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Genetics 214

INTRODUCTION TO GENETICS - study unit 1
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Genetics: study of how genes bring about traits in living things/how it is inherited
and how biological information is stored, transmitted, translated and expressed
Genes: specific sequences of nucleotides that code for a particular protein
Genes transmitted between generations through meiosis and sexual
reproduction
As data grows, scientists access more information through age of information
and big data
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Began around 12 000 BCS in middle
East
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Beginning of agriculture & animal domestication
Foundation for advancement of human society allowing for development of civilisations
(Ancient Egypt & Mesopotamia)
First permanent villages & emergence of social complexity
Development of agriculture changed = eating habits, interactions with environment &
organised societies
Humans produced food in abundance allowing settlement into permanent villages
instead of being nomadic = emergence of social complexity, power & class
Domesticated animals for transportation, source of energy & entertainment

Ancient Greek philosophy = spontaneous generation
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Diseased = appearance of new-borns with congenital
disorders/deformities
Humors altered in individuals before passed onto offspring explaining how new-borns
inherit traits that parents has acquired in response to environment
Aristotle = male semen contained vital heat with capacity to produce offspring of the
same form as the parent, heat cooked & shaped menstrual blood produced by female
(physical substance giving rise to offspring)
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Darwin &
Mendel = traits inherited from combination of existing traits (not a process of blending)
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Cell theory proposed (Schwann): all organisms are composed for basic
structural united called cells derived from pre-existing cells
Spontaneous generation disproved: cell theory & Louis Pasteur disproved
spontaneous generation (Greek philosophers)

Genetics 214

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Gregor Mendel’s (Australian monk) postulates

Pea plants laid foundation of heredity: traits were inherited through discrete units
(genes)
Introduced concept of genes (building blocks of heredity)
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Darwin’s theory of natural selection (independent; unknown of Mendel’s work)

Origin of species by Charles Darwin: explanation of mechanisms of evolutionary
change but lacked understanding of genetic basis of variation & inheritance
Therefore theory open to reasonable criticism in 20th century
Heredity & development were dependant on genetic information residing in genes
contained in chromosomes contributing to each individual by gametes= chromosome
theory
Chromosomal theory of inheritance: Walter Sutton 1902
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Chromosomes are the carriers of genetic material & genetic factors are located
on loci on chromosomes
Transmission of traits from parent to offspring due to transfer of chromosomes
between generations
Diploid number (2n): each species has a set number of chromosomes
Humans diploid number = 46 chromosomes comprised of 23 homologous pairs
(n)
Chromosome exists in pairs as homologous chromosomes
Chromosomes undergo meiosis & mitosis
Segregation and exchange of chromosomes (crossing over) between the 2 sets of
chromosomes in a parent cells
Meiosis halves the chromosome number so that each gametes (egg/sperm cell)
receives 1 copy of each chromosomes = haploid (n) genetic information passed
between generations
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Adenine pairs with uracil (U) in RNA
Sugar-phosphate backbone: DNA is a polymer made of nucleotide units
DNA strands held together by hydrogen bonds between bases on adjacent
strands

Replication errors cause mutations = new variations
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DNA: chemical name for the long, stringy chromosomes inside of cells
Genes: segments of DNA coding for proteins
Alleles: different version of the same gene with small differences in their
nucleotide sequences

Central dogma: how genes create phenotypes from DNA to proteins
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DNA sequence - transcription - mRNA sequence (amino acid coded in 3 codons
identical to coding strand with a U base not T base) - translation - protein
Bases on a single strand act as a code
3 letter codons form coding for amino acids (building blocks of proteins)
DNA stores information to run the cell
RNA polymerase enzyme transcribes DNA into mRNA (messenger ribonucleic
acid)
mRNA splits apart the 2 strands that form the double helix & reads the template
strand copying the sequence of nucleotides from the coding strand
Only difference between the mRNA and original DNA is that uracil (U) with a
similar structure replaces thymine (T)
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species (humans, plants and animals) =
insight in relationships between populations, genetic structure & investigate genetic
basis of traits (foraging behaviour, growth rate and disease resistance)
Population genetics: study of genetic variation within & between population and the
evolutionary forces shaping it to understand the dynamics of species & population
characters with how they change over time
Investigates patterns of variation in populations & help with evolutionary history
Quantitative genomes + population genetics = study genetic architecture of quantitative
traits in natural populations, understand evolutionary history & dynamics of species,
identify underlying genes & genetic basis of variation in traits = conservation of species
& develop effective management strategies
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Fundamental underpinning of modern biology’s research agenda

Cloning and gene transfer technology
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Recombinant DNA (1970s): specific restriction enzymes used by bacteria to cut
& inactive DNA of invading viruses is used to cut any organisms DNA at specific
nucleotide sequences = reproducible set of fragments
Discovered ways to insert DNA fragments produced by restriction enzymes into
carrier DNA molecules (vectors) = recombinant DNA molecules
When transferred into bacterial cells, copies/clones of the combined vector &
DNA fragments are produced during bacterial reproduction
Large amounts of cloned DNA fragments isolated from bacterial host cells used
to isolate genes, study organisation/expression/nucleotide sequence & evolution
Genomic libraries: collections of clones that represent an organisms genome
defined as the complete haploid DNA content of an organism
Recombinant DNA technology = biotechnology industry

Omics revolution
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Genomics, proteomics & bioinformatics are derived from recombinant DNA
technology

Genetics 214

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They combine genetics with information technology = exploration of genome
sequences, structure & function of genes, protein sets in cells and evolution of
genomes i
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Human Genome Project
Genomics (study of genes): studies the structure, function and evolution of
genes and genomes [gene, epigenetics, mRNA & post-transcriptional
modification]
Proteomics (set of proteins in a cell under a set of conditions): studies their
functions and interactions [post-translational modifications]
Bioinformatics (information technology): stores, retrieves & analyses massive
amounts of data generated by genomics & proteomics
Develops hardware and software for processing nucleotide and protein data

Gene editing
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Recombinant DNA technology (methodology used in molecular genetics) allows
genes from one organism to be spliced into vectors & cloned producing many
copies of specific DNA sequences
Cas 9 enzyme with guide RNA by attaches to the PAM sequence on the genomic
DNA, Cas 9 enzyme cuts the gene as it is a restriction enzyme and that same
enzyme cuts another gene to get the donor DNA, repair takes place where the
targeted donor gene undergoes genome editing

Approaches in research
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Known phenotype
Forward/classical genetics
Naturally occurring/intentionally induced mutations (chemicals, X-rays or UV
light) = altered phenotypes & worked through lab-intensive and time consuming
processes to identify genes causing new phenotypes leading to the identification
of genes of interest/the gene = gene sequence determined
o Crossing experiments & pedigree analysis
o Population analysis & genetic mapping
Gene discovery
Known gene
Reverse genetics
DNA sequence for a gene of interest is known but role & function of gene is not
understood
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In a knockout organism, look for apparent
phenotype changes & those at cellular/molecular level to determine the function
of the gene
o Gene knock out or knock in & gene editing
Resulting phenotype

Model organisms (yeast, zebrafish, Drosophila fruit fly & Mus musculus mouse)

Genetics 214

Large catalogue of mutant strains for species were created & mutations were
carefully studied, characterised & mapped for their well-characterised genetics
▪ After Mendel’s work 1900, research using wide range of organisms confirmed
that principles of inheritance is universal significance to plants & animals
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Organism has characteristics that made them suitable for genetic research
Easily to grow, short life cycles, many offspring & genetic analysis straightforward
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Impact on society
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Biotechnological applications
o Food security & agriculture industry
o Medicine and health & pharmaceutical industry
Mass production of medically important gene products
o Environmental suitability
o Forensics and law enforcement
o Genetic testing = detection of genetic disorders & those at risk of having
affected children
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Only then get to see recessive trait
yy cross-fertilisation YY = F1 100% Yy yellow - self-fertilisation = F2 25% YY & 50%
Yy yellow and 25% yy green = 1:2:1 genotype and 3:1 phenotype
1 dominant/homozygous: 2 hybrid/heterozygous: 1 recessive/homozygous
genotype
3 dominant: 1 recessive phenotype

Dihybrid crosses: more than 1 trait considered & in combination
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Pure bred/homozygous genotype for each of the 2 traits cross-fertilised with pure
bred for alternative phenotype of the 2 traits
YYRR cross-fertilisation yyrr = F1 100% YyRr yellow & round - self-fertilisation =
9

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F2 16 yellow; round, 16 yellow; wrinkled, 16 green; round and 16 green/wrinkled

Genetics 214

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F1 creates all dominant alleles of the 2 traits which self-fertilise where F2 creates
more variation 9:3:3:1 phenotype

Genetic locus: location of a particular gene on a chromosome
Each genetic locus: 2 alleles, 1 on each homologous chromosome/1 on each
homologue
Pair of homologous chromosomes: 1 inherited from male parent and 1 from female
parent
Homozygous dominant - A allele on paternal homologue & A allele on maternal
homologue
Homozygous recessive - bb (1 gene pair at 1 locus)
Heterozygous/hybrid - Cc (1 gene pair at 2nd locus)
Mendel postulates
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Gene pairs: genetic traits controlled by genes existing in pairs in an individual
therefore there are 2 sets of chromosomes in a cell (2n)
Dominance and excessiveness: when2 unlike alleles for a trait present in an
individual, 1 allele is dominant to the other being recessive
F1 only dominant phenotype (genotype Yy) and F2 can be recessive phenotype
(yy)
Independent segregation: in gamete formation, alleles from same gene
separate/segregate randomly so that each gamete receives one of the alleles
with equal likelihood (MONOHYBRID)
Metaphase
Independent assortment: in gamete formation, different genes assort
independently of each other, so all possible combinations of gametes form with
equal frequency
Genes at 1 locus does not influence how alleles at the 2nd locus segregate into
gametes (DIHYBRID)
Crossing over/recombination & anaphase
Genetic variation & random fertilisation

Test cross
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Cross individual with dominant phenotype (unknown genotype & gametes) to
homozygous recessive individual (known gametes) as recessive only has 1
genotype
Cross between homozygous recessive (yy) and a dominant?? (YY or Yy)

Genetics 214

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Test cross = all dominant offspring indicates parent is homozygous dominant
(YY) as the recessive (yy) was known
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Offspring is heterozygous
with dominant phenotype & homozygous recessive

Produce rule
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Product rule: when 2 independent events occur, the combined probability of the
2 outcomes is equal to the product of their individual probability occurrences
If yellow is dominant in P = ¾ yellow & ¼ green
If round is dominant in P = ¾ round & ¼ wrinkled
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For yellow and round = ¾ × ¾ = 16 yellow & round = 9

Combination (yellow & wrinkled etc
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05/5%
Degrees of freedom = n-1 = no
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05
Subtract 1 as you make assumption/observations that the population reflects real
population: lose a degree of freedom from assumption
2 phenotypic classes (yellow/green) monohybrid - 1 = 1 degree of freedom & 4
phenotypic classes (yellow/green & round/wrinkled) dihybrid - 1 = 3 degrees of freedom

Genetics 214

X2 chi-square value < critical value = fail to reject null hypothesis as the deviations
between o & e results due to chance and are not significant
Probability of results occurring by chance is p>0
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05/5%

