Exam 1 Flashcards
(165 cards)
1
Q
- Positive Eugenics
A
Encourage the ablest and healthiest people to have more children
2
Q
- Negative Eugenics
A
Advocated culling the least able from the breeding population to preserve humanity’s fitness;
Defectives” should be prevented from breeding through means like:
Compulsory sterilization
Custody in asylums
3
Q
Discuss at least one flaw in the research methods in the scientific origins of eugenics
A
Complex behavioral traits are hard to study with simple mendelian laws of inheritance and complex traits are often poorly defined
There was a tendency to treat the traits as if they were a single entity (not polygenic)
4
Q
Describe one of the eugenics laws enacted in the United States
A
Virginia’s Eugenical Sterilization Act: Based off of Laughlin’s Model Eugenical Sterilization Law that authorized sterilization of “socially inadequte” (those who were “feeblminded”, “insane”, “epileptic”, “diseased”, “blind/deaf”, etc
5
Q
Describe Ernst Rüdin’s role in psychiatric genetics and Nazi Germany’s policies
A
Rüdin’s work argued for the sterilization of people suffering from mental illness and was based on the idea that metal illnesses were inherited by passing down a single gene. When Hitler took office in Germany he pushed through: The Law to Prevent Hereditarily Sick Offspring by using forced sterilization, this idea was based on Rudin’s work in psychopathology.
6
Q
Describe one finding suggesting significant environmental influences on IQ
A
Murray’s assertion that it is hard to raise the IQs of disadvantaged children leaves out the most important data point. Adoption from a poor family into a better-off one is associated with IQ gains of 12 to 18 points
7
Q
Describe the two major theories of inheritance before Mendel
A
One parent (the male) contributes the majority of the inherited features in a child (ex: homonuculus theory)
Blended inheritance: the parental genes become mixed and are forever changed in the child (like if you were to mix blue and yellow to get green)
8
Q
Describe the reasons Mendel was so successful
A
- He did matings within the same species
- Studied non-complex traits
- Looked at simple, quantatative traits
- Looked at traits with dominance
9
Q
What are true breeding plants?
A
A true-breeding plant is one where it will always yield the same result when crossed with the same kind of plant
Ex: Peas with smooth seeds bred with other peas with smooth seeds the offspring will always have smooth seeds
10
Q
What are F1 and F2 generations
A
F1: The F1 generation are the offspring of two true-breeding parents (two smooth seed true-breeding parents have offspring with smooth seeds)
F2: The offspring of the F1 generation
11
Q
Mendel’s results when he crossed true breeding parents (resulting in the F1 generation), and self-fertilized the F1 generation (resulting in the F2 generation)
A
F1 generation with true-breeding parents: smooth(P1) x smooth(P2) = all offspring are smooth seeded
F2 generation: ¾ of offspring were smooth, ¼ were wrinkled
Showed us that there is dominance of a particular trait (smooth seeds in this case)
12
Q
The two hypotheses Mendel came up with to explain his results from F1 and F2 generations
A
- Each parent has two alleles (can be either same or different), but transmits only one of the alleles to each offspring
- When an individual’s two alleles are different, one allele could dominate the other.
13
Q
What is the difference between heterozygous and homozygous
A
Homozygous: Individuals with two same alleles
Heterozygous: Individuals with two different alleles
14
Q
What is the difference between dominant and recessive
A
Dominant: A particular allele dominates (controls phenotypic expression) over another
Recessive: Need two copies of the allele for a phenotypic trait to be expressed
15
Q
Name and describe Mendel’s first law of heredity
A
Law of segregation: States that there are two genes for a trait, offspring inherit one allele (copy of the gene) from each parent because the parent alleles separate during gamete formation and during fertilization the offspring has one allele from one parent and another allele from the other parent. One of these alleles can dominate another
16
Q
Describe disease and inheritance pattern of Huntington’s Disease
A
Huntington’s disease: neurodegenerative disorder that general shows up in adulthood and affects males/females of all ethnic backgrounds
Inheritance pattern: Caused by a single gene and dominant allele (H), affected individuals are generally heterozygous because the H allele is rare it is unlikely both parents will pass on the dominant allele. Because the parent generally is heterozygous and usually only one parent is affected the risk for offspring with one affected parent is 50%
17
Q
Describe disease and inheritance pattern of PKU
A
PKU: Affected individuals have a deficiency in Phenylalanine hydroxylase that results in the buildup of phenylalanine in the body. Too much build up can lead to toxic levels and cause brain damage (results in mental disability, movement disorders, etc.) Can be treated though by screening for it before birth and then avoiding foods containing phenylalanine (proteins, diet soda, etc.)
