Genetics 3

Question 1

Genotype vs phenotype?

Answer 1

-Each individual has two copies (alleles) of each gene, one inherited from each parent (genotype)
Exception: if the gene is on the X or Y chromosome.
-All somatic cells in the same animal carry identical genes, but different cells have different functions and develop inside different organs.
-The expressed phenotype of cells/organisms influenced by: genotype (the alleles present), inherited epigenetic factors (e.g., DNA methylation), and non-inherited environmental factors (e.g., temperature, cytokines).

Question 2

Genetic mutations/variants?

Answer 2

-Alterations to the genetic code can increase, modify, or stop the production of a specific protein.
-Consequences of alterations to the coding sequence of the genetic code on the consequential amino acid sequence can be easily predicted, and to a lesser extent, the effect on the individual.
-Consequences of alterations to the non-coding sequence are much more difficult (to impossible) to predict.

Question 3

Basic mammalian genetics?

Answer 3

-Dominance: Dominant allele (usually) produces a functional protein. Only one copy is required to affect the phenotype. May be complete or partial (incomplete). A recessive allele (usually) produces a reduced/ faulty protein. Both recessive copies are required to affect the phenotype. Some alleles are ‘co-dominant’
-Allele b may be recessive to allele B, but dominant to third allele b
-Dominance does not indicate whether an allele is beneficial, detrimental, or neutral.

Question 4

Example of application: Basic coat colour in dogs?

Answer 4

Dominance/recessive allele interaction: Black/brown controls the type of eumelanin produced.
B is dominant and produces black eumelanin
b is recessive and produces brown eumelanin.
-Extension: controls the proportion of eumelanin and pheomelanin.
E extends the amount of eumelanin.
e reduces eumelanin and increases pheomelanin (yellow/red)
-Epistasis (e.g., gene interaction): absences of extension results in yellow coat but the nose colour will reflect the type of eumelanin produced.

Question 5

Single nucleotide polymorphisms (SNPs)?

Answer 5

-At any single position within the genome sequence, nucleotide differences may occur.
-These differences can be passed down from one generation to the next and their presence used as (linkage) markers for the presence of nearby genes controlling traits of interest: Genome-wide association studies (GWAS) using SNPs can focus the search for target genes. Analysis of SNPs can be massively parallel, automated and high-throughput.

Question 6

Genome mapping in animals?

Answer 6

-Identify the gene(s) responsible for a genetic trait of economic or clinical importance: model for human disease. Map genes in animal pedigrees. Eliminate affected individuals and carriers from the gene pool.
-Marker-assisted selection: use in conventional breeding programmes to increase the frequency of desirable traits. Introgression from one population into another (e.g., introducing fecundity genes from Chinese pigs into European breeds).

Question 7

DNA sequencing using Sanger technique?

Answer 7

-Based on properties of dideoxynucleotide (ddNTPs): Modified nucleotides that contain a hydrogen group of the 3’ carbon instead of a hydroxyl group (OH). When integrated into a DNA sequence, they prevent the addition of further nucleotides, thus stopping the elongation of the DNA chain.
-Reactions containing individual ddNTPs (i.e., ddGTP, ddATP, ddTTP, and ddCTP) can be run separately on a gel or labeled with different fluorophores and processed in the same reaction.

Question 8

DNA sequencing using pyrosequencing?

Answer 8

A combination of biochemistry, physics, optics, and phenomenal computing power.

Question 9

Example of application: Sensory neuropathy in the Border Collie?

Answer 9

GWAS study in Manhattan plot:
-Two populations compared. One with the target trait, the other without.
-Each dot signifies an SNP
-Genomic coordinates are on the X-axis, with the negative logarithm of the association P-value for each SNP on the Y-axis.
-Investigation can then be focused on candidate genes in this region.
-Region contained 27 genes
-One was FAM134B (associated with hereditary sensory neuropathy in humans): candidate gene.
-Sequencing of FAM134B exons (coding sequence) was unremarkable.
-A 6.47MB inversion was identified with breakpoints in intron 3 of FAM134B and in an upstream intergenic region.
-All affected dogs (no controls) were homozygous for this inversion.

Question 10

Point mutations/variants?

