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Animal Management Y2 > Genomics > Flashcards

Genomics Flashcards

1
Q

Define thymidine and cytidine.

A

Thymine + deoxyribose = deoxythymidine/thymidine.

Cytosine + deoxyribose = cytidine.

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2
Q

Why does uracil in DNA increase stability of DNA and efficiency of replication?

A
  • Cytosine can deaminate spontaneously to produce uracil through hydrolytic deamination – and were this to happen there could be confusion as to which nucleobase in the pair were correct.
  • Uracil-DNA glycosylases excises uracil bases from double stranded DNA, replacing them with cytosine again – to correspond to the opposite base.
  • Because RNA is shorter-lived than DNA and any potential uracil-related errors do not lead to lasting damage – uracil remains instead of the more complex thymine.
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3
Q

What are the clinical applications of predictable pairs?

A
  • Sequence of the complementary strand of DNA predicted from the coding sequence and vice versa
  • Primers can be designed for use in combination with DNA polymerases to amplify specific DNA fragments for use in diagnostic assays.
  • Targeted DNA amplification can be used to determine presence/absence of target gene and starting quantity of target gene.
  • Both coding and complementary sequences can be used in genetic assays.
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4
Q

What can sequencing be used to do?

A
  • Predict function of target gene
  • Predict amino acid sequence of transcribed gene
  • Determine species present within the sample
  • Identify variants/carriers within a species
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5
Q

What are 2 ways that chemotherapy works?

A

Limiting cells cell turnover – immune mediated disease and neoplasia

Interference with host cell replication and transcription by anti-metabolites

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6
Q

Name 5 anti-metabolites.

A

Leflunomide
Mycophenolate mofetil
Azathioprine
Cytarabine
Fluorouracil

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7
Q

Describe the effect of leflunomide.

A

Interfere with enzymes required in the de novo synthesis of nucleobase uracil.

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8
Q

Describe the effect of mycophenolate mofetil.

A

Interfere with enzymes required in the de novo synthesis of nucleobase guanine.

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9
Q

Describe the effect of azathioprine.

A

Inhibits the de novo synthesis if guanine, but also masquerades as guanine, being incorporated into DNA halting replication and inducing lymphocyte and monocyte apoptosis.

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10
Q

Describe the effect of cytarabine.

A

Another anti-metabolite. It masquerades as cytidine, being incorporated into DNA halting replication. It also inhibits DNA and RNA polymerases and nucleotide reductase enzymes needed for DNA synthesis.

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11
Q

Describe the effect of fluorouracil.

A

Inhibits synthesis of thymidine from uracil.

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12
Q

Which anti-metabolites are particularly effective in lymphocytes and why?

A

Lymphocytes in particular rely on nucleobase de novo synthesis rather than recycling. Thus, use of leflunomide and mycophenolic acid leads to a relatively selective inhibition of DNA replication in T cells and B cells.

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13
Q

Describe the effect of doxyrubicin.

A

Evicts histones, inhibits enzymes required for replication.

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14
Q

Describe the effect of cyclophosphamide.

A

Cross links guanine and prevents replication. Chlorambucil cross links DNA.

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15
Q

Describe the effect of vincristine.

A

Prevents chromosomal separation during replication.

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16
Q

How to antiviral and antibiotic drugs take advantage of differences in pathogen cell machinery?

A
  • Interference with viral replication
  • Interference with bacterial DNA folding
  • Interference with bacterial translation
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17
Q

Describe the effect of erythromycin.

A

Blocks movement of eth tRNA from the A site to the P site.

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18
Q

Describe the effect of chloramphenicol.

A

Inhibits peptidyl-transferase

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19
Q

Describe the effect of streptomycin.

A

Binds to 30S bacterial ribosomal subunit and blocks transition to a chain elongation complex

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20
Q

Describe the effect of tetracycline.

A

Blocks binding of amino acid tRNA to the A site.

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21
Q

What is recombination?

A

Typically there are 2-3 recombinations per chromosome. Recombination fraction is the proportion of alleles in one parent that could only have resulted from crossing-over during meiosis.

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22
Q

What is linkage?

A

Because whole segments move, genes that are close together are less likely to get separated than those further apart. This is known as linkage and is why some characteristics appear to be inherited together (e.g. those with red hair are more likely to be fairer skinned).

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23
Q

What are pseduoautosomal regions?

A
  • Three small regions of homology the pseudoautosomal regions (PAR1, PAR2 and PAR3) exist between X and Y chromosomes
  • These allow homologous pairing between X and Y chromosomes during meiosis
  • Pseudoautosomal genes exhibit an autosomal, rather than sex-linked, pattern of inheritance.
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24
Q

Describe sequencing using the Sanger technique.