Probability rule
Conversion graph/table showing the basis on which to reject/fail to reject (H0)
Probability<0
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05 fail to reject null hypothesis (H0)
Fail to reject (NOT ACCEPT HYPOTHESIS) - nothing is 100% known
Probability value (cut off for significance): p = 0
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05: probability less than 5%, observed deviation not due to chance therefore
deviation significant & reject null hypothesis
p>0
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Work out X2
2
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Use df (going down), match X2 to appropriate critical cell value & read off p
value (row across at the top)
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05 = reject or p > 0
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2 parents autosomal dominant trait: children might be unaffected; children might be
affected, and no children may be affected
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Yellow pea plant with round seeds GgWw crossed with itself, proportion green with
round seeds? 3/16 as you have a recessive dominant
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Child has an autosomal dominant phenotype, parents: at least 1 of the parents should
have the trait
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Based on which ultimate value do I reject or fail to reject the null hypothesis in chisquare test: P value (set number of class = df & use chi-square test value to get p value
which is comparing the observed & expected value
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05 fail to reject with no
difference and p<0 reject as it is not due to chance but a biological entity)

Hardy Weinberg
Frequencies of alleles remain constant in a population over time if no evolutionary
influences occur
No selection, no mutation, no migration, large population and random mating
Compare constant population to those evolving populations & the deviation
Frequencies of alleles/genotypes causing diseases compared
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p + q = 1 USED FOR ALLELES
p is dominant allele and q is recessive allele
0
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4 = 1 (60% and 40%)
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Get the missing value from the total in a population if given 1 value
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Use the recessive genotype frequency as there is only 1 combination [q]
Don’t use dominant genotype frequency as there is more than 1 combination [p]
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recessive/total) = q2
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Get p2 from solving for p for equation p2 + 2pq + q2 = 1
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Y is yellow so Yy or YY dominant and y is blue so yy recessive (dominant p = Y and
recessive q = y)
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Therefore 988 yellow feet (or given yellow and need to find blue to solve further)
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012 = q2
Number of penguins with blue is q2 as there are 2 alleles per individual
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11 recessive
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89 dominant

Genetics 214

5
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892; Yy = 2pq = 2(0
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11) and yy = q2 = 0
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Pigment forms in absence of A therefore aa
If there is an A allele then white = no pigment as it is repressed
o DOUBLE RECESSIVE EPISTASIS
A-B-; A-bb; aaB-; aabb = 9:3:3:1 = epistasis = (9 dark: 3 albino: 3 albino: 1
albino) = 9:7
B gene converts colourless to dark pigment (dark only there is A & B as it
is dominant)
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Non-lethal when no allele for lethal is present = normal
Recessive lethal: lethal when homozygous recessive & 2 alleles are needed
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Reference to coat colour = dominant action = heterozygous displays mutant
phenotype
2
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In hypercholesterolemia, individuals homozygous for the allele causing disorder lack
receptors on liver cells taking up cholesterol
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Not co-dominance as not 2 completely different receptors
contributing, there is ½ a receptor therefore an intermediate (incomplete)
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Any man can be father, not enough genotypic
information as A & B can be homozygous or heterozygous for dominant allele

Gene interactions
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Genes interact with one another with several genes influencing a particular
characteristic = developmental biology epigenesis (epistasis/epistatic to a gene)
Gene 1, gene 2, gene 3, gene 3 = character 3

Genetics 214

Epistasis
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Expression of 1 gene pair masks or modifies the effect of another gene pair
Ratio out of 16 = dihybrid cross but 1 trait investigated not the 2 different traits
together as it would be in the classical dihybrid cross (expect 9:3:3:1)
Ratio’s expressed in 16 parts in the study of a single character indicates that 2
gene pairs are interacting during the expression of the phenotype
Homologous pairs of chromosomes: 1 chromosome = A & B locus with the
homologue = A & b locus
Gene B influences the effect of gene A therefore gene B is epistatic to the gene A
Gene A with enzyme A changes X white to Y brown but then gene B with enzyme
B changes that Y brown to Z black (gene B influences the effect that gene A
caused & gene B is epistatic to the gene A)
AaBb x AaBb = white: brown: black = 9:3:4 (not 9:3:3:1 that would have occurred
if gene B didn’t change the brown effect to black)
Epistasis has 2 genes influencing a trait but only 1 trait is investigated (black on
the brown trait) looks like dihybrid ratio but 3 combinations (white, brown & ->
black) not 2 combinations (white & brown)

Recessive epistasis (focus on a gene with homozygous recessive influencing ratio)
Case 1 - mouse on coat colour
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A- aa
Second gene pair bb gives albino regardless of genotype at locus A
Recessive condition for gene B impacts the phenotypic expression
If homozygous (bb) at albino locus gene B = no pigmentation

bb genotype masks or supresses the expression on A allele = epistasis
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P: AABB x aabb with F1: AaBb
A-B- 9/16, A-bb 3/16, aaB- 3/16, aabb 1/16
Therefore albino is 3/16 + 1/16 = 4/16
9 agouti: 3 black: 4 albino
9:3:3:1 = 9:3:4

Genetics 214

Dominant epistasis (focus on gene with homozygous dominant/heterozygous influencing ratio
therefore other gene shown when the epistatic gene (dominant) is not there i
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homo recessive)

Case 2 - squash on colour
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Dominant allele at A locus = always white fruit
aaB = yellow and aabb = green
B locus only expressed in plant that has the aa genotype (no dominant allele)
If recessive allele combination at 1st locus A (aa but B) = yellow
DOMINANT AT 1 LOCUS and if not appearing & recessive = other gene appears
Recessive at both genes/loci (aa bb) = green

A- genotype marks or supresses the expression on B allele = epistasis
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P: AABB x aabb with F1: AaBb
A-B- 9, A-bb 3, aaB- 3, aabb 1
9:3:3:1 = 12 (epistatic gene white): 3 (yellow): 1 (green)

Complimentary epistasis (duplicate recessive)
Case 3 - pea on flower colour
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At least 1 dominant allele at each gene pair (A AND B loci) needed for purple
flowers
Dominant for BOTH genes/loci = purple flowers
Homozygous recessive or at least 1 dominant allele = white flowers
P: Aabb x aaBB with F1: AaBb
A-B- 9: aaB- 3: A-bb 3: aabb 1
9 (purple): 3:3:1 (white) = 9:7

Cooperating epistasis (additive) (novel phenotype)
Case 4 - squash on fruit shape
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Cross disc shaped fruit with long fruit = all F1 disc shaped
F1 self-fertilising both parental types occur in F2 & new shape sphere
BOTH dominant alleles at both genes = disc
1 dominant for either of the genes = sphere
Recessive alleles at both genes = elongated
Both gene pairs influence the fruit shape equally
P: AABB x aabb with F1: AaBb
A-B- 9 disc: aaB- 3 sphere: A-bb 3 sphere: aabb 1 elongated
9:3:3:1 = 9:6:1

For epistasis: the classical ratio is modified & need P1 pure breeding for both = F1
interbreed/self-fertilised = F2 so for deductions on modes of inheritance you need a
classical experimental cross all the way to F2
Complementation analysis

Genetics 214

If both mutant phenotypes are recessive (dd x dd) then offspring will be wild type (Dd)
2 mutations
Wild type is a MUTATION (loss of function) + refers to non-mutated allele that is
unchanged = normal phenotype
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2 of the same mutation/different mutations across genes have the same impact
on phenotype = same phenotype

Mutations in separate/different genes
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Homozygous recessive for mutant allele at gene 1 (mama) & Dominant
homozygous wild type allele at gene 2 (++) x Dominant homozygous wild type
allele at gene 1 (++) & Homozygous recessive for mutant allele at gene 2 (mbmb)
F1 = wings as offspring is heterozygous at both genes and wild type
F1: Gene 1: ma+
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Gene 2: ++
2 different null mutations = complementation as 2 non-functional copies of gene
therefore still wingless
Gene 1 is mutant in all cases while gene 2 is normal therefore noncomplementation

Pleiotropy
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1 gene influences or impacts multiple traits (more than 1 trait)

Genetics 214

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Gene 2 = character 1, character 2, character 3, character 4
1 gene causing variegate porphyria = multiple symptoms viewed as multiple
phenotype
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With codominance, a likely ratio resulting from single trait cross would be 1: 2: 1
Classical ratios see with experimental cross: P1 pure breeding, self-fertilise F1 =
1:2:1
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Mutant allele can be lethal in recessive form but
dominant in phenotypic trait)
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Dominant epistasis is 12: 3: 1
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Fox in winter = white coat & summer = brown coat
Nutrition: lactose intolerant from dairy products = bloating, pain & rumbling

Some genes are internally environmentally sensitive
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Age: different age cause different changes in physiological conditions
o Prenatal age = blood group antigens

Genetics 214

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o 1 year = rickets from lack of vitamin D
o 2-5 years = muscular dystrophy
o 20-30 years = baldness triggered by change in internal physiological
environ
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o DNA methylation inhibits gene activity
o An allele is differentially methylated in males vs female
o Same allele inherited but how allele from maternal or paternal behaves is
different due to mechanism of methylation related to silencing of genes
(stopping expression of gene) in each sex
o Environmental exposures, stress, diet & lifestyles induce epigenetic
changes that determine where’re genes are turned on or off

Women who smokes while pregnant induced epigenetic changes in 3 generations (self, unborn,
germline reproductive cells of unborn) = epigenetic modification on a functional gene

Variation in phenotypic expressed due to who is the parental origin of the chromosome
carrying the particular gene -> selective gene silencing -> epigenetically
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Different genes are epigenetically silenced in egg (chromosome from maternal)
and sperm (chromosome from paternal) -> formation of adult
In all adult sperm: all imprints are erased & re-written with paternal imprint
pattern, even genes from maternal mom
In all adult eggs: all imprints are erased & re-written with maternal imprint
pattern, even genes from paternal dad

Genetics 214

SUMMARY: altered phenotypic expression

Extranuclear inheritance
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The nuclear genome of an individual is not the only source of genetic material
that influences its phenotype
Transmission of genetic information through cytoplasm (and mitochondria), not
the nucleus (nucleic DNA)
Inherited from one parent

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Infectious heredity
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In duplication you restore functional allele

Antibiotic resistant in Chlamydomonas (algae)
Reciprocal cross (sr mt+ x ss mt- and ss mt+ x sr mt-) showing uniparental inheritance of
non-chromosomal genes and lack of segregation (sr = streptomycin resistant and ss =
sensitive to streptomycin) with secretion of nuclear genes (mt+ and mt-)

Mutant allele (disorder) is mating type - (mt-) & resistant allele (no) is mating type +
(mt+)
Mating type mt+ (resistant normal allele) sr x mating type mt- (mutant sensitive) ss
Maternal phenotype is resistant (mt+) therefore
50% +: 50%- but all offspring are streptomycin resistant mt+ therefore maternal is mt+
so all offspring phenotypically show the maternal phenotype of not being
affected/resistant IRREGARDLESS of its own offspring genotype (++--) and father
Mutant allele (no) is mating type + (mt+) & resistant allele (disorder) is mating type - (mt)
Mating type mt- (resistant normal allele) sr x mating type mt+ (mutant sensitive) ss
Maternal phenotype is sensitive (ss) therefore
50%+: 50%- but all offspring are sensitive mt+ therefore the maternal is mt+ so all
offspring phenotypically show the maternal phenotype of being affected/sensitive
IRREGARDLESS of its own genotype (++--) and father
Mitochondrion
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Energy creation site therefore defected copy = not grow as fast
o Mutation in petite yeast = small colonies due to not growing fast in same
time span as normal copies as there is a defected copy due to less
energy
Depends on where you inherit defective copy which is the maternal line

