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GED L16 Notes Flashcards

1
Q

Contrast Mendelian & Complex traits

A 
Mendelian Vs. Complex Traits
-	Mendelian:
	Single gene (monogenic)
High penetrance
	Predictable inheritance
	Simple relationship 
Genotype & phenotype
                   Eg. Cystic fibrosis
-	Complex:
(Multifactoral / quantitative)
	Multiple genes (polygenic) 
Low penetrance
	Familial clustering 
	Unpredictable inheritance
	Complex relationship 
Genotype & phenotype
	Strongly influenced by environment (Multifactoral)
	Prevalence -> ~600/1000
Eg. Alzheimers, Autism, Crohn’s, Athsma, Cleft lip, Coronary Heart Disease, Diabetes, Neural Tube defects / spina bifida
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2
Q

Describe Mendelian traits

A 
-	Mendelian:
	Single gene (monogenic)
High penetrance
	Predictable inheritance
	Simple relationship 
Genotype & phenotype
                   Eg. Cystic fibrosis
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3
Q

Describe complex traits

A 
-	Complex:
(Multifactoral / quantitative)
	Multiple genes (polygenic) 
Low penetrance
	Familial clustering 
	Unpredictable inheritance
	Complex relationship 
Genotype & phenotype
	Strongly influenced by environment (Multifactoral)
	Prevalence -> ~600/1000
Eg. Alzheimers, Autism, Crohn’s, Athsma, Cleft lip, Coronary Heart Disease, Diabetes, Neural Tube defects / spina bifida
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4
Q

Give an example illustrating genetic & environmental influences on complex traits

A

• Genetic & Environmental Influences -> Complex Traits:
Eg. Type 2 Diabtetes
~ 6-8% population
Genetic: One affected parent -> 15%
Both parents affected -> 75%
Environment: BMI >30 -> 20%

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5
Q

Describe analysis of quantitative, complex traits

A 
-	Quantitative Traits:
>Quantitative / continuous traits
	Complex traits 
Eg. Height, Blood Pressure, Serum, Cholesterol, BMI, Crop yield
                     Mean:
	Arithmetic Average (Centre of Distribution)
                     Variance:
	Spread of values around mean.
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6
Q

What are quantitative traits?

A
  • Quantitative Traits:

>Quantitative / continuous traits

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7
Q

Give an example of quantitative complex traits?

A

Eg. Height, Blood Pressure, Serum, Cholesterol, BMI, Crop yield

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8
Q

What is the mean?

A

 Mean:

 Arithmetic Average (Centre of Distribution)

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9
Q

What is the variance?

A

 Variance:

 Spread of values around mean.

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10
Q

What kind of graph / chart can be used to analyse quantitative, complex traits?

A

Normal distribution curve

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11
Q

Describe analysis of discontinuous, complex traits

A
  • Discontinuous Traits:
    >Distcontinuous / Discrete traits
    -> Disease / Trait is either present or not present
     Complex traits
     Threshold Model:
     Underlying continuous liability (genetic & environmental factors)
    » Threshold above which disease / trait is present
     Families with incr. risk of trait / disease
    » Distribution of liability shifted towards threshold.
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12
Q

What are discontinuous traits?

A
  • Discontinuous Traits:
    >Distcontinuous / Discrete traits
    -> Disease / Trait is either present or not present
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13
Q

What is used to analyse discontinuous, complex traits & describe how?

A

 Threshold Model:
 Underlying continuous liability (genetic & environmental factors)
» Threshold above which disease / trait is present
 Families with incr. risk of trait / disease
» Distribution of liability shifted towards threshold.

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14
Q

What is used to analyse discontinuous, complex traits?

A

Threshold model

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15
Q

Describe the way in which discontinuous & continuous traits are analysed

A
  • Continuous & discontinuous traits

 Same underlying aetiology.

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16
Q

What is the polygene hypothesis?

A

Genetic basis of complex traits

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17
Q

How is the polygene hypothesis measured?

