genetics:

Question 1

classical genetics (mendelian genetics):

Answer 1

Focusses on inheritance of traits.
-Transmission of genes as outcome of fertilization. Named after Mendel.
Methods used prior to ‘molecular’ techniques.
-Trait selection.
-Crossing experiments in agriculture and livestock.
Basis of modern genetics.

Question 2

population genetics:

Answer 2

Study of evolution from a genetic point of view.
-Change in collective genetic material in a population.
-Basis of species adaptability in response to selection.
Provides evidence into other processes.
-Mutation, selection, gene flow, genetic drift.
-Application in Forensic Match Probabilities.

Question 3

molecular genetics:

Answer 3

DNA structure varies among organisms.
-Gene regulation and gene expression.
-Explanation of biomolecular processes.
Needs a basis for detecting these variations.
-Molecular techniques used in laboratories.
-PCR, VNTR and STR analysis, DNA sequencing.

Question 4

developmental genetics:

Answer 4

How genes control development of organism.
-From single cell to fully formed organism.
-Uses both quantitative genetic and genomic methods.
Shows genes controlling development are ancient.
-Similar genes exist between species.
-Often studied using model species (fly, zebrafish, mouse).

Question 5

genomics:

Answer 5

Focus is on genomes not simply genes.
-Structure, function, evolution, mapping, editing genes.
-Considers all genes not just targeted genes.
Patterns that explain physical traits, disease.
-Looks at multiple contributing factors across genome.
-Can also infer ancestry based on genome.

Question 6

quantitative genetics:

Answer 6

Study of quantitative traits (phenotypes) not genes.
-height, mass, hair colour.
-These vary on a continuum and have a heritable basis.
Emphasis on statistical analysis to support findings.
-Quantitative geneticists have maths or stats backgrounds.
-Theoretical support in the absence of direct proof.

Question 7

hardy weinberg equilibrium:

Answer 7

Contrived example of model population.
-Infinitely large population.
-Random mating.
-No mutation, no migration, no selection..
Under these conditions allele and genotype frequencies attain equilibrium.
Used as a basis for testing assumptions about real populations.
-If allele and genotype frequencies are not in equilibrium, then 1 (or more) of these forces mut be acting upon it.
-Allows scientists to explore population genetic theories.

Question 7

hardy weinberg equation:

Answer 7

Describes the frequency of genotypes in a two-allele state.
p2 + q2 + 2pq = 1
-The sum of all the homozygotes (pp and qq) + the sum of all the heterozygotes (pq and qp) is equal to 1.
Real populations may have values close to 1 or deviate significantly from 1.
-Could mean that the population is under selection pressure, is inbreeding, is small (or under sampled) or you’ve not observed all the alleles (null alleles etc).

Question 8

heterozygosity (obs and exp):

Answer 8

The average proportion of loci that are heterozygous can be presented as Observed Heterozygosity (Ho).
-Based on determining the number of heterozygotes in a data set.
-calculate meam.
-Can be presented as Expected Heterozygosity (He).
Based on expected numbers under HWE

If HO deviates significantly from HE then the population is not in HWE.

Question 9

polymorphism:

Answer 9

Polymorphism describes the existence of 2 or more alleles at a locus- A1 and A2.
The proportion of polymorphic loci-
-Number of polymorphic loci/total number of loci, i.e. 3 loci polymorphic and 7 monomorphic= 3/10.
Polymorphism Information Content (PIC).
-Tells you how useful the loci is in a panel of markers.

Question 10

population structure (fst):

Answer 10

The proportion of the total inbreeding in a population due to population differentiation or the reduction in heterozygosity that is due to the structure of the population.
Ranges in value from 0 to 1. 0 = no genetic differentiation between populations, 1= complete genetic differentiation.
Can be calculated using measures of heterozygosity.
-HS = Average heterozygosity in the subpopulation.
-HT = Average heterozygosity in whole population.
fst= ht-hs/ht

Question 11

inbreeding (fis):

Answer 11

Proportion of the total inbreeding in a population due to inbreeding within sub-populations or the probability that two alleles at a locus are identical by descent.
Ranges in value from 0 to 1. 0= no inbreeding observed in the population, 1= clonal reproduction.
Can be calculated using measures of heterozygosity.
-HO = Observed heterozygosity in population.
-HE = Expected heterozygosity in population.
fis= 1- Ho/He.