Inheritance of segregational petties (mendelian)

Genetics 214

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n homozygous petite x n homozygous normal (dominant) = 2n heterozygous
normal with gametes 50% petite: 50% normal

Inheritance of neutral petties (non-mendelian)
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Petite x with normal = normal masks effect of defective mitochondrial = all
normal therefore the masks mutant due to the normal mitochondria of the
mother
n homozygous petite x n homozygous normal = 2n heterozygous normal with
gametes all normal as the maternal must have been homozygous normal
(express maternal phenotype IRREGARDLESS of own genotype)

Inheritance of suppressive petties (non-mendelian)
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Petite x with normal = offspring petite as the mother had the defective petite
mitochondria = mutant copy becomes the dominant copy in cell therefore
suppresses normal copy
n homozygous petite x n homozygous normal = 2n heterozygous petite with
gametes all petite as the mutant allele suppresses the normal allele
Mendelian vs mitochondrial inheritance of same phenotypic characteristic

Mitochondrial disorders = mutations in MITOCONDIRAL DNA = defective mtDNA =
defective mitochondria = defective phenotypes
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Inheritance is maternal there mother if mother is affected = all offspring affected
irrespective of father
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mutant mitochondria = normal function with a
small number of mother mitochondria into each early egg cell =
bottleneck effect there the number of mitochondria increases due to split

Genetics 214

of mitochondria in mitosis (gamete formation depends on amount of
defectiveness inherited)
o Mature egg cells = 80%, 50% or 20% mutant = severe, mild & no disease
o Mature egg cells may inherit more defective mitochondria than other cells
in a certain organs i
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if in muscle tissue = more defective mitochondria
inherited = defective muscular issues not neutral

Infectious heredity
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Symbiotic or parasitic association of a microorganism within a host organism
affects hosts phenotype = transmission of altered phenotype to offspring
Invading infectious microorganism exists in symbiotic relationship with host
organism so = invader to maternal egg cytoplasm (ooplasm) = affects offspring
phenotype when reproduction occurs
o Strain of microorganism = strain of offspring
Rhabdovirus Sigma in Drosophila
o Causes carbon dioxide sensitivity
o Provide carbon dioxide & then decrease the levels to recover
o Affected flies with virus with not recover normally from CO2 anaesthesia
Protozoan in Drosophila
o Affected flies produce predominantly female offspring if reared at
21°C/lower therefore temperature dependant
o When ooplasm from affected individuals/protozoan itself is injected into
oocytes of normal individuals = temperature sensitive altered sex ratios
Bacterial endosymbionts (intracellular)
o Head lice: bacteriome exists outside of ovaries but the bacteria migrates
to ovaries & infects individual eggs
o Rice weevil: bacteriocytes inside the weevil migrating to the ovary to be
symbiotic with caecum

Genetics 214

Endosymbiosis theory
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Mitochondria originated from endosymbiotic purple bacterium = evolution =
animals fungi & protists containing mitochondria
Chloroplasts originated from endosymbiotic cyanobacteria = purple bacterium
into cyanobacteria host cell = exist with each other = evolution = plants & algae
containing mitochondria & chloroplasts
o Purple bacterial cell gave advantage to host cell & become dependent on
host cell with cyanobacteria so neither
cell exists independently
o Cyanobacteria lost some genes &
became incorporated into nuclear
genome with the additional mitochondria
genes therefore mitochondria and
chloroplast do not exist without host cell

Maternal effects
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mRNA transcripts stored in the egg before fertilisation influences the phenotype
of offspring after fertilisation
o Used to initiate metabolism of the newly formed individual before new
genome is used
o mRNA transcripts occur from the mother in egg -> used in offspring
coding for the formation of the new offspring genome -> influence
phenotype
Maternal genotype strong influence on embryonic development therefore every
embryo has maternal environmental impacts
Gene products in egg in cytoplasm & sperm contributes nothing to zygote
cytoplasm
o Embryo egg is derived from cytoplasm donated by the maternal line
Initial embryonic activity regulated by genes of maternal (not embryonic own
genes only) therefore embryonic developed dependant on material genotype

Non-permeant effect (flour moth): red eyes = mutant aa and brown eye = wildtype (Aa) =
offspring expected mendelian brown & red
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Brown male Aa x red female aa = offspring Aa & brown and aa & red as maternal
pigment in individual irrespective of genotype of offspring
If larvae has right phenotype homozygous for pigment it will phase out and go red

Permanent: coiling in snail shell Limnaea

Genetics 214

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Dextral dominant DD x sinistral recessive dd = dextral donates egg and sinistral
sperm = heterozygote which is dextral
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e
...

Cytokinesis splits cytoplasm to complete dividing of cells
o END: 2 identical diploid cells - 2 sets of chromosomes one from each
parent (organism growth or replacing damaged cells)
Both processes and cytokinesis: involved in making new cells and undergo
prophase, metaphase, anaphase and telophase but meiosis goes through it
twice
Meiosis:

Sperm and egg cells/gametes/primary spermatocyte and primary oocyte
Begins with diploid cell duplicating genetic material in interphase stage to form 2
identical sister chromatids at centromere - chromosome division - 1st meiotic - 2nd
meiotic cycle
o Prophase 1: chromosome visible and condensing/thickening,
chromosomes match with homologous pairs (1 father & 1 mother)
forming tetrads/bivalent where chromosomes are the same size & same
type of gene at the same locus on each homologue
...
Cytokinesis
splits cytoplasm to complete dividing of cells
END: 2 haploid cells that are not identical because of crossing over in prophase and
random assortment in anaphase
o Prophase 2: chromosomes condense in both cells (no homologous
pairs/crossing over) and no replication
o Metaphase 2: nuclear membrane/envelope already broken down and
chromosomes line up along metaphase plate in single file no pairs
o Anaphase 2: sister chromatids are pulled away at the centromere by the
spindle fibres that are attached to centrioles
o Telophase 2: chromosomes uncoil, nuclear envelope forms and
chromosomes decondense forming nuclei on each side creating new
cells
...
Heterozygous: allele are different
(Yy)

Genetics 214

▪
▪

Combination of genes is the genotype resulting in a visual aspect being the
phenotype
Punnett square: different alleles on both axes Yy across top and yy down
o Yy x Yy = YY, Yy, Yy and yy (only green)
o 1st generation: each homozygous dominant will be given to each offspring
with a recessive trait giving but the Y will mask making them all dominant
(100%)
o 2nd generation: 2 heterozygous self-fertilisation leads to YY, Yy, yy with the
phenotypes in a 3:1 proportion (75%:25%)
o RRYY, RRYy, RrYY, RRyy, rrYY, rryy etc (wrinkled/ not and yellow/green)
▪ RY, Ry, rY, ry x RY, Ry, rY, ry = 9:3:3:1

Recap on pedigrees
▪
▪
▪
▪
▪
▪
▪
▪

▪

▪
▪

▪

Pedigree: family tree showing information of an inherited trait passed along
generations
Circle = female and squares = males
I and II across show generations
Horizontal line = marriage line and vertical line leads to = offspring
Shaded shape = the trait in the pedigree expressed
You can get autosomal dominant/recessive traits and sex-linked on X
chromosome
Half shaded = carriers
Trait: trait tracked is recessive as dominant alleles lead to dominant traits and
recessive traits only expressed when dominant allele not expressed and 2
recessives
Trait: is autosomal recessive therefore chromosome that is not sex chromosome
o 46 chromosomes, 44 (22 pairs) autosomes 2n and last pair is gametes n
o Not sex linked therefore no coefficients
Free earlobe dominant E and attached earlobes recessive e
Genotypes - shaded ones have to be autosomal recessive therefore ee
(attached) and unshaded ones have to either be Ee or EE and decided on - if the
offspring are only ee and ee and 1 parent is ee the other can’t be EE as there is
distribution getting an allele from each parent and EE has no little e to give = ee
Ee is a carrier if all the offspring are Ee and can show up in another generation
1
...
Unshaded = 2 dominant or 1 dominant and 1 recessive
3
...
List both Ee and EE of the offspring is they are all unshaded, if a parent is ee,
the other parent will have to be EE as if it is Ee it will donate its e with the
other e = ee and there are no ee offspring

Genetics 214

▪
▪
▪
▪
▪

If offspring are ee and Ee and a parent is ee, the other parent can’t be EE as
there is no donation from that parent of an e allele to create a recessive
offspring of ee therefore that parent is a carrier only
If offspring are Ee and a parent is ee the other parent can be EE or Ee (include
both)
Sex-linked recessive traits/pedigrees: colour-blindness and baldness
All circles have XX as they females and squares have XY as they males
Track sex-linked recessive therefore shaded is recessive
Shaded is r for recessive and R for dominant, the trait can only occur on the X
chromosome
...
If she is unshaded then it is XrXR (carrier) or XRXR but has to be
the carrier one as the Xr needs to go the male and there are some dominant
offspring needed and R
...
Metacentric: centromere in the centre where P and Q are equal in length
2
...
Acrocentric: centromere close to the end where Q arm is longer P arm is shorter
4
...
After meiosis, haploid cells contain paternal set or maternal set of
every homologous pair of chromosomes
✓ Process of crossing over (prophase 1) reshuffles alleles between maternal and
paternal members of each homologous pair
✓ Segregation and assortment independently into gametes = genetic variation
Genetic locus: location of a particular gene on a chromosome
Allele: variation of the gene at that locus
▪
▪

At each genetic locus there are 2 alleles with 1 on each homologous
chromosome
AA - homozygous (2 same alleles) dominant, bb - homozygous recessive and Cc heterozygous (2 different alleles) with the different pairs at 3 different loci
o There is an C allele on a paternal homologue inherited and an c on the
maternal homologue inherited both lining up in the same plane

Diploid (2n): all somatic cells derived from members of the same species contain an
identical number of chromosomes; two sets of chromosomes one from each parent
▪
▪

2 copies of each chromosome one being maternal and the other maternal
Diploid set crosses with another diploid set as homologous chromosomes

Haploid (n): sex cells/gametes with 1 copy/single set of chromosomes with nonhomologous chromosomes not double stranded (one member of each homologous
pair) that cannot line up and exchange genetic information in gametogenesis
✓ Ploidy = number of sets of chromosomes that are found within the nucleus
✓ Zygotes = multicellular diploid organisms from a single-celled fertilized egg
o Mitosis for would healing and cell replacement in tissues

Genetics 214

Growth: diploid 2n - mitosis - diploid 2n
Reproduction: 2n diploid - meiosis - n haploid gametes/spores - fertilisation of 2
gametes - diploid re-established - zygote has 2 complete haploid sets
Consistency/maintaining genetic continuity throughout generations
Mitosis
✓ Production of 2 cells, each with the same number of chromosomes as the parent
cell
Unreplicated chromosomes consists single chromosome & 1 sister chromatid
...
Reductional division - meiosis 1 (divide genetic material in half)