Answer 10

-Insertion, deletion, or substitution of base pairs (NB: not all are SNPs).
-Alterations of the exonic sequences can result in: silent (no change to the amino acid sequence), missense (a different amino acid is encoded), nonsense (a premature STOP codon is introduced), frame shift (change in amino acid sequence from that point on).
-Alteration on the intronic sequence can result in: no changes, altered splicing sites (change in the amino acid sequence), or altered gene expression (increased or decreased).

Question 11

Autosomal dominant?

Answer 11

Characteristics:
-Defect seen in every generation
-Every affected offspring has at least one affected parent
-Normal offspring from affected parents will produce normal offspring
-An equal number of males and females will be affected.
-If the defect is rare, but not lethal: most matings that produce an affected individual will be aa x Aa. The segregation ratio will be 0.5.
Example:
-Feline polycystic kidney disease: nonsense point mutation in PKD1
-Scottish fold osteodystrophy (incomplete penetrance): missense point mutation in TRPV$

Question 12

Autosomal recessive?

Answer 12

Characteristics:
-Defect may skip a generation
-All offspring of affected parents (where both are affected) will be affected.
-“Normal” parents of an affected individual must be carriers.
-An equal number of males and females will be affected.
-If the defect is rare, but not lethal: Most matings which produce an affected individual will be Aa x Aa. The segregation ratio will be 0.25.
Examples:
-Hyperuricosuria in Dalmatian dogs: missense point mutation in SLC2A9.
-Avermectin sensitvity in Collie dogs: Frame shift mutation (4bp deletion) in ABCB1

Question 13

X-linked dominant?

Answer 13

Characteristics:
-Every affected offspring has at least one affected parent.
-Affected males (i.e., XAY-) mated to normal females (XaYa) will transmit the defect to all their daughters and none of their sons.
-Unless the defect is very common: affected females will be heterozygous (i.e., XAXa) and therefore when mated to a normal male the defect will be transmitted to half of the offspring.
Example:
-X-linked dominant hereditary nephritis in Samoyed: nonsense point mutation in COL4A5

Question 14

Why perform genetic screening?

Answer 14

-Identify carriers: recessive disorders, gradually remove from the gene pool, and avoid mating two carriers (reduce risk of breeding affected individuals).
-Identify affected: with late onset presentation, remove affected cats from the gene pool/counsel owners, and avoid mating affected/carriers (reduce the risk of breeding affected cats).
-Select for desriable traits that are recessive/polygenetic: direct mating to increase likelihood of obtaining desirable offspring.

Question 15

X-Linked recessive?

Answer 15

Characteristics:
-May skip a generation.
-Incidence in males» female.
-When the defect is rare: affected individuals will be males and will have inherited the gene from the dam.
Example:
-Haemophilia A (Factor VIII deficiency) in Havanese dogs: SINE (short interspersed nuclear element) insert in F8.
-The causative mutations of haemophilia in most other breeds (including cocker spaniels) is unknown.

Question 16

How to perform genetic screening?

Answer 16

-Collect a sample containing DNA from the patient: collected by vet or owner. Cheek swab (contamination risk if suckling) or blood sample (invasive). Might need to record microchip ID at time of sampling (official record).
-Submit sample to laboratory: currently, most tests are for individual mutations/variants (specific to species and may be specific to breed. NB: may be more than one mutation/variant per phenotype. Some (most) mutations/variants are unknown).
-Interpret results.

Question 17

Interpreting results?

Answer 17

Results may be reported as:
-Normal or homozygous ‘wild type’ alleles.
-Carrier (heterozygous for recessive allele).
-Affected (homozygous for recessive allele; heterozygous or homozygous for dominant allele).
Forward planning:
-Affected individual should be monitored for disease progression; prospective owners should be warned.
-Affected individuals should not be bred from (potentially difficult if heterozygous affected for dominant allele -> reduced gene pool).
-Carrier animals can be used cautiously in breeding programmes.

Question 18

Predicting offspring?

Answer 18

-Normal (AA) x Carrier (Aa)
-Predicted ratios: 50% predicted to be normal, while 50% predicted to be carriers.

Carrier (Aa) X Carrier (Aa)
-Predicted ratios: 25% predicted to be normal, 50% predicted to be carriers, 25% predicted to be affected.