Inheritance pattern: Caused by a single gene and is recessive (requires two copies of the allele. If two parents are unaffected (but are carriers, meaning they each have a recessive allele) then the chance their offspring will get the disease is 25%
18
Q
What is the difference between monohybrid cross and dihybrid cross?
A
Monohybrid: Mating between individuals that differ in only one trait (ex: mating between wrinkled and smooth seeded parents)
Dihybrid: Mating between individuals who differ in two traits (ex: yellow vs. green seed color AND wrinkled vs. smooth seeds)
19
Q
Mendel’s results from his dihybrid crosses (F1 and F2 generations)
A
YYSS (true breeding dominant parent for yellow, smooth seeds) x yyss (true breeding recessive parent for green, wrinkled seeds)— NOTE: for the true-breeding parents yellow always come with smooth and green always with wrinkled
F1 generation: YySs (all offspring were yellow, smooth)
F2 generation: YySs x YySs: result was new recombinant types not seen in parental generation (new types: green, smooth and yellow, wrinkled)
20
Q
What is the difference between parental types and recombinant types
A
Parental types: phenotypic combinations seen in the true breeding parents (i.e., the yellow smooth peas and green wrinkled peas)
Recombinant types: new phenotypic combinations not seen before in the true breeding parents (i.e., the yellow wrinkled peas and the green smooth peas)
21
Q
Name and describe Mendel’s second law of heredity
A
Law of independent assortment – During gamete formation, different pairs of alleles segregate independently of each other.
22
Q
Know the fact that Mendel’s second law of heredity is violated when genes for two traits are very close together on the same chromosomes
A
If genes influencing different (or same traits) are close together on the chromosome then during recombination they may always end up crossing over together just do to the fact they are close together.
23
Q
How likely are recombinant types when two traits are caused by genes close together on the same chromosome?
A
Unlikely; if the inheritance pattern of a trait violates Mendel’s 2nd law then it is evidence that the genes are close on the same chromosome
24
Q
What is phenotypic frequency and how to calculate
A
Proportion of individuals in a population that have a particular phenotype (e.g., a particular disease)
To calculate:need to know whether a disease is dominant or recessive, then need to know the number of individuals in a sample, then need to know the genotypes of those individuals, count the # of individuals who have a particular trait given their genotype
25
What is genotypic frequency and how to calculate
Proportion of total individuals in a population that have a particular genotype
To calculate: need to know whether a disease is dominant or recessive, then need to know the number of individuals in a sample, then need to know the genotypes of those individuals, count the # of individuals who have a particular genotype
26
What is allelic frequency and how to calculate
Proportion of all copies of a gene in a population that are of a given allele type
To calculate: need to know number of people in population, need to know their genotypes, then need to breakdown into each allele, then multiply each allele by the number of copies of each allele there is and divide by total number of alleles in population
EX:
R : 8 people with RR (16 copies) and
2 people with Rr (2 copies)
= 8+8+2 = 18/32 = .56
r : 2 people with Rr (2 copies) and
6 people with rr (12 copies)
= 2+6+6 = 14/32 = .44
27
Know why we’re interested in the Hardy-Weinberg equilibrium
Many inherited diseases result from disease-causing alleles that are recessive to the wildtype alleles (non-disease-causing allele) whose allele whose frequency is greater than 1%
28