Homologous chromosomes separate
Each chromosome still 2 sister chromatids therefore chromosome separates
Meiosis 1 = 2n to n
2
...
Meiosis 2 = n to n
Interphase double stranded, homologous chromosomes pair with crossing over
creating recombinants and non-recombinants, meiosis 1, 2 daughter nuclei still double
stranded, meiosis 2, 4 daughter nuclei single stranded chromosomes (half the number
started with and start, and end has single chromosomes)
4 single chromosomes = 4 double stranded chromosomes so 8 chromatids = 2 cells
each with 2 double stranded chromosomes = 4 cells each with 2 single stranded
chromosomes
New gene combinations
Crossing over: 2 chromosome break and re-join with homologues leg for exchange and
shuffling of genetic material
...
Meiosis results in continuity of amount of genetic material from one generation
to next: diploid 2n - haploid n + haploid n - diploid 2n so that there is no
tetraploidly
2
...
Each parent passes an allele at random to their offspring =
diploid organism so that each organism contains 1 allele from each parent
...
Chromosomes per sex cell? There are
96/2 = 48 chromosomes and sex cells are haploid therefore half = 24
✓ 32 chromosomes in each somatic cells
...
78 chromosomes = 156 sister chromatids
...
5 bivalents in telophase 2 as total chromosomes/4 is 78/4 = 19
...
Only exchange of segments between 2 non-sister chromatids = 2
chromatids recombinant chromosomes and 2 chromatids not recombinant
chromosomes
▪

▪

Incomplete linkage: 2 genes on single pair of homologs with crossovers
occurring
o Exchange occurring between 2 non-sister chromatids
o P1 with AABB [AB sister chromatid to AB] x aabb where 1 of the AB sister
chromatid overlaps with a non-sister chromatid of 1 of the ab sister
chromatids at chiasma/synapse (other AB and ab is unchanged) =
crossover gametes Ab single chromosome and aB & non-crossover
gametes AB and ab = heterozygous AaBb F1 with crossovers
Test cross: AaBb (2AB and 2ab gametes) F1 x aabb (ab) -> 1:1:1:1 (¼ aabb: ¼
aaBb: ¼ Aabb: ¼ AaBb)
o Genotype crossed to recessive parent (aabb)
o >50% parental phenotypes
o <50% recombinant phenotypes: only 2 of the 4 adjacent non-sister
chromatids exchange genetic material
o Genes are linked with crossing over and recombination

Interlocus distance
Number of crossover events between 2 loci is directly proportional to the distance
between them: more recombination occurs, and the number of crossover events
increases when the 2 genes are further apart from each other
▪
▪

▪
▪

% of crossover events = distance between 2 genes
Distance and frequency of recombination between 2 points: many cross-overs
with a distance increase between gene A & B and fewer cross-overs with a
distance decrease between gene A & B
Distance between genes increases (cM) = more % recombinant gametes in the
offspring form and less (non-crossover) gametes form
Closer the 2 genes are located next to each other (decrease in distance) = less %
likelihood of the formation of chiasma and cross-over recombinants between
them

INDEPENDENT ASSORTMENT
Different chromosomes
50% parental phenotypes and 50% recombinant phenotypes
LINKAGE NO CROSSING OVER -> complete linkage
2 genes on the same chromosome

Genetics 214

Genes are closer with no crossing over and no recombination
100% parental phenotypes with the gametes in the same equal proportional 50%: 50%
of parental gametes
LINKAGE CROSSING OVER -> incomplete linkage
2 genes on the same chromosome
Recombination occurs = linkage breaks = some recombinants and some nonrecombinants
Depends on distance: 2 genes further apart recombination likelihood of recombinants
increases and if genes are closer together the less likelihood of recombinants
decreases
>50% parental phenotypes
<50% recombinant phenotypes: not more than 50% as 2 adjacent non-sister
chromatids exchange genetic material the other 2 are original parent (2/4 chromatids =
50%)
Linkage group
▪
▪
▪

▪

Group linked genes are represented on the same single chromosome
No
...
2
B-C distance = 30% = 30cM = probability 0
...
2 x 0
...
06 = 6% = 6% of offspring to show double crossover phenotype
Percentage DCOs always much lower than either of the single crossover classes
Mapping in Drosophila (3 genes)
▪

Y chromosome never combines so X chromosome acts as double

Yywwecec female x y+w+ec+ male = gametes ywec X from female XX & y+w+ec+ X from
male XY as the male Y allele does not contribute
F1: yy+ww+ecec+ female (from maternal and paternal = XX) x ywec male (from maternal
and other Y chromosome not contributing = XY)
▪
▪
▪
▪
▪

Non-crossovers = most abundant phenotype in F2 offspring
Look for 2 phenotypic class with most offspring = parental non-crossover
phenotypes
Double cross-over = least observed phenotype therefore 2 phenotypic classes
that you don’t often see
Reciprocal classes = same numbers
Single cross-over events = intermediate frequency

Genetics 214

NCO (non-crossovers): combination of parental gametes
▪
▪

No crossing over during formation of F1 gametes
Parentals easily distinguished -> largest percentage of F2 phenotypes

DCO (double crossovers)
▪

Have the lowest probability of occurrence = offspring which contain them occur
in lowest frequency

SCO (single crossovers)
▪

More common -> highest probability and high frequency occurring in offspring

SCO per each distance + DCO distance of all combined
▪
▪

Total no
...
5% + 0
...
56% = 1
...
56 cM
Total no
...
06% = 4
...
06 map units = 4
...
3% map units/cM distance between v & pr
pr-bm = 43
...
223 x 0
...
097 = 9
...
8% DCO obs

DCO exp ≠ DCO obs therefore crossing over in 1 locus interferes with the crossing over
at another nearby loci creating the coefficient of coincidence (C) which is the observed
probability (actual double crossovers) of the expected double crossovers and the rest of
the probability is due to interference at chiasma not creating a certain number of double
crossovers of the expected double crossovers
Coefficient of coincidence (C) = DCO obs/DCO exp = 0
...
097 = 0
...
804 = 0
...
e
...
e
...
Linkage group: represents all genes located on the same chromosome (crossing
over can occur and can produce recombinant gametes)
2
...
2 genes fail to assort independently = linkage
4
...
e
...
18 and 820 parental genotype (parentals)/1000 progeny in F2
Parentals offspring have 2 reciprocal classes -> 2 different genotypes
Recombinant offspring have 2 reciprocal classes - > 2 different genotypes
CONCEPT 5
Single crossovers are more frequent -> 2 genes A and B have a crossover at 2 adjacent
non-sister chromatids with 1 chiasma point = recombinants (other 2 non-sister
chromatid untouched = parentals)
Double crossovers can be observed -> 3 genes A, B and C have a crossover at 2
adjacent non-sister chromatids with 2 chiasma points = recombinants (other 2 nonsister chromatid untouched = parentals)

20% A + B (between A/a and B/b) = 0
...
3
SCO therefore 0
...
3 = 0
...
/DCO exp
...
3 gene mapping for x-y-z, following observed: NCO 65%; SCO x-y 15%; SCO y-z
17% and DCO 3% (simultaneous x-y and y-z crossovers)
...
How is C value determined when calculating interference: DCO obs
...

Fraction of observed DCO expressed compared to expected DCO
3
...
2, what is the interference? Negative interference
C>1 = negative interference has more DCO observed than expected
...
Deduce about allele combination in heterozygous F1 parent based on: a+c // +b+
75%, ++c // ab+ 10%, a++ // +bc 13% and abc // +++ 2%
a+c // +b+ is the phenotypic allele combination between it shows the highest
frequency (most observed class) therefore heterozygous
Test cross
2 linked genes
▪
▪
▪
▪

4 types of offspring phenotypes
Non-recombinant: 2 represent parental phenotypes (NCO) = highest frequency
Recombinant: 2 represent crossing over between 2 genes (SCO) = lowest
frequency
If there are 2 genes with 2 recombination events (crossing over at 3 chiasma)
between A and B & gametes look like parentals
...
observed
148
1828
227
ONLY 6 CLASSES WITH GENE S, B AND C (3) = NO

DCO

sBc
sbc
sbC

217
1779
139

1
...
Place classes that occur in highest frequency first to the lowest frequency
Parental combinations (NCO) determined by identifying 2 classes in highest frequency

3
...
1% -> (1828+1779)/4338 as NCO grouped
2 SCO1 classes
6
...
2%

% recombination = 6
...
2%
Distance between the genes = 6
...
2cM
4
...
Correct order is BSC because both SCO1 and SCO2 is theoretically possible to
get the genes in relevant combinations
6
...
1% -> (1828+1779)/4338 as NCO grouped

SBc = BSc
sbC = bsC

148
139

2 SCO1 classes
6
...
2%

7
...
2cM

S

6
...
1% ->
BSc SCO1 6
...
Use frequency of this crossover/recombination to determine distance
6
...
6cM
To produce SCO2, a crossing over has to occur between B and S [BSC NCO 83
...
2%]
...
2 so B-S 10
...
The crossover recombinants originate from BSC gametes & bsc gamete x bsc
gamete & bsc gamete: heterozygous x homozygous test cross due to there only
being 6 classes and no DCO
2 LINKED GENES (just SCO)
▪
▪
▪

4 classes
Only determine distance
Use SCO = lowest frequency

3 LINKED GENES (no DCO)
▪
▪
▪

6 classes
Determine order by comparing NCO (highest frequency) with both SCO1 and
SCO2
Write everything in correct order & determine distances by comparison of NCO
with both SCO1 and SCO2

3 LINKED GENES (with DCO)
▪
▪
▪

8 classes
Determine order by comparing NCO (highest f) with DCO (lowest f)
Write everything in correct order & determine distances by comparison of NCO
with both SCO1 and SCO2 as well as DCO

Mapping of 2 genes: maize
R - coloured aleurone
rr - colourless aleurone
Y - green plants
yy - yellow plants
RrYy x rryy -> heterozygous x testcross double homozygous
coloured green x colourless yellow
Phenotype
Coloured yellow
Coloured green
Colourless yellow
Colourless green

No
...
Determine order of genes
Coloured yellow and colourless green are parental phenotypes NCO as highest in
frequency
Sort according to highest -> lowest frequency = 2 top classes NCO and 2 bottom
classes SCO
Phenotype
Coloured yellow

No
...
Calculate distances between genes
For NCO: 88 + 92/total 200 = 90% NCO
For SCO: 12 + 8/200 = 10% SCO
Therefore R 10cM
Y -> distances refer to crossovers & the 10% is the SCO =
10cM
Coloured yellow 88 and colourless green 92 due to parental Ry, rY gamete homolog
(RrYy) x ry, ry (rryy) gamete homolog = Rryy & rrYy = coloured yellow & colourless
green
Mapping of 3 genes: maize without DCO
1
...
If no DCO occurs, both SCO1 and SCO2 compared to NCO
3
...
Draw gene map of 3 genes
Mapping of 3 genes: maize with DCO
In maize the recessive mutant genes (without + as + is wild type):
bm - brown midrib
v - virescent seedling
pr - purple aleurone
bmbm+vv+prpr+ x bmbmvvprpr -> heterozygous x test cross
*Reciprocal classes occur
Most = NCO and least = DCO