A
  • Nature vs. nurture:

 Proportion of phenotypic variation in quantitative trait that is genetic or environmental.

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18
Q

What is nature vs. nurture in regards to the polygene hypothesis?

A
  • Nature vs. nurture:

 Proportion of phenotypic variation in quantitative trait that is genetic or environmental.

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19
Q

What is the equation used to find the total phenotypic variance?

A

 VP = VG + VE
Total Phenotypic Variance = Genetic Variance + Environmental Variance
» VP can also be shown as VT

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20
Q

Describe how to calculate the total phenotypic variance?

A

 VP = VG + VE
Total Phenotypic Variance = Genetic Variance + Environmental Variance
» VP can also be shown as VT

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21
Q

What is heritability?

A

 Proportion of phenotypic variance due to genes

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22
Q

Describe heritability

A

 Heritability
 Proportion of phenotypic variance due to genes
 H2 = VG / VP
 H2 -> value between 0 – 1
 High heritability
&raquo_space; Genetic differences in population
&raquo_space; High proportion of phenotypic variation
-» Easier to identify genetic variants associated with trait than using
traits with low heritability.
 Estimating Heritability:
» Twin Heritabilty Studies:
1. Identical (monozygotic twins ; MZ)
Share same environment & all alleles
Relatedness ; r = 1
2. Non-identical (dizygotic twins ; DZ)
Share same environment & half of their alleles
Relatedness ; r = 0.5
 If MZ twins resemble more than DZ twins ; genes contribute to
variation in trait .
> Assumes equal environmental effects
&raquo_space; Twin Concordance Studies:
 Concordance:
Probability a twin is affected by a particular trait if their twin is
affected.
> Concordance = 1.0 -> Other twin always affected
= 0.6 -> 60% chance other twin affected.
 Difference in concordance between MZ & DZ twins used to estimate hertitability.
Eg. MZ concordance = 0.50 (50%) ; DZ = 0.08 (8%)
Heritability = 0.85
 Heritability Estimates in Humans:
 Autism -> 0.4 - 0.9
 BMI -> 0.5 – 0.9
 Crohn’s -> 0.75
 Height -> 0.6 – 0.8
 IQ -> 0.3 – 0.8
 T1 Diabetes -> 0.9
 T2 Diabetes -> 0.4 – 0.8

                             High heritability does not mean genetic determination. 
                             Differences between groups with high heritability for a trait are not  
                                 the result of genetic differences.
23
Q

What is the equation used to calculate heritability?

A

H2 = VG / VP

24
Q

Describe how the value of heritability can be interpreted

A 
 H2 = VG / VP
 H2 -> value between 0 – 1
 High heritability 
   >> Genetic differences in population 
   >> High proportion of phenotypic variation
25
Q

Describe the way in which it is easier to identify genetic variants associated with a trait and why.

A

By using population with higher heritability value because this indicates genetic differences in population are due to a high proportion of genetic variation

26
Q

Describe the use of twin studies to study heritability

A

 Estimating Heritability:
» Twin Heritabilty Studies:
1. Identical (monozygotic twins ; MZ)
Share same environment & all alleles
Relatedness ; r = 1
2. Non-identical (dizygotic twins ; DZ)
Share same environment & half of their alleles
Relatedness ; r = 0.5
 If MZ twins resemble more than DZ twins ; genes contribute to
variation in trait .
> Assumes equal environmental effects
&raquo_space; Twin Concordance Studies:
 Concordance:
Probability a twin is affected by a particular trait if their twin is
affected.
> Concordance = 1.0 -> Other twin always affected
= 0.6 -> 60% chance other twin affected.
 Difference in concordance between MZ & DZ twins used to estimate hertitability.
Eg. MZ concordance = 0.50 (50%) ; DZ = 0.08 (8%)
Heritability = 0.85

27
Q

Describe identical twins

A
  1. Identical (monozygotic twins ; MZ)
    Share same environment & all alleles
    Relatedness ; r = 1
28
Q

Describe non-identical twins

A
  1. Non-identical (dizygotic twins ; DZ)
    Share same environment & half of their alleles
    Relatedness ; r = 0.5
29
Q

What does it mean if MZ twins resemble more than DZ twins?