Know the basics of Hardy-Weinberg equilibrium – i.e., what p and q stand for, the two formulas (p+q = 1; p2 + 2pq + q2 = 1), and the frequency of different kinds of individuals (homozygous dominant, heterozygotes, and homozygous recessive)
* p= the frequency of dominant alleles; the chance that any particular egg or sperm has the dominant allele
* q= the frequency of recessive alleles; the chance that any particular egg or sperm has the recessive allele
* p+q= 1
* p2= frequency of offspring with two dominant alleles (homozygous dominant)
* q2 = frequency of offspring with two recessive alleles (homozygous recessive)
* 2pq= Probability of the heterozygous genotype (heterozygous)
* p2+ 2pq + q2= 1 ; The frequency of the offspring genotypes
* EX: Frequency of PKU individuals (homozygous recessive) in the population is 1/10,000 = q2= .0001
o this means q = .01
o p + q = 1
o p + .01 = 1
o p = .99
o Frequency of individuals who are carriers for PKU (heterozygous)= 2pq = 2 ( .99) (.01) = .02
29
What is incomplete dominance
The F1 generation resembles neither purebred parent (blended)
Ex:
True-breeding red flowers (AA) x True-breeding white flowers (aa)
F1 generation= Aa= Pink flowers (intermediate phenotype for heterozygotes)
Easy way to tell ratios is that the phenotypic ratio is an exact reflection of genotypic ratios
30
What is codominance?
F1 generation resembles both true-breeding parents
Ex: Lentils
True-breeding spotted lentils (SS) x True-breeding dotted lentils(ss)
F1 heterozygous generation= Ss= both spots and dots for the heterozygote intermediate genotype
31
Know the examples involving more than 1 allele we discussed (e.g., human blood types) and be able to complete Punnett squares involving these traits
Blood type has 3 possible alleles (BUT EACH PERSON ONLY HAS TWO ALLELES)
IA= dominant over i and results in adding sugar A to blood (blood type A)
IB= dominant over i and results in adding sugar B to blood (blood type B)
i= recessive and results in carrying neither sugar (blood type O, universal donors)
IAIB= codominant such that an individual with this genotype carries both A and B sugar (can receive either A or B blood type)
32
What are mutations? What is the difference between monomorphic and polymorphic?
Mutations – chance alterations of genetic material; Arise spontaneously in nature and lead to new alleles. Once they occur in gametes they then get inherited and passed down to other generations
* Monomorphic: Gene with one wild-type allele
* Polymorphic: Gene with more than one wild-type allele
33
What is pleiotropy?
Pleiotropy – phenomenon of a single gene determining a number of distinct and seemingly unrelated characteristics (Ex: Mendel noticed that specific seed coat colors are always associated with specific flower colors)
34
What is epistasis?
Example with Labrador coat colors
Epistasis – A gene interaction in which the effects of an allele at one gene hides the effects of alleles at another gene
Bb or BB: B dominant to b = black coat
bb: recessive = brown coat
Second gene (E or e)
EE or Ee= no effect
ee= overrides what is happening at b/B locus and = yellow
This is an example of recessive epistasis
35
What is complete penetrance? Incomplete penetrance?
Complete penetrance: 100% of population with a particular genotype show the expected phenotype
Ex: Huntington’s disease
Incomplete penetrance - < 100% of population with a particular genotype show the expected phenotype
Ex: Retinoblastoma (75% penetrance)
36
What is expressivity? Varying expressivity? Unvarying expressivity?
Expressivity: the degree or intensity with which a particular genotype is expressed in a phenotype
Varying:the degree or intensity with which a particular genotype is expressed in a phenotype differs in the population
Ex: retinoblastoma
Unvarying:the degree or intensity with which a particular genotype is expressed in a phenotype does not differ in the population
Ex: pea color
37
What are the factors that contribute to expressivity?