Genetics 214

1
...
Crossovers of NCO are:
++bm
prv+

++v
prbm+

vprbm
+++

3
...
Interlocus distance: v-pr-bm
Determine total number of crossing overs (%/cM) between genes as:
SCO1 + DCO and SCO2 + DCO
To produce SCO1 = crossing over has to occur between v and pr therefore the frequency
of this crossover/recombination class determines the distance as 14
...
8% or
cM = 22
...
3 cM
NCO to SCO1 as 14
...
3 cM

pr

bm

To produce SCO2 = crossing over has to occur between pr and bm therefore the
frequency of this crossover/recombination class determines the distance as 35
...
8% or cM = 43
...
4 cM
NCO to SCO2 as 35
...
3 cM pr 43
...
Draw the gene map of the 3 distances with accurate distances apart
bmbm+vv+prpr+ x bmbmvvprpr -> heterozygous x test cross
Gametes as v+bm, +pr+ x vprbm, vprbm
Creates crossover of v+bm = 230 and +pr+ = 237 as NCO

Genetics 214

In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly x a
male which is claret (dark eyes) & hairless (no thoracic bristles) -> [P = F1]
Phenotypically wild type F1 female progeny were mated to fully homozygous (mutant)
males and the following progeny (1000 total) were observed [F1 x F1 = F2]
a
...
WRT the three genes, what are the genotypes of the
homozygous parents used in making the
phenotypically wild F1 heterozygote?
c
...
What is the coefficient of coincidence?
*Don’t mention phenotype of spinless mutant = wild type +
*Grouped high -> low frequency in reciprocal phenotypes
1
...
The groups represent correlation to NCO, SCO and DCO
Group 1 = most = NCO, Group 2 = SCO, Group 3 = SCO, Group 4 = least = DCO
3
...
Genotypes of homozygous parents: look at NCO with h in the middle:
s++ and +hc
s++, s++ x +hc, +hc (sHC, sHC x Shc, Shc)
5
...
Final map
s

10cM

h

30cM

c

7
...
/DCO exp
...
03/(0
...
3) = 1
I = 1-C = 1-1 = 0
No interference
RECAP
Genes
Linkage
Same

Assortments
Separate

Chromosomes
Special types

<->

Mutations

Closer genes = less likely for recombination
CONCEPT 1: observed phenotypic classes
2 genes: only distance and 4 classes of 2 NCO and 2 SCO

Genetics 214

3 genes: order & distance and 6 classes of 2 NCO and 4 SCO (2 SCO1 and 2 SCO2)
3 genes: order & distance and 8 classes of 2 NCO, 4 SCO (2 SCO1 and 2 SCO2) and 2
DCO
CONCEPT 2: genetic distance
RF (A-B) = 13
...
4% -> B to C closer
RF (A-C) = 18
...
e
...
5%/cM -represents the directly measured A-C RF is smaller than map distance
calculated from A-B and B-C RFs due to double crossovers
AB + BC = 13
...
4cM = 19
...
5%/cM (difference due to
DCO)
Further apart distance (distance increase cM) = increase in recombinant frequencies (r)
= straight line linear graph

CONCEPT 3: cis vs trans configuration
Arrangement of linked loci in dihybrids as AaBb
Trans: parental/non-cross over and recombinants not the same
Ab Ab x aB aB P1 = F1 and the F1 undergoing a test cross (ab ab) = >1/4 Ab ab >1/4 aB
ab, <1/4 AB ab and <1/4 ab ab F2
> = parental type and < = recombinant
Cis: parental/non-cross and recombinants can be the same
AB AB x ab ab P1 = F1 and the F1 undergoing a test cross (ab ab) = >1/4 AB ab, >1/4 ab
ab, <1/4 Ab ab and <1/4 aB ab F2
> = parental type and < = recombinant

Genetics 214

CHROMOSOME MUTATIONS - study unit 2

1
...
Molecular mutations: point mutations altering amino acid position & coding for a
protein (contributes to allelic variation) and sequence repeats
Point mutations more likely in males: lethal point mutations more with advanced male
age due to males constantly producing gametes therefore more opportunity of build-up
of point mutations in germline cells whereas females born with all eggs

Variation in chromosome number
Nondisjunction in meiosis -> chromosomes not separate -> daughter cells extra
chromosome/set
Aneuploid: 2n +- a extra chromosomes -> normal diploid with extra/less copy
chromosome
Monosomy (2n - 1), trisomy (2n + 1) and tetrasomy (2n + 2)
Euploid: complete haploid sets of chromosomes
Diploid (2n), triploid (3n), tetraploid (4n) and polyploid (3n, 4n, 5n etc
...
e
...
Terminal deletion: deletion at end of a chromosome
2
...
Paracentric inversion: inverted segment does not include centromere
A breakage occurs, and the arm ratio is unchanged
2
...
Dicentric 2
centromeres on the end except for 1
▪

Pericentric (centromere & ratio changed) inversion heterozygote
o Inversion loop including crossovers: duplication and deletions

Genetics 214

o NCO = normal sequence & no recombination in 1 chromosome -> AB
(centromere) CDEF
o SCO = duplication and deletion in 2 & 4 chromosome -> AB (centromere)
Cda
o NCO = inverted sequence in 3 chromosome -> abc (centromere) def
o SCO = duplication and deletion in 2 & 4 chromosome -> FEDc (centromere)
bef

Translocation
▪
▪
▪

Movement and crossover/recombination of genetic material in the genome
No genetic material is lost or gained but abnormal bivalent chromosomes
Cruciform or tetravalent (4 chromosomes)

1
...
Reciprocal translocation: 2 non-sister chromosomes exchange fragments of genetic
material therefore exchange fragments to another place on the same chromosome

N1 and N2 that is normal non-homologous 2n but different
T1 and T2 that is translated (segments of the other chromosome) chromosomes
2n
1
...
Adjacent segregation 1: adj
...
non-homologous centromeres to same pole ->
seldom each contains duplications & deletions inviable gametes -> T1/N1 & N2/T2
3
...
Chromosome 13, 14, 21 and 22
Breakage occurs in the 2 non-homologous chromosomes
o Centric fusion: larger chromosomal segments fuse at centromeric region
= larger submetacentric or metacentric chromosomes
o Acentric fragments: small acrocentric fragments are lost

Familial Down Syndrome
▪
▪

▪

Individual with chromosome 14 or 21 translocation = translocation carrier 14/21
45 chromosomes = normal even though only 45 and not 46 chromosomes
Translocation carrier (small fragment lost)
o Normal -> 46 chromosome
o Translocation carrier -> 45 chromosomes -> Robertsonian
o Trisomy 21 (down) -> 46 chromosomes
o Monosomic (lethal) -> 45 chromosomes
Translocation carrier gametes fertilised with normal gamete -> normal 46
chromosome cell, translocation/Robertsonian phenotypic & abnormal carrier 45
chromosome cell, extra 2n + 1 set = 46 trisomy 21 (down) cell and a missing a
chromosome 2n - 1 set = 45 monosomic (lethal)

Genetics 214

Fragile sites
▪
▪

Fragile X: region on X chromosome where chromosome can break forming a gap
o Region on chromosome does not stain due to gap on chromosome
Caused by repetitive elements

1
...
Rare fragile site: present in few people and association with genetic disorders

Martin Bell Syndrome: fragile-X
▪

▪
▪
▪

More in males due to only having 1 X chromosome therefore if heterozygous i
...

in females you are normal therefore less common due to the masking effect
o 1/1500 males and 1/8000 females
Men: mental retardation, enlarged testes and abnormal facial features (long
faces and large ears)
Fragile X syndrome: 2nd most common genetic cause of mental retardation after
Down syndrome
Trinucleotide (CGG) in FMR1 gene (familial mental retardation)
o FMRP widely expressed cytoplasmic protein abundant in brain & testis
o CGG 6-55 (normal): 55-200 (pre); >230 (full)
o Genetic anticipation

SEX CHROMOSOMES AND SEX DETERMINATION - study unit 2

Genotypic influence: sex determined by genotype
Environmental influence: sex determined by internal and external environmental
conditions

Genetics 214

Sex chromosomes is a pair of heteromorphic chromosomes that characterises 1 sex or
other
▪
▪
▪
▪

XX/XO Protenor system: C
...
/more than 1 X
chromosome) & Y chromosome -> 47 XXY, 48 XXXY, 48 XXYY, 49 XXXXY and 49
XXXYY
o Develop as male but has female phenotype -> wide hides, narrow
shoulders, poor beard growth, fewer chest hairs and female pubic hairs

X chromosome ratio does not determine maleness (mammals)
▪

Turner syndrome: only 1 X chromosome -> 45 X
o Lack a 2nd X chromosomes
o Lack of Y chromosomes
o Develop as female but lacks female phenotypes -> no menstruation,
poor breast development, widely spread nipples, small fingernails and
elbow deformity with brown spots

47 XXX Syndrome and 47 XYY syndrome

Genetics 214

▪

▪

Female
o Symptoms vary from normal to underdeveloped sexual organs, sterility
and mental retardation
o 48 XXXX and 49 XXXXX
o Symptoms more pronounced
Male
o Above average height and subnormal intelligence
o Multiple Ys

Y chromosome
▪
▪

▪

Pseudoautosomal regions: PARS at both ends of a chromosome
Homology with X chromosome = synapse & recombination during meiosis
Nonrecombining region: male specific region MSY in middle of
chromosome
Heterochromatin and euchromatic
Sex-determining region of P arm Y chromosome: SRY below PAR region at
top
Gene absent on X chromosome
TDF (testis determining factor)

Sex differentiation in humans
▪
▪
▪

XY and functional SRY present -> Mullerian duct (female organs) deteriorates and
Wolffian duct develops = male
XX and functional SRY absent -> Wolffian duct deteriorates and Mullerian duct
develops = female
Due to SRY being the sex determining region on the Y chromosome & absent on X

Dosage compensation
▪
▪
▪

XX vs XY
Occurs in interphase, test for DNA positive and not seen in males as they have 1
X
XX has double the amount of X chromosomes = double the amount of genes on X
chromosome? DOSAGE COMPENSATION

Genetics 214

▪
▪

▪
▪

▪

Male functional with 1 X chromosome
Female cells contain darkly stained bodies/regions in nucleus = Barr body DNA
o Barr body -> sex chromatin body that is an inactivated X chromosome
o Dosage compensation inactivates 1 of the X chromosomes in the double
XX
Male cell or in syndrome with 1 X (Turner syndrome) = no Barr body as only 1 X
1 Barr body for each extra X chromosome (inc
...
Barr
bodies)
o Only contain 1 activated functional X chromosome
o N-1 = # X Barr bodies per cell where N = # X chromosomes
Some genes are activated, and some are inactivated

Lyon hypothesis
▪
▪
▪

Deactivation of 1 homologous X chromosome happenings in early
embryogenesis during blastocyst formation
In embryonic stem cells, the X chromosome that is activated is random and all
daughter cells receive same copy of X chromosome but deactivated
Black & orange X-linked genes -> random X chromosome inactivation for black or
orange -> all daughter cells inherit the same X chromosome inactivated as in the
mother cell which represent Barr bodies & further development takes place