A

 If MZ twins resemble more than DZ twins ; genes contribute to
variation in trait .
> Assumes equal environmental effects

30
Q

Describe twin concordance studies

A

> > Twin Concordance Studies:
 Concordance:
Probability a twin is affected by a particular trait if their twin is
affected.
> Concordance = 1.0 -> Other twin always affected
= 0.6 -> 60% chance other twin affected.
 Difference in concordance between MZ & DZ twins used to estimate hertitability.
Eg. MZ concordance = 0.50 (50%) ; DZ = 0.08 (8%)
Heritability = 0.85

31
Q

How is heritability estimated using twins?

A

Twin concordance studies

 Difference in concordance between MZ & DZ twins used to estimate hertitability.

32
Q

What does a concordance value of 1 mean in relation to concordance twin studies?

A

Concordance = 1.0 -> Other twin always affected by a particular trait if their twin is
affected.

33
Q

Give examples of Heritability Estimates of Humans in regards to different factors

A 
 Heritability Estimates in Humans:
	Autism -> 0.4 - 0.9
	BMI -> 0.5 – 0.9
	Crohn’s -> 0.75
	Height -> 0.6 – 0.8
	IQ -> 0.3 – 0.8
	T1 Diabetes -> 0.9 
	T2 Diabetes -> 0.4 – 0.8
34
Q

What are two important corrections to misconceptions surrounding heritability?

A

 High heritability does not mean genetic determination.
 Differences between groups with high heritability for a trait are not
the result of genetic differences.

35
Q

What are two important facts to note regarding heritability?

A

 High heritability does not mean genetic determination.
 Differences between groups with high heritability for a trait are not
the result of genetic differences.

36
Q

Describe genetic variance & human Disease

A

• Genetic Variance & Human Disease
 Common Disease – Rare Variant Hypothesis (CD-RV)
 Common Disease – Common Variant Hypothesis (CD-CV)
 Disease-Associated Alleles should be common:
- Late onset diseases
 Little effect on fitness (Weak Purifying Selection)
- Advantageous / Neutral alleles in the past
 May confer disease susceptibility in modern societies

37
Q

Describe why Disease-associated alleles should be common

A
  • Little effect on fitness (Weak Purifying Selection)
  • Advantageous / Neutral alleles in the past
     May confer disease susceptibility in modern societies
  • Disease causing alleles
     Maintained at high frequency by balancing selection.
38
Q

Describe Genome wide Association studies

A

• Genome Wide Association Studies (GWAS)
 Population – level approach to disease gene mapping.
 Identification of SNPs:
- Cases: C Allele -> 62%
T Allele -> 38%
- Controls: C Allele -> 49%
T Allele -> 51%
- Odds ratio for C allele:
 (Odds of disease with C) / (Odds of disease with T)
= [A/B] / [C/D] = [62/49] / [38/51]
= 1.7
- Allele odds ratio ; >1.0 -> Allele gives Higher risk of disease
< 1.0 -> Allele is protective
 Process:
 Coloured dots represent single SNP on DNA chip
 Odds ratio (OR) for each variant calculated
 Probability (P) OR is significantly higher in cases compared to controls i
illustrated -> Manhattan Plot.
 Manhattan plot
Illustrates the probability -> Odds ratio of an allele is significantly higher in cases compared to controls.
Variants of high significant associations with trait visible as skyscrapers.
 Haplotype Blocks:
Groups of closely linked haplotypes (SNPS on the same chromosome) which are inherited together.
 Each haplotype block -> defined by small no. tag SNPs
 Linkage Disequilibrium:
Co-inheritance of SNPs in a haplotype block

39
Q

What are Genome wide association studies used for

A

 Population – level approach to disease gene mapping.