Modifier genes: alter the phenotypes produced by the alleles of other genes
Ex: mouse tail lengths. Mutant allele makes tail short, but not all mice have same amount of shortness because there is a modifier gene that affects how short it gets
Environmental effects on expressivity:
Ex: alleles that account for dark coat color in cats; “Siamese” allele for coat color produces enzyme that doesn’t function at normal body temperature, so dark pigment only happens in cat’s extremities (dark extremities and light body color)
Chance and expressivity: Some people are less affected just by chance (ex: retinoblastoma only affecting one eye)
38
What is a phenocopy?
A change in phenotype arising from environmental agents that mimics the effects of a mutation in a gene (not heritable because they do not arise from a change in a gene)
Ex: A sedative called “thalidomide” was being taken by pregnant women and mimicking the effects of a rare dominant trait called phocomelia where limb development is interrupted
39
Know what the frequency of the disorder is for males and females when the allele frequency is known
Ex: If a disorder is X-linked recessive and frequency of disorder= 10%
Expected frequency of disorder in males= 10%, same as disease frequency
Expected frequency of disorder in females= 10% squared (0.1)2= frequency of disorder squared because they need two alleles that are disordered to get the disease (females have two X chromosomes)
40
What is nondisjunction? Be able to discuss the examples of nondisjunction we discussed in class
Nondisjunction: failure to apportion the chromosomes equally
Ex 1: Down syndrome- 3 copies of chromosome 21 and is consider trisomy (where there are three, instead of two, copies of a chromosome)
Ex 2: Turner syndrome- example of monsomy (only one copy of a chromosome) and occurs when only one sex chromosome is inherited (XO)--- leads to learning problems, people are unusually short, infertile
Ex 3: Klinfelter syndrome: another example of trisomy, but in the sex chromosomes so inherits an extra X (XXY)—leads to learning problems, unsually long limbs, infertile
41
Why is Down syndrome is more common for children of older mothers?
More common because all female eggs are present at birth, but undergo the last step of cell division each month. Older eggs are then more prone to errors in cell division
42
What are expanded triplet repeats? What are examples of this discussed in class?
A special form of mutation that involves repeat sequences of DNA
Ex 1: Huntington’s disease: most cases are caused by a repeat sequence of three nucleotide bases, CAG, on chromosome 4
CAG codes for glutamine and this repeat sequence leads to a greater number of glutamines in the middle of a protein
Most people have a repeat sequence at this location of about 11-34
People with Huntington’s disease have a repeat of more than 40 copies of this CAG sequence
Note: expanded triplet repeats are unstable and often increase in subsequent generations make the disease progression faster or earlier onset
Ex 2: Fragile X syndrome: Expanded triplet repeat on the X chromosome that makes the X chromosome fragile. It is more common in males than females, but doesn’t follow traditional X-linkage inheritance patterns because it is do to an expanded triplet repeat, which are often unstable and increase in subsequent generations
A parent with the fragile X expanded triplet repeat may not show signs of mental retardation because they may only have 40-200 copies of triplet repeat.
However, because it is unstable the repeats increase in subsequent generations so offspring may get 200+ copies of this repeat sequence and that does manifest as mental retardation
43
What is premutation?
When a parent produces expanded repeats that don’t cause an effect (like mental retardation) in their offspring, but then that repeat is unstable and expands further in the offspring’s offspring (and they then have mental retardation)
44
What is a morbidity risk estimate?
Morbidity risk estimate: Chance of being affected (having a particular disease) during entire lifetime
45
Be aware of examples of complex traits we discussed in class
* Schizophrenia
* Pea size
* Height
46
How can multiple genes acting in accordance to Mendel’s laws lead to a complex trait?
Each gene of a polygenic trait is inherited according to Mendel’s laws
But the alleles don’t act in a totally dominant or recessive manner and are additive, where each allele contributes something to the phenotype.