Genetics 214

Transmission pattern of X-linked genes
▪
▪
▪
▪
▪

Pure breeding red female X+X+ and hemizygous white male XwY
F1: female gives X+ chromosome to male and female offspring -> male and
female red eyes but female is heterozygous -> X+Xw and X+Y
F1 x F1 = F2: red females X+X+ & X+Xw and red & white males X+Y & XwY
o Red females > red males > white males
Females red as red allele from father and mutant white from mother
Males only X chromosome from female so if mutant only inherit mutant allele

RECIPROCAL CROSSES between A and B: F1 phenotypes differ
▪
▪
▪

Pure breeding white female XwXw and hemizygous red male X+Y
F1: female gives Xw chromosome to male and female offspring -> male white and
female red due to paternal being dominant -> X+Xw and XwY
F1 x F1 = F2: red & white females X+Xw & XwXw and red & white males X+Y & XwY
o 1:1 ratio of red eyes to white eyes is observed in each sex

The difference in phenotype according to sex not anticipated according to Mendel’s
laws -> transmission of the X-chromosome: female X and X sex chromosome
(homologous i
...
same size/length) and male X sex chromosome with a SMALLER Y sex
chromosome (not homologous)
...
0
o Ratio of X chromosomes: number of haploid sets = sex
Intersex 3A + XXY
o X/A = 2/3 = 0
...
0
Diploid male 2A + XY
o X/A = 1/2 = 0
...
67
Metamale 3A + XY -> supermale
o X/A = 1/3 = 0
...
5
Diploid female 2A + XXY
o X/A = 2/2 = 1
...
5
o Y chromosome does not determine maleness
o Genes on autosomes determines maleness
Female -> XX : 2A = 0
...
5 - 1
Supermale -> close to 0 (0
...
5

▪

Gene Balance Theory: threshold for maleness reached when X:A ratio is 1:2 = 0
...
0
Regulatory cascade leads to female sexual development
Dosage compensation suppressed -> not transcribed
Presence of Sxl gene product prevents translation of msl2 message & assembly MSL
Creation of DsxF by Tra and Tra 2 -> transcribed
Non-activation of Sxl if sex ratio from X: autosome ratio in males = < 1
...

2
...

4
...


Chromosome counting = early Sxl expression
Sxl protein promotes alternative splicing of own transcript
Sxl protein promotes female splicing of Tra
Tra and Tra 2 promotes female splicing of dsx
DsxF stimulates development of female sexual traits

▪

Lower X chromosome dosage in XY flies

1
...

3
...

5
...
0) therefore
transcription of male X linked genes double and does not inactive X chromosome
Dosage compensation

Chromosomal genotypic sex determination (CSD and GSD)
Protenor XX/XO and Lygaeus XX/XY & ZZ/ZW
Reptiles, amphibians and fish
▪

GSD
o Snakes -> ZZ/ZW
o Lizards -> ZZ/ZW and XX/XY

Environmental sex determination
Temperature dependent sex determination (TSD)
Crocodiles, turtles and lizards
▪
▪
▪

Sex of determination depends on temperature the egg is incubated at
Sex governed by internal and external environmental conditions
Steroids and enzymes are involved in their synthesis

Genetics 214

▪
▪

Effect of temperature on androgen, estrogen and synthesis enzymes affect sex
differentiation
Aromatase enzyme
o Androgens (male steroid hormones) to estrogen (female steroid
hormones)
o Aromatase gene transcription mediated by temperature sensitive factors

Undifferentiated gonad with low aromatase activity therefore not affected by
temperature
High temperature -> switches aromatase enzyme on -> high aromatase activity in ovary
due to androgens converted to estrogen -> female
Low temperature -> androgens remain androgens -> low aromatase activity in testis due
to only androgens -> male

Macroclemys temminckii
Low temperature = less males and more females
High temperature (inc
...
) = more males and less females
Back to normal when temperature further increases
Trachemys scripta
Low temperature = more males and less females
High temperature (inc
...
) = less males and more females
Alligator mississippinesis
Low temperature = less males and more females
High temperature (inc
...
) = more males and less females
Back to normal when temperature further increases
Socially dependent sex determination (SSD)
▪
▪
▪
▪
▪

Change in sex over time
Polyphonic system: sequential hermaphrodites
Strict social hierarchy determines sex ratio in population
Protogynous: female -> social environment optimal -> dominant male -> only
when die does it go back to female
Protoandrous: male -> social environment optimal -> dominant female -> only
when dies does it go back to male

Genetics 214

RECAP!
Genes
Linkage
Same

Assortment
Separate
Chromosomes

Special types
Autosomes

Mutations
General structure
Sex chromosomes

Dosage comp
...
= female and XY sex chromo
...
5 -> male (0
...
67 -> intersex (between 0
...
5 -> metafemale (1
...
33 -> metamale (between 0-0
...
B+B+ and B+Bb female beardless
& BbBb female beard
...
75 males beard and 0
...
Among species, sex determined by: differences in sex chromosomes, differences in the
number of sets of chromosomes and environmental factors
2
...
Dosage compensation of X chromosome in fruit flies is by: increases expression of X
chromosomes of males
If mammal -> to express same amount of gene products, females deactivates an X
chromosome and formation of Barr bodies in females
4
...
Use the proportion (fraction) of the extreme phenotype of the parent as red or
white
2
...
Therefore 1/4n = 1/16 = ¼2
4
...
6cm and 36cm plants are crossed
...
In F2, continuous variation observed with most being 21cm and 3/200 are short
as the 6cm P parent
How many gene pairs contribute to the phenotype?
1/4n number of F2 expressing P phenotype = 3/200 (as that proportion represents at
least 1 of the P phenotype as 6cm) = 1/64 (¼3)
Therefore n = 3 as ¼3
Therefore 3 genes pairs -> A, B and C each with 2 alleles
2n + 1 = 2(3) + 1 = 7 distinct classes
How much does each additive allele add to the phenotype?
Determine range -> 36-6 = 30cm
3 gene pairs = 6 alleles
Range/(gene pairs x alleles) = range/(gene pairs x 2) = range per additive alleles which
is the average effect per allele contributing to each phenotype
30/(3 x 2) = 30/6 = 5cm contributing each additive allele to phenotype
Base length with no additive alleles is 6cm
...
effect of each allele x number add
...
7% of the data occurs within 3 standard deviations about the mean
∑ 𝑋𝑖
Mean: 𝑋̅ =

▪

Variance: 𝑠 =

▪
▪

Standard deviation: 𝑠 = √𝑠 2
𝑠
Standard error: 𝑆𝑋̅ = 𝑛

▪

Covariance: covXY =

▪

Correlation: r = (covXY)/(𝑠𝑋 𝑠𝑌 )

▪

𝑛

2

∑(𝑋𝑖 − 𝑋̅)2
𝑛−1

√
∑[(𝑋𝑖 −𝑋̅)(𝑌𝑖 −𝑌̅)]
𝑛−1

Genetics 214

QUIZ
✓ Polygenic inheritance: additive alleles quantitative to phenotypic development
✓ Qualitative genes contain discrete class and quantitative genes quantified on a
scale
✓ Quantitative traits influenced by genes and the environment
✓ Cross polygenic inheritance, 2/125 of offspring F2 extreme as one of the P1
parents, how many gene pairs involved: 2/125 = 1/62
...
formed) due to non-homogenous
environment effect so -> VG = 0 & VE ≠ 0 -> no genetic variation but environmental
variation so VP = VE
Genetic heterogeneous between breeds due to homogenous controlled environment
effect so -> VE = 0 & VG ≠ 0 -> genetic variation but no environmental variation so VP = VG

Controlled/greenhouse: inbred line A all individuals homogenous and line B all individuals
homogenous but not in terms of one another with inbred phenotypic variation differences due
to genetic variation as the environment is constant -> VE = 0 & VP = VG
Not controlled: inbred differences due to environmental variation -> VG = 0 & VP = VE

EXAMPLE
In a wild strain of tomato plants the phenotypic variance for weight is 3
...
In another
strain of highly inbred tomatoes raised under the same environmental conditions the
phenotypic variance is 2
...
With regard to the wild strain, estimate the genetic
variance and the broad sense heritability:
Inbred line: genetic variation VG = 0 as all individuals are the same within the inbred line,
therefore phenotypic variation will be due to the environment VE
VP = VG + VE = 2
...
2g2
Wild population: environmental variation is the phenotypic variance for the inbred line,
therefore solve for genetic variation VG using the full equation for normal population
VP = VG + VE = 3
...
2g2
VG = 3
...
2 = 1g2
H2 = VG/VP = 1/3
...
313
Components of genetic variation
VG = VA + VD + Vl (VI ~ 0)
VP = VE + VA + VD + VI (VI ~ 0 and VGXE ~ 0) -> VP = VE + VG where VG = VA + VD
A = additive and D = dominance
Heritability index: narrow sense of heritability or realised heritability as h2
h2 = VA/VP
h2 indicates how a population responds to selection evolving overtime

Genetics 214

Response to selection depends on h2 where selection experiments can be used to
estimate realised heritability
0 <- h2 heritability -> 1
~1 -> variation from additive effects
~0 -> variation less from additive effects
▪
▪

Trait means before (M), during (M1) and after selection (M2)
M is the base population -> M1 is the selected individuals -> M2 is after selection

h2 = (M2-M) / (M1-M)
▪

▪

Response to selection: (mean of next generation that evolves after selection mean base population) / (mean of selected parental group - mean of base
population)
M2-M = response R and M1-M = selection differential S

h2 = R/S where R = h2S (breeders equation)
▪

Know heritability -> predict response to selection -> R = h2S (breeders equation)
EXAMPLE 1

Maize kernel diameter
M = mean diameter of base population 22 mm
M1 = mean diameter of selected sample 10mm
M2 = mean diameter after selection evolved 13 mm
h2 = (M2-M) / (M1-M) = (13-22) / (10-22)
M2 - M = R response = 13-22 = -9
M1 - M = S selection differential = 10-22 = -12
h2 = R/S = -9/-12 = 0
...
They chose a group with average weight 600g to be
the parents of the next generation
...
What is the genetic variance?
VG = VA + VD + VI = 870 + 410 = 1280g2
b
...
What is the broad sense heritability?
H2 = VG/VP = 1280 / 2620 = 0
...
What is the narrow sense heritability?
h2 = VA/VP = 870 / 2620 = 0
...
What is the mean weight of the next generation?
h2 = (M2-M) / (M1-M)
0
...
e
...
allele = avg
...
effect of each allele x number additive alleles) = F1
phenotype
V P = V G + VE
VG = VA + VD
VP = VE + VA + VD -> VP = VG + VE where VG = VA + VD
H2 = VG/VP
h2 = VA/VP
h2 = R/S where R = h2S
h2 = (M2-M) / (M1-M)
Proportion add
...
alleles: H2 - h2 or VD/VP because (VD + VI)/VP & VI = 0 where VD = VG VA due to VG = VA + VD + VI
__________________________________________________________________________________

Oil content in maize due to realized heritability:

Genetics 214

Milk yield in cattle due to realized heritability:

Growth rate in broilers due to realized heritability:

Limits to selection response in realized heritability:
▪
▪
▪

If variation no longer has a response, there is no variation to select from for M1
due to over selection in populations and if heritability = 0, there is no variation
Heritability low for traits essential to species survival -> conception rate & litter
size
Heritability high for traits non-essential to species survival -> tail length & body
mass

Resembled amongst relatives: parent offspring regression
P = A (additive genetic effect) + D (dominance effect) + I (epistatic effect) + E
(environmental effect) + e (residual effect)
Additive genetic effect is transmitted to offspring -> reliably predicted based on
additivity
Non-additive genetics as dominance & epistasis effects are proper to individual and is
not transmitted -> depends on combinations formed across all loci in zygote therefore
cannot be known

Genetics 214

Environmental effects are not transmitted
...
e
...
e
...
e
...
& VP =
VG)

Genetics 214

If 1 of identical twins has disease, the other identical twins will as well: inc
...