A
  • Based on properties of dideoxynucleotide/ddNTPs
  • Modified nucleotides that contain a hydrogen group on the 3’ carbon instead of a hydroxyl group
  • When integrated into a DNA sequence they prevent the addition of further nucleotides thus stop the elongation of the DNA chain
  • Reactions containing individual ddNTPs can be run separately on a gel or labeled with different fluorophores and processed in the same reaction
25
Q

Describe sequencing using pyrosequencing.

A

Combination of biochemistry, physics, optics and phenomenal computing power. Limitations: expensive, amount of data generated. Interpretation and storage of the data. In part reliant on genome scaffold being present upon which the data can be placed.

26
Q

What are SNPs?

A

Single nucleotide polymorphisms

  • 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
  • Analysis of SNPs can be massively parallel, automated and high-throughput
27
Q

What is SNP array analysis?

A

Method based on hybridisation of different SNP alleles and detection using fluorescence. SNPs uniformly distributed across the entire genome and are typically species specific.

28
Q

What are the applications of SNP array analysis?

A
  • Genome-wide association studies (GWAS) enabled selection
  • Identification of quantitative trait loci
  • Evaluation of genetic merit of individuals
  • Comparative genetic studies
29
Q

What is a Manhattan plot?

A
  • Two populations compared – one with target trait, the other without
  • Each dot signifies a 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
30
Q

Describe X linked dominant characteristics?

A
  • Every affected offspring has at least one affected parent
  • Affected males mated to normal females will transmit the defect to all their daughters and none of their sons
  • Unless the defect is very common:
  • Affected females will be heterozygous and therefore when mated to a normal male the defect will be transmitted to half the offspring
31
Q

Describe X linked recessive characteristics.

A
  • May ‘skip a generation’
  • Incidence in male&raquo_space; female
  • When the defect is rare:
  • Affected individuals will be males and will have inherited the gene from the dam
32
Q

Define heritability.

A

The proportion of the phenotypic variation that id due to genotypic variation. Can be predicted by analysis of population statistics

33
Q

What are 6 methods of selection?

A
  • Performance test (selection based on phenotype)
  • Pedigree analysis and ancestor evaluation
  • Progeny testing
  • Performance of siblings
  • Marker assisted selection
  • Calculations: estimated breeding value, true breeding value, selection intensity
34
Q

What are the limitations of performance test selection?

A
  • Phenotype may not reflect genotype/influence of dominant/recessive genes and epistatic effects
  • Only effective for high heritability characteristics
  • If autosomal recessive trait select against
  • Recessive allele prevalence will eventually stabilise at 0.001% within the population
  • Recessive allele cannot be eliminated due to presence of carriers
35
Q

What is pedigree analysis/ancestor evaluation selection method?

A

Estimated breeding value is calculated from records of ancestors. Useful for late manifesting traits, where parental records are the only indication of a young animal’s potential performance.

36
Q

What are the properties of progeny testing selection method?

A
  • Assessing non-additive gene effects
  • Assessing low heritability characteristics
  • Assessing sex-limited characteristics
  • Assessing carriers of a recessive gene
  • Useful where semen is banked
37
Q

What is performance of siblings selection method?

A
  • The closer the relative the more reliable the data. Litter mates are best
  • Provides additional data to an individual’s performance
  • Useful for characteristics that cannot be seen and assessed on the live individual
38
Q

What is marker assisted selection in cattle?

A
  • Selection index – Best Linear Unbiased Prediction (BLUP)
  • A number of marker sites have been mapped in the bovine genome with loci (QTLs) linked to certain production traits
  • Originally developed for selection using phenotypic data (traits that are governed by a number of genes also identified)
  • With use accuracy improving
39
Q

What is selection intensity?

A
  • Genetic gain is greater the greater the selection intensity
  • Selection intensity is usually male&raquo_space; female
  • The larger the population the greater the possible selection intensity
40
Q

What are the factor affecting the rate of improvement of a selection intensity?

A
  • Heritability of the trait
  • Intensity of selection
  • Generation interval
  • Number of traits required for improvement
41
Q

Name 3 methods available for selection for more than 1 trait.

A

Tandem selection

Independent culling

Selection index – marker assisted selection

42
Q

What is tandem selection and an example of its application?

A

For tandem selection there is a tendency for the last trait to be lost as the next is being selected. Example: when we look at the apparent dichotomy between selecting for growth rate and the strength of legs in chickens.

43
Q

What is the basis of blood classification?