 Identification of SNPs -> Odds ratio of allele

40
Q

Describe how SNPs can be identified using genome wide association studies

A

 Identification of SNPs:
- Cases: C Allele -> 62%
T Allele -> 38%
- Controls: C Allele -> 49%
T Allele -> 51%
- Odds ratio for C allele:
 (Odds of disease with C) / (Odds of disease with T)
= [A/B] / [C/D] = [62/49] / [38/51]
= 1.7
- Allele odds ratio ; >1.0 -> Allele gives Higher risk of disease
< 1.0 -> Allele is protective

41
Q

Describe what an allele odds ratio of >1 means

A

> 1.0 -> Allele gives Higher risk of disease

42
Q

Describe what an allele odds ratio of <1 means

A

< 1.0 -> Allele is protective

43
Q

Describe the process in which genome wide association studies are used

A

 Process:
 Coloured dots represent single SNP on DNA chip
 Odds ratio (OR) for each variant calculated
 Probability (P) OR is significantly higher in cases compared to controls i
illustrated -> Manhattan Plot.

44
Q

What does a Manhattan plot illustrate & how is info analysed?

A

 Manhattan plot
Illustrates the probability -> Odds ratio of an allele is significantly higher in cases compared to controls.
Variants of high significant associations with trait visible as skyscrapers.

45
Q

What is a Manhattan plot used for?

A

Genome wide association studies for identification of SNPs

46
Q

What are haplotype blocks?

A

 Haplotype Blocks:
Groups of closely linked haplotypes (SNPS on the same chromosome) which are inherited together.
 Each haplotype block -> defined by small no. tag SNPs

47
Q

What is linkage disequilibrium

A

 Linkage Disequilibrium:

Co-inheritance of SNPs in a haplotype block

48
Q

Describe haplotype blocks

A

 Haplotype Blocks:
Groups of closely linked haplotypes (SNPS on the same chromosome) which are inherited together.
 Each haplotype block -> defined by small no. tag SNPs

49
Q

Describe linakge disequilibrium

A

• Linkage Disequilibrium:
- Linkage Equilibrium:
 Each haplotype present in expected frequency based on allele frequency calculations.
Eg. EF(AB) = F(A) x F(B)
- Linkage Disequilibrium:
 Some haplotyoes present at higher / lower expected frequencies based on allele frequency calculations.
- Causative Role in Disease
 SNP 1 -> Linkage disequilibrium with SNP 2
»But is not causative
 Association between SNP 1 & disease illustrates haplotype block carrying causative agent (SNP)

50
Q

Describe what linkage equilibrium is

A
  • Linkage Equilibrium:
     Each haplotype present in expected frequency based on allele frequency calculations.
    Eg. EF(AB) = F(A) x F(B)
51
Q

Describe linkage disequilibrium in regards to calculation of haplotype frequency

A
  • Linkage Disequilibrium:

 Some haplotyoes present at higher / lower expected frequencies based on allele frequency calculations.

52
Q

Describe causative role in disease in relation to linkage disequilibrium

A
  • Causative Role in Disease
     SNP 1 -> Linkage disequilibrium with SNP 2
    »But is not causative
     Association between SNP 1 & disease illustrates haplotype block carrying causative agent (SNP)
53
Q

Describe the role / contributions of GWAS in human disease genetics

A

• Role of GWAS in Human Disease Genetics:
- Majority of disease-associated variants have small incr. risk of disease.
- >90% variants present in non-coding DNA
»Difficult to identify causal variant.
- Generate new biological hypotheses about causes of disease
Eg. Inflammatory Bowel Disease & Autophagy.
- Studies of rare SNPs & other types of DNA seq. variation - eg. Copy Number Variants (CNV) – necessary to further understand complex traits.

Genes Evo & Dev Y1 (23 decks)

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