlation between the
number of times the triplet of bases is repeated
and the age of onset of the condition: the more
stutters, the earlier the condition appears.

82. Mutation in the gene for the enzyme tyrosinase
either the absence of tyrosinase or the presence of
inactive tyrosinase
the first two steps of the conversion of the amino acid,
tyrosine into melanin cannot take place.
Tyrosine cannot be converted into DOPA and dopaquinone.

83. Gene control in prokaryotes
• Transcription
• Controlled by transcription factors
– These are proteins that bind to a specific DNA
sequence and control the flow of information
from DNA to RNA by controlling the formation
of mRNA.

84. • Structural Genes - that code for proteins
required by a cell
• Regulatory Genes - that code for proteins that
regulate the expression of other genes.
• The synthesis of a repressible enzyme can be
prevented by binding a repressor protein to a
specific site, called an operator, on a bacterium’s
DNA.
• The synthesis of an inducible enzyme occurs
only when its substrate is present.
• Transcription of the gene occurs as a result of
the inducer (the enzyme’s substrate) interacting
with the protein produced by the regulatory gene.

85. Operon
• An operon is a length of DNA making up a
unit of gene expression in a bacterium.
• It consists of one or more structural genes
and also control regions of DNA that are
recognised by the products of regulatory
genes.
• The different roles of structural and
regulatory genes can be seen by looking
at the control of gene expression in a
prokaryote using the lac operon.

89. • lacZ, coding for β-galactosidase
• lacY, coding for permease (which allows
lactose to enter the cell)
• lacA, coding for transacetylase.

90. No Lactose in the medium
• the regulatory gene codes for a protein
called a repressor
• the repressor binds to the operator region,
close to the gene for β-galactosidase
• in the presence of bound repressor at the
operator, RNA polymerase cannot bind to
DNA at the promoter region
• no transcription of the three structural
genes can take place.

91. Lactose in the medium
• Lactose binds to the repressor protein
• Alters its shape
• Prevents repressor protein binding to the
operator region
• RNA polymerase can bind to the promoter
and the structural genes can be
transcribed.
• Lactase is made

92. Eukaryotes
• the number of diff erent proteins that act
as transcription factors increases with
increasing size of the genome.
• Th is means that eukaryotes have many
more ways of regulating gene expression
than have prokaryotes.

93. • General transcription factors are
necessary for transcription to occur.
• They form part of the protein complex that
binds to the promoter region of the gene
concerned.
• Other factors activate appropriate genes in
sequence, allowing the correct pattern of
development of body regions.

94. • A transcription factor is responsible for the
determination of sex in mammals.
• Transcription factors allow responses to
environmental stimuli, such as switching
on the correct genes to respond to high
environmental temperatures.
• Some transcription factors, including the
products of proto-oncogenes and tumour
suppressor genes, regulate the cell cycle,
growth and apoptosis (programmed cell
death).
• Hormones have their effect through
transcription factors.

95. Gibberellin
• The plant hormone, gibberellin, controls
seed germination in plants such as wheat
and barley by stimulating the synthesis of
amylase.
• It is a good example of how a hormone
can influence transcription.
• It has been shown that, in barley seeds,
application of gibberellin causes an
increase in the transcription of mRNA
coding for amylase.

97. • Gibberellin causes breakdown of DELLA proteins .
• A DELLA protein inhibits the binding of a transcription factor, such as
phytochrome interacting protein (PIF), to a gene promoter.
• By causing the breakdown of the DELLA protein, gibberellin allows PIF to
bind to its target promoter.
• Transcription of the gene can then take place, resulting in an increase in
amylase production.