A
  • Based on the presence and absence of antibodies and inherited antigenic substances on the surface of red blood cells (RBCs)
  • 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 cell surface antigens can stem from the effects of one gene and collectively form a blood group system
44
Q

What are some examples of significant reactions to even minor antigenic differences in transfused blood?

A

Anaphylactic response – reaction to foreign plasma proteins; marked histamine release > cardiovascular shock. Seen within minutes of starting transfusion, life-threatening.

Haemolysis – reaction to foreign RBC antigens. Intravascular or extravascular destruction of donated RBC. Occurs within minutes, to hours, to days.
Potentially life-threatening (and shortens lifespan of transfusion).

Pyrexia – antigenic response to other blood components. Seen within minutes to hours of starting the transfusion. Common.

45
Q

How is blood grouped according to genotype?

A
  • Performed using DNA from cheek-swab or whole blood
  • Known alleles for blood type determined
46
Q

How is blood grouped according to phenotype?

A
  • Typically performed using antibodies to the target surface antigen.
  • Antibodies may be fixed to a surface or free in solution
  • A positive test line or agglutination indicates the presence of antigen on the RBC surface
47
Q

How is blood typing limited?

A

Current commercial genotyping (feline)
- Only available for common feline blood group
- Can only test for known variants for known genes
- Additional blood types not tested for

Limitations of current phenotyping
- Point of care tests often identify only the major antigens (DEA1 for dogs, AB group in cats; CA in horses)
- Even large commercial laboratories might not be able to differentiate some blood groups

48
Q

What is cross matching of blood?

A
  • Where time allows patients should have a cross-match performed
  • Often not practical
  • Essential if patients has received blood products > 4 days prior
  • Commercial point-of-care cross-matching kits are available for dogs
49
Q

How can cross matching of blood be done in practice?

A
  • Mix washed donor RBCs with recipient plasma (major) and vice versa (minor)
  • Examine under microscope for agglutination or haemolysis
  • Can take up to 1 hour to develop
50
Q

Describe the feline blood groups.

A

AB blood group system (completely independent of human ABO system)

  • One locus – cytidine monophospho-N-acetylneuraminic acid hydroxylase (CMAH) – converts ’B’ antigen (acetylneuraminic acid) to ’A’ antigen (glycolylneuraminic acid)
  • Two alleles – A and B (null; recessive)
  • Rarely group AB cats – both A and B antigens expressed

A second feline blood group known (Mik red cell antigen4). Cats are either positive or negative for the antigen.

51
Q

Describe canine blood groups.

A
  • 13 different loci, 8 DEA (dog erythrocyte antigen) types are recognised as international standards
  • Not all can be diagnosed by anti-sera
  • Naturally occurring antibody if found against DEA 3, 5 and 7
52
Q

Describe equine blood groups.

A
  • Complex; 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
53
Q

Describe cattle blood groups.

A
  • Complex; 12 different systems and each system can have many alleles
  • Groups of alleles act as multiple antigenic determinants and are called phenogroups
54
Q

What is neonatal isoerythrolysis?

A
  • Dam produces antibodies against the offspring’s RBC
  • New-born absorbs these antibodies from the colostrum within the first 24-48hrs following birth
  • Life-threatening haemolysis occurs
  • Mainly a problem in the horse and cat
55
Q

Distinguish neonatal isoerythrolysis in horse and in cats.

A
  • Horse – mating a group Aa negative dam to a group Aa positive sire producing a group Aa positive foal
  • Cat – mating a group B queen to a group A stud cat producing group A kittens (can happen at first mating)
56
Q

How can neonatal isoerythrolysis be avoided?

A
  • Assess potential – blood type breeding animals if history of neonatal isoerythrolysis or (in cats) if breed has a high incidence of group B
  • Reduce risk – avoid combinations with potential to mismatch: mate Aa negative stallions with Aa negative mares. Mate group B queens to group B stud cats
  • Assess risk – test offspring at birth: at risk offspring hand-reared for 24-48 hours or fostered
57
Q

Why perform genetic screening?

A

Identify carriers
Identify affected
Select for desirable traits that are recessive/polygenetic

58
Q

Describe the process of genetic screening for a known mutation.

A
  1. Collect a sample containing DNA from patient. Collected by vet or owner. Cheek swab can be contaminated and blood sample can be invasive.
  2. Might need to record microchip ID at time of sampling.
  3. Submit sample to laboratory.
  4. Interpret results.
59
Q

Describe the process of screening for an unknown mutation.

A
  1. Determination of carrier status of normal animal
  2. Backcross to a known homozygous recessive: if any affected offspring are born then the test animal is a carrier.
  3. Backcross to a known heterozygous carrier: if any affected offspring are born, the test animal is a carrier.

Animal Management Y2 (16 decks)

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