Interpretation
Disorder genetic
Disorder
Disorder genetically caused
Disorder environmental

MZ TWINS WITH THE SAME GENES
Healthy person -> gene A
Environmental factors A
Clear that disease caused by environmental factor B

Disease person -> gene A
Environmental factors B

Resembled amongst relatives: adoption studies
Adopted individuals compared with biological parents and adoptive parents where the
adopted child and biological parents have shared genes and adopted child and
adoptive parent shave shared environment
Relation between individual & biological parents therefore genes caused with different
environments, disease in adoptive child & biological parents where biological parents =
adoptive child same similar rates of disease -> strong genetic component
No relation between individual & biological parents therefore environment caused,
disease in adoptive child & adoptive parents where adoptive parents = adoptive child
same similar rates of disease -> strong environment component

Genetics 214

Problems: child adopted so incomplete information on biological families & adopted
placements is not random (same community/culture as biological families)
Twin & adoption studies
Identical twins for adoption and adopted by different families -> 2 genetically identical
individuals exposed to different environments
If gene caused, similar rate of disorder therefore identical twins raised together =
identical twins raised apart
If environment caused, not the same rate for disease between 2 populations therefore
identical twins raised together ≠ identical twins raised apart
Problems: adoption not random as families that adopt can be similar due to both
wealthy & wanting child therefore raised by different families but similar parenting styles
so not so diff
...
Identical twins ≠ fraternal twins
2
...
Adopted child = biological parents
Environmental rates:
1
...
Identical twins ≠ identical twins raised apart
3
...
Marker C correlation with different genotypes
not same phenotype therefore QTL on same chromosome as marker C (vertical
proportional line)
...
Individuals from an inbred cavefish line, with
small eyes, crossed with an inbred lakefish line, with big eyes
...
Offspring has intimidate
or large eyes

Potential QTL on chromosome 5 with marker loci A, B and D mapped on chromosome
...
Marker phenotype
distribution among backcrossed offspring:

Genetics 214

F1 and F2 (F1 x F1) -> discrete classes small eyes and large eyes
F1 intermediate with P1 at each extreme
F1 intermediate eyes due to the 2 extreme small and big eyed P1
Intermediate (I) F1 backcrossed to lakefish line with big eyes (L) = F2 intermediate or big
eye
Known chromosome 5 map as:
A
B
D
A, B and D are the marker loci
Which marker loci are liked to QTL phenotype?
A-L = 875 -> 9
B-L = 760 -> 4
a-l = 898 -> 9
LINKED
b-l = 778 -> 4
LINKED
INDEPEN
...
9 + 0
...
87% = 9
...
84 + 0
...
81% = 21
...
87 cM Q 21
...
0: multifactorial trait & polygenic
environmental effects
(Aa)n x (AA)n -> (AA)n & (Aa)n ~1:1 -> not bimodal DUE TO TRANSGRESSIVE
SEGREGATION
Transgressive segregation: P1 not all fixed for all additive/non-additive alleles therefore
combination with more extreme additive/non-additive allele phenotypes then original
extreme phenotypes = smoothing of curve

Genetics 214

AABBcc x aabbCC -> AaBbCc
Cavefish AABB more additive alleles than lake fish CC
Smaller eyes P1 cavefish AABBCC
...
7 and g = 3/10 = 0
...
7
f g = 0
...
7

f g = 0
...
49
f Gg = 0
...
21
f gg = 0
...
49 + ½ (0
...
21) = 0
...
09 + ½ (0
...
21) = 0
...
5)
f of one allele is higher -> most individuals are homozygous

Allele frequencies: gene pool of each allele -> p + q =1
▪

Represent dominant allele & recessive allele where p is dominant and q is
recessive

(p + q)2 = 1 fusion

Genotypic frequencies: fraction of individuals in a population -> p2 + 2pg + q2 = 1
▪

Represents homozygous dominant, heterozygous and homozygous recessive

f G = p = 0
...
3 -> recessive
p + q = 1 where 0
...
3 = 1
Implications
▪

▪

▪
▪

Dominant phenotype with dominant allele should more frequent over time
recessive phenotype with recessive allele less in a population which is not the
case
o If dominant phenotype has no selective/evolutionary advantage, it does
not become more frequent
Genetic variation maintained in population if no evolutionary forces to increase
or decrease variation therefore maintained since allele frequencies remain
unchanged in an ideal population
Genotypic frequencies to determine allelic frequencies and visa versa
Use one genotype frequency to calculate all the other genotype frequencies
(rec
...
Calculate allele frequencies from observed genotype frequency in parental
generation as p & q
f G = f GG + ½ f Gg
f g = f gg + ½ f Gg
2
...
p2, hetero 2pq & homo rec
...
5
f MN = 0
...
3
Genotypic frequencies occur after 1 generation under conditions assumed HW
equilibrium
Allele frequencies:
f M = p = f MM + ½ f MN = 0
...
2) = 0
...
3 + ½ (0
...
4
Genotypic frequencies:
f MM = p2 = 0
...
6 = 0
...
6)(0
...
48
f NN = q2 = 0
...
4 = 0
...
5
f MN = 0
...
3
After 1 generation under HWE:
f MM = 0
...
48
f NN = 0
...
05 -> reject null hypothesis Ho therefore less than critical X2
p > 0
...
05: observed deviation is significant and not due to chance alone
o Reject null hypothesis and population not in HWE

Genetics 214

X2 = ∑(𝑜 − 𝑒)2/e
Df = number of classes (phenotype/genotype) - 1 sampling error - 1 parameter
estimation (genotypic frequencies are counted from allele frequencies in sample =
assumptions therefore room for error)
Population sample of 100
Obs
...

X2 = ∑(𝒐 − 𝒆)2/e

MM
50
36
5
...
33

NN
30
100
16
100
12
...
02

Df = 3 - 1 - 1 = 1
p > 0
...
05 fail to reject
X2 = 34
...
05 so p <<< 0
...
001
Reject null hypothesis
Case study
Resistance to HIV-1 infection
▪
▪
▪

CCR-5 receptor protein occurs on immune cells
Deletion in gene makes protein non-functional
Homozygotes are resistant to HIV-1

Do frequencies of mutant alleles in population remain constant over generations?
How resistant is population to HIV-1?
Observed genotypic numbers & frequencies (count):
+/+
582 f = 0
...
162
-/8
f = 0
...
827 +/+ + ½ (0
...
908
q = f (-) = 0
...
162) +/- = 0
...
908 x 0
...
423
+/- = 2pq = 2(0
...
092) = 117
...
092 x 0
...
959
Statistical test:
++
+--

Observed
582
114
8

Expected
580
...
619
5
...
00428
0
...
69906

Genetics 214

Df = 3 - 1 - 1 = 1
X2 = 0
...
30 so p > 0
...
f MM = p2, f MN = 2pq and f NN = q2
✓ Statistically test if locus/population is in HWE using X2 chi-square test of the
observed and calculated expected values and use df = number classes - 1 - 1 to
compare to p < 0
...
05 fail to reject
✓ Some loci in HWE and some not in HWE depending on evolutionary force on
locus which relates to the phenotype
✓ Allele frequencies: p + q = 1
✓ Genotypic frequencies: p2 + 2pq + q2 = 1
Extensions of Hardy-Weinberg law: multiple alleles
▪
▪

Some loci have more than 2 alleles: expand formula to include more alleles
Extensions of HW law to multiple alleles -> ABO blood group system and albino
series (C dominance series)

Blood groups
▪
▪
▪

Antigen phenotype: A, B, AB and O
Antibody in plasma: Anti-B for A therefore not receive B type blood, Anti-A for B,
no antigen for AB and Anti-B & Anti-A for O
Genotype: IAIA or IAi, IBIB or IBi, IAIB, ii

Genetics 214

3 allelic variants segregating in population in frequencies: p + q + r = 1
fr IA = p
fr IB = q
fr i = r
4 phenotypes A, B, AB and O BUT 6 genotypes IAIA, IAi, IBIB, IBi, IAIB and ii
IA and IB are dominant over i but co-dominant with respect to each other
Genotypic frequencies (frequency of population with a certain genotype):
p2 + q2 + r2 + 2pq + 2qr + 2pr = 1
2
is for the homozygous for each of the 3 alleles and 3 heterozygote possibilities
2pq (dom
...
) + 2pr (dom
...
)
Each individual still only has 2 copies of the gene therefore diploid with 2 alleles even if
there are more than 3 alleles in a population
EXAMPLE
Phenotypic frequencies are known for population in HWE for ABO locus:
A = 0
...
13 and O = 0
...
26 = r2
r = i = 0
...
51

p2 + q2 + r2 + 2pq + 2qr + 2pr = 1
p2 + 2pr + r2 = A blood type + O type = 0
...
26
(p + r)2 = (0
...
26)2 = 0
...
38 allele frequency for A allele
For B as q ->
p+q+r=1
q = B = 1-0
...
51 = 0
...
51
fr IA therefore A = p = 0
...
38, q = 0
...
51
IAIA -> p2 = (0
...
14
IAi -> 2pr = 2(0
...
51) = 0
...
11)2 = 0
...
11)(0
...
11
IAIB -> 2pq = 2(0
...
11) = 0
...
51)2 = 0
...
Conly contains: 38 full colour, 144 intermediate and 18
albino
Allelic frequencies:
fr C = p -> full colour as any combination with C
fr cch = q -> intermediate as cchcch or cchc
fr c = r => albino only as cc
p = fr C -> full colour, q = fr cch -> intermediate and r = fr c -> albino
Frequency Genotype
Phenotype
p2
2pq
2pr
q2
2qr
r2

CC
Ccch
Cc
cchcch
cchc
cc

Full
Full
Full
Inter
Inter
Albino

p = full
2pq is full colour & intermediate
2pr is full colour & albino
q = intermediate
2qr is intermediate & albino
r = albino

c homozygous recessive albino (r) as it is the only certain phenotype in genotype ->
Albino cc = 18
r2 = 18/(18 + 144 + 38) = 0
...
3
fr c = r = 0
...
09 = 0
...
09 = 0
...
81
(q + r)2 = 0
...
9
q = 0
...
3 = 0
...
6 allele frequency for intermediate