Question 19

Geentic screening-unknown mutation?

Answer 19

-Determinations of carrier status of normal animal (i.e., AA or Aa). Backcross to a known homozygous recessive (aa). If seven or more normal (A) offspring are born, 99% likelihood that the test animal is not a carrier Backcross to a known heterozygous carrier (Aa) if sixteen normal (A) offspring are born 99% likelihood that the test animal is not a carrier. If any affected (homozygous recessive, aa) offspring are born to either combination, -> test animal is a carrier.

Question 20

Blood groups?

Answer 20

A blood type is a classification of blood:
-Based on the presence and absence of antibodies and inherited antigenic substances on the surface of red blood cells.
-Antigens may be proteins, carbohydrates, glycoproteins, or glycolipids (which may also be present on the surface of other cell types).
-Several of these red blood cells surface antigens can stem from one allele and collectively form a bllod group system

-Mixing of blod with even only minor antigenic differences has the potential to result in signiciant reactions: Transfusion reactions, neonatal isoerythrolysis.

Question 21

Feline blood groups?

Answer 21

AB blood group system:
-Cytidine monophospho0N-actylneuraminic acid hyroxylase (CMAH)
-B antigen (acetylneuraminic acid) -> A antigen (glycolylneuraminic acid)
-Two known alleles: A and b (null; recessive).
-Rarely type AB cat: both A and B antigen expressed
-Alloantibodies: All type B cats have strong anti-A antibodies (without prior exposure) ~1/3 type A cats have weak anti-B antibodies.
-Milk red cell antigen: cats are either positive or negative. Appears the be rare in the UK.

Question 22

Feline blood groups prevalence?

Answer 22

-Amongst domestic shorthairs: 77-95+% of cats are type A. <5-17% of cats are type B. <1-6% of cats are type AB.
-Frequency of blood types can vary between breeds and countries: Siamese are all type A. High frequency of type B in British Shorthair and Devon Rex. ~20% of Ragdolls have type AB in some studies.

Question 23

Feline blood groups genotyping?

Answer 23

-Recessive variants in CMAH result in type B cats: a number of SNPs associated with type B, including G139A (common) and C136T (pix mutation). These SNPs can be detected using sequencing.

Question 24

Canine blood groups?

Answer 24

-13 different loci, eight DEA (Dog Erythrocyte Antigen) types are recognized as international standards (not all can be diagnosed by anti-sera).
-These groups include DEA 1,3,4,5,7, and Dal.
-Naturally occurring antibody is found against DEA 3,5, and 7.
-Antibody-antigen interactions: DEA 1 (acute hemolytic transfusion reactions), DEA 3, 5, and 7 (permanent red blood cell sequestration, loss in 3-5 days), DEA 4 (no effect in vitro).

Question 25

Blood groups in other species?

Answer 25

Horse: Comeplex (7 systems- A, C, D, K, P, Q, and U). The two most potent antigens are Aa and Qa. Horses do not normally have alloantibodies from birth. Cattle: Complex ( 12 different systems, and each system can have many alleles. Groups of alleles act as multiple antigenic determinants and are called phenogroups).

Question 26

Neonatal isoerythrolysis?

Answer 26

-Mechanisms: The Dam produced antibodies against the offspring's RBC. A newborn absorbs these antibodies from the colostrum within the first 24-48 hours following birth. Life-threatening haemolysis occurs. -Mainly a problem in the horse and cat: horse (mating a group Aa positive foal (usually requires prior exposure). Cat (mating a type B queen to a type A stud cat, producing type A kitten(s). NB: this can happen at first mating).

Question 27

Avoiding neonatal isoerythrolysis?

Answer 27

-Assess potential- blood type breeding animals if: history of neonatal isoerythrolysis. In cats, if the breed has a high incidence of type B (e.g., BSH; Devon Rex). -Reduce risk- avoid combinations with potential to mismatch: Mate Aa negative stallions with Aa negative mares. Mate type B queens to type B stud cats (but this is a faff, makes type B cats 'less desirable', and limits gene pools). -Assess risk-test offspring at birth: safe offspring treated as normal. At risk offspring hand-reared for 24-48 hours or fostered.