This then leads to continuous variation at the phenotypic level (many varying degrees of a phenotype due to each alleles contribution)
47
Know the basics of a liability-threshold model
Assumes that the risk for a disorder is normally distributed (due to the additive nature of complex traits), however, most people are below the “threshold” where the disorder occurs
Those who are related to someone with a disorder increase their liability and thus move their distribution curve further to the right
48
What is DNA?
Deoxyribonucleic acid and is responsible for heredity
49
What are the four bases in DNA?
* C: Cytosine
* G: Guanine
* T: Thymine
* A: Adenine
50
Which bases pair together in DNA?
* C and G
* T and A
51
What are the two functions of DNA?
* Replication
* Synthesis of proteins
52
Know the steps involved in the replication of DNA
* Occurs during cell division
* DNA double-helix unzips forming two single strands with exposed nucleotide bases
* The now exposed nucleotide bases (that were previously paired) attract their complementary pairs (ex: C attracts G and T attracts A)
* Now you have two double-stranded DNA helices where there was previously one
53
Know the steps involved in the synthesis of proteins
* Transcription: turns DNA into RNA or mRNA (messenger RNA)
* Translation: mRNA turns the sequence into amino acids and then into proteins
54
Know steps involved in transcription
DNA double helix unzips and then copying occurs to mRNA which is similar to DNA replication, however, the mRNA strand formed is single-stranded and any thymine gets substituted for a uracil (so all adenines are paired with uracils on mRNA)
Newly formed mRNA strand now leaves the nucleus and enters the cytoplasm and binds with ribosomes where the translation process will occurs
55
Know steps involved in translation
a. In the ribosome there is a strand of mRNA that is waiting for transfer RNA (tRNA) to come in
b. tRNA enters the cell and has a complementary code of three nucleotides that bonds to the complementary code of three nucelotides in the mRNA
i. The three nucleotide bases on mRNA are called codons
ii. The complementary three nucleotide bases on tRNA are called anticodons
c. The tRNA sequence also has its associated amino acid
d. The ribsome then moves along the mRNA strand while tRNA pieces (and associated amino acids) come in to bond to the complementary bases in the mRNA
e. Once the tRNA anticodon bonds to the complementary mRNA codon it adds its associated amino acid to a growing chain of polypetides (these are what make up a protein)
i. Proteins are generally made up of polypetide chains that are 100-1000 amino acids in length
ii. The order of the amino acids determines the shape and function of the protein
56
Know the “central dogma” of molecular genetics
* Genetic information flows from DNA to mRNA to protein
* A gene is a DNA segment that is a few thousand to several million base pairs in length
* A DNA molecule contains a linear message of four repeating bases (A, T, C, G)
57
Understand the difference between DNA, mRNA, and tRNA
* DNA: Contains 4 bases: C, G and A,T ; double-stranded; encodes the sequence to make up amino acids
* mRNA: messenger RNA; 4 bases C, G and A, U (uracil); single-stranded, mRNA leaves the nucleus where DNA lives to allow for translation of genetic sequence into amino acids
* tRNA: transfer RNA; only three bases and these three bases are complementary to three on mRNA; transfer RNA contains its associated amino acid that it then transfers to the ribosome (where the amino acids are assembled into polypeptide chains that make up proteins)
58
Know what codon and anticodon are
Codon: all 20 amino acids are specified by a list of three sequential mRNA bases (A, C, U, G). This list of three sequential mRNA bases are what is called a codon
Anticodon: each mRNA codon attracts its complementary anticodon that is on the tRNA. This pairing allows the amino acid associated to detach from the tRNA and add to the polypeptide chain
59
Know the difference between introns and exons
Introns: Pieces of a gene that are transcribed into mRNA, but are spliced out before the mRNA leaves the nucleus
o Vary in length from 50-20,000 base pairs
o Can help regulate transcription of other genes
Exons: Pieces of a gene that are transcribed and spliced together after removing introns. The exons are the code of mRNA that exits the nucleus and gets translated into proteins
o Usually only a few hundred base pairs
60
Know what alternative splicing is and its role in biological complexity
During the process of removing introns and reassembling exons the