Genetics 214

C full colour (p)
Full colour = 38
p+q+r=1
p + 0
...
3 = 1
p = 0
...
1 allele frequency for full colour (dominant)
Genotypic frequencies:
p = 0
...
6 and r = 0
...
01
2pq = 0
...
06
q2 = 0
...
36
r2 = 0
...
64% females

Locus is in Hardy-Weinberg equilibrium
fr XBY -> p = 0
...
08 due to 8% colour-blind as mutant allele
fr XBXB -> p2 = (0
...
64%
fr XbXb -> 2pq = 2 x 0
...
92 = 14
...
08)2 = 0
...
50 of 100 hens have the barring phenotype
and the rest do not display any barring
...
5
fr ZbW -> q = 50/100 = 0
...
3
fr ZBZb -> 2pq = 25/50 = 0
...
2
pm = 0
...
5) = 0
...
2 + ½ (0
...
45 -> recessive q + ½ heterozygous
Average allele frequencies in population:
pave = (p + pm)/2 = (0
...
55)/2 = 0
...
5 + 0
...
47
Expected genotypic NUMBERS:
fr ZBW -> p = 0
...
47 x 100 = 47
fr ZBZB -> p2 = (0
...
53 x 0
...
47)2 x 50 = 11
Testing for HWE through X2 test:

O

Z BZ B

Z BZ b

Z bZ b

Z BW

Z bW

15

25

10

50

50

Genetics 214

E

14

25

11

53

47

X2

0
...
09

0
...
19

0
...
05
Therefore fail to reject H0 and population is in HWE at this locus
Extensions of Hardy-Weinberg law: estimating heterozygous frequency
▪
▪
▪

Only recessive phenotypes (aa) are known
Not possible to distinguish between AA and Aa due to dominance
Use HW law and determine allele and genotype frequencies

Recessive disease carriers -> Cystic fibrosis
▪
▪
▪
▪
▪

Autosomal recessive disorder
1 in 2000-3000 affected
Affects mostly the lungs but also pancreas, liver and intestine
Abnormal transport of chloride and sodium
Thick and viscous secretions

Phenotype frequency:
1/2500 = 0
...
004
q = 0
...
98
Homozygote recessive -> 1/2500 with 2% affected with Cystic fibrosis
Heterozygotes carries with mutant allele but not affected:
2pq = 2 x 0
...
02 = 0
...
If 60% of the seeds in a randomly mating population
are able to germinate in contaminated soil, what is the frequency of the resistance
allele and what percentage of the surviving seeds are carriers?
Resistant alleles are dominant:
60% germinate (resistant) in contaminated soil
Therefore p2 + 2pq = 0
...
6 + q2 = 1
q2 = 0
...
63

Genetics 214

p+q=1
p + 0
...
37
p2 = (0
...
14
2pq = 2 x 0
...
63 = 0
...
63)2 = 0
...
46/0
...

In a population there are deviations in: mutations & migrations (mating system)
and selection & genetic drift (life history)

What happens if yellow flies are more successful at reproducing compared to wild type?
Over time, yellow allele frequencies increases across generation and therefore the
yellow phenotype increases = changes in allele frequencies = population not in HWE
equilibrium
Finite (large) population sizes
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Real populations are not infinity large -> most occur in small populations
In a small population there are a limited number of individuals that can breed
Some population are small such as endangered species and new populations
arising
Population bottlenecks and founder effects occur in small populations
o Founder effect reduces a population as a segment within a population
moves
Therefore random genetic drift occurs which causes the fixation or loss of alleles

Population bottleneck: original population generation 1 with large genetic diversity ->
population reduction generation 2 in a bottleneck event with a few surviving ->
generation 3 with a smaller surviving population -> final population generation 4 with
small genetic diversity & recovery can take place
Original population -> bottleneck catastrophic reduction occurring within a population
with chance survivors -> new population

Genetics 214

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Large population with rare alleles & individual dies = not impact gene pool
In a large population the allele frequencies constant over time
o More generations are required before allele is eliminated or fixed
In a small population the allele frequencies fluctuate between generations
o Alleles increasingly or decreasing fluctuate to dominant/recessive alleles
Alleles can become fixed therefore all individuals have only one genotype in the
population -> not due to selection of 1 allele or another just due to chance
o Reduce genetic diversity occurs by chance in small populations
Need genetic diversity for evolutionary potential to respond and adapt

Founder effects occurs in small population with a high allele frequency due to chance
(1 in 300 Dutch settles in SA)

Mutations
Independent segregation and assortment creates variation but only acts on variation
already existing in the population -> generates new gene and allele contributions but
only uses standing genetic variation (crossing over at synapsis = recombinant
chromatid)
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Mutations is the ultimate source of new genetic variations
Mutations increase allele frequency which increases new novel allele &
decreases allele of the existing wild type allele according to a fixed mutation rate
in generations
o Over time allele & genotypic frequencies will change within each
generation if there are constant mutations
Mutations are random: advantages or deleterious effects due to impacts of
selection and neutral effects due to impacts of drift
In a small population there is a loss of new mutant alleles
In a large population the mutant alleles persists due to chance
Mutation rate: fraction (number) of gametes in a gene pool with new mutant
alleles
Decrease in frequency of A (prob
...
a ->
mutant recessive allele due to mutations) over generations occurs as a function
of the mutation rate µ therefore a decrease in AA and increase in Aa & aa
∆q = μp -> decrease in dominant p = increase in recessive with mutation rate

Genetics 214

Neutral theory
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Natural theory is new mutations with little or no effect on survivability
Allele frequencies are influenced only by mutations rates and genetic drift
Natural selection preserves or removes new mutations that influences fitness

Selection
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Phenotypic and genotypic variation occurs within a population
Some phenotypes are more successful in a population by adapting to stresses
therefore produces more offspring
Some genotypes increase in next generation and some decrease in next
generation
Certain traits are selected for in favourable conditions and some are selected
against with those having favourable traits flourishing and breeding therefore
dominating the population and the unfavourable traits that cannot adjust die off
Genotypes do not have equal survivability

Population of 100 with fr (A) as p = 0
...
5
Under HWE:
Survivability:
2
AA: p = 0
...
5 = 50
Aa: 90%
2
aa: q = 0
...
53
fr a = (45 + 40)/180 = 0
...
9 as 90% and w (aa) = 0
...
e
...
e
...
e
...
6 = 16% & wild type q1 = 0
...
0 = 0% & wild type q2 = 0
...
8 = 4% & wild type qfused = 0
...
Asian elephants = different species = not
interbred
Beef vs dairy cattle
Race vs draught horses

Mainland-island model of migration
Population 1: mainland large therefore movement of alleles do not impact allele
frequencies
Population 2: island is smaller therefore susceptible to introduction of migrant alleles
Mainland allele to island population = no impact on the mainland population (large
population) therefore allele frequencies are constant
On island = impact with allele frequencies changing (smaller population therefore
deviation)
Mainland: fr (A) = pm = 0
...
4
Migrants: m = 0
...
1) x 0
...
1 x 0
...
42
ABO with a B allele gradient -> consequence of migration
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Original B allele mutation in Eastern Europe with high %
Follow gradient of B allele migration to West

Non-random mating
Positive assortative mating: individuals choose to mate frequently with other individuals
that have a similar phenotype to themselves -> similar genotypes
Genes are associated with specific trait = increase homozygote genotypic frequencies
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Strong facilitating speciation
Cichlid fish and poison-dart frogs

Negative assortative mating: produce offspring with a mate that has an opposite
phenotype to themselves -> opposite genotypes
Genes are associated with specific trait = increase heterozygote genotypic frequencies
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Flower has male and female reproductive organs therefore the pollen and
stamen occurs in the same flower

Genetics 214

Plant self-incompatibility system: if pollen has a similar allele to the seed allele
occurring in the flower, the pollen is unable to fertilise egg in flower
...
If the wild type allele has a frequency of p = 0
...
What is the fixation index at this
locus?
380 sample - 291 homozygotes = heterozygotes & wild type frequency is the dominant p
H0 = (380-291) / 380 = 0
...
6 x (1-0
...
48 -> expected is under HWE therefore 2pq
F = He - H0 / He = (0
...
234) / 0
...
513
High fixation index therefore less heterozygotes in population than expected under HWE
More homozygotes than heterozygotes are observed -> inbreeding by self-fertilising
If locus was in HWE -> fixation index ~ 0 as observed and expected would be the same
Processes that define microevolution: departure from HWE due to
Selection, mutation, migration, finite populations and non-random mating
Within population: stabilizing selection, mutation with homozygosity, migration
diverging populations and types of natural selection -> increases genetic variation &
disruption selection, genetic drift and types of natural selection -> decreases genetic
variation
Between populations: mutation, genetic drift and types of natural selections ->
increases genetic variation & migration and types of natural selection -> decreases
genetic variation

Formation of new species
Population

Genotypic and allele frequencies remain constant across generations HWE
Generation 1

Gene pool

Generation 2

Micro-evolutionary processes (gene pool)
Genotypic and allele frequencies change across generations -> population evolves not
HWE
Reduced gene flow
Mutation
Differential selection

Genetics 214

Random genetic drift
MICRO & MACRO EVOLUTION
Population -> geographical barrier splitting into two populations reducing gene flow ->
each population acquires different genetic differentiation over time due to differential
selection, genetic drift and mutations -> evolution of reproductive isolating
mechanisms (RIMs) -> populations come into contact through secondary contact the
RIMs prevent gene flow between them and cannot interbreed -> selection of prezygotic
RIM and if postzygotic RIMs evolve then selection strengthens prezygotic RIMs ->
different species A and B -> speciation
Speciation through reproductive isolation (formation of new species)
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Interbreeding or potentially interbreeding populations can be reproductively
isolated from all groups -> no longer produce viable offspring -> new species
evolve
Reproductive isolating mechanisms are biological barriers preventing or
reducing interbreeding between populations

Biological barriers
Prezygotic: egg and sperm fertilisation prevented and zygote not produced
1
...
Temporal isolation (seasonal)
Early summer frogs phenotypically different to late summer frogs = speciation = new
species

3
...
Mechanical isolation: shaped a certain way so that certain species gain access
Honeybee nectar from foxglove flower and hummingbird nectar from trumpet flower
Flower & stigma shape or only releasing pollen when wings vibrate at certain frequency

5
...
Non-viable hybrid
2
...
F2 breakdown: hybrids are fertile = individuals F2 from hybrid not fit as not
adapted
Speciation through physical separation (formation of new species)
Darwin HMS Beagle -> Darwin Finches -> Galapagos islands -> Allopatric speciation
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Original flock from mainland (South America) spread over various islands
becoming geographically isolated where on each island there is reproductive
isolation
Contact with original species -> no interbreeding = viable offspring -> speciation
Number of islands increase -> over time -> number of species of finches
increases

Rift Valley lakes -> Nile and Congo Cichlid fish -> Sympatric speciation
Lake Vitoria with species; Lake Tanganyika with species & Lake Malawi with species etc
...

sapiens
African branch basal is to the origin of modern humans due to no inbreeding with other
branches therefore modern humans evolved from Africa -> migrated to Europe and Asia
Ancestry of European or Asian means there is a low frequency of genetic evidence for
interbreeding between H

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