Question 12

genetic diversity:

Answer 12

Genetic diversity is required for populations to adapt to environmental change.
-Species face changing environments (climate, disease, parasites).
-Species must evolve (adapt) to over-come environmental stresses.
-Genetic diversity reflects evolutionary potential.
Loss of genetic diversity is linked to a reduction in survival.
-Positive correlation between genetic diversity and average fitness.
-Level of genetic diversity can be monitored over time.
-Genetic diversity can be managed in captive and wild populations.

Question 13

mutation:

Answer 13

Only de-novo (new) source of genetic variation.
-Occurs more frequently in non-coding regions.
-Can be single base pair (point mutation).
-Can be loss/gain of repeat unit.
Mutations in coding regions can impact fitness.
-Can result in change to expressed amino acid.
-Can be ‘silent’
Can impact the heritability of the mutation.
-May not be inherited due to death of organism.
-May have positive impact on fitness and be selected.

Question 14

meiosis and mating:

Answer 14

Meiosis causes variation in sex cells.
-Chromosomal recombination.
-Homologous chromosomes separate differently.
Random pairing of gametes.
-The random fertilization of eggs with sperm.
-Variation of siblings even though 50% shared DNA.
Random mating in population.
-Each female gamete has equal chance to pair with each male gamete.

Question 15

migration:

Answer 15

Migration can add genetic diversity.
-Gene flow from other populations.
-Population have different allele frequencies.
-Population prone to different selection pressures.
If barriers prevent gene flow….
-Genetic isolation of populations.
-Over millennia genetic variation increases.
-May eventually evolve into 2 species.
-The point at which this happens is a grey zone.
Therefore barriers can lead to loss of diversity.

Question 16

random genetic drift:

Answer 16

Changes in a population’s genetic diversity.
-Through random fluctuations NOT selection pressure.
-Allele frequencies vary and lead to more/less variation.
Occurs in small finite populations.
-Due to fertilization of few gametes (not all gametes).
-Over time individuals become ‘fixed’ as homozygous at a single locus (A1A1).
-Fixed alleles are useful (species ID).

2 events typically contribute to genetic drift…

Question 17

events leading to genetic drift:

Answer 17

Population Bottleneck.
-Sudden loss of genetic variation in the population.
-Often result of catastrophic event (mass fatality).
-Example being Cheetahs.
Founder Effect.
-Migrating populations into new areas.
-Represent subset of original population variation.
-Seen in both animal and human populations.

Both these effects lead to increased inbreeding.

Question 18

marker selection:

Answer 18

Extent of genetic diversity.
-Nuclear DNA is more variable than mtDNA.
Single Nucleotide Polymorphisms (SNPs).
-Occur ~every 1000 bases.
-Average locus contains 126 SNPs.
-Occur in coding and non-coding regions.
Microsatellites (Short Tandem Repeats).
-Occur less frequently than SNPs.
-Have a greater diversity than SNPs.

Question 19

use of stats to measure locus discrimination power:

Answer 19

There are a number of common metrics (measures) that are used to describe the power of a profiling kit.
-Probability of Identity (PI) - the probability that 2 individuals selected at random will have an identical profile at the tested locus.
-Matching probability (MP) – the same as PI.
-Power of Discrimination (PD) = 1 - PI.
-Probability of a Match (pM) = Inverse of PI = 1/PI.
Importantly, these can be calculated in the absence of a specific profile and only need allele frequencies to calculate.

Question 20

likelihood ratio:

Answer 20

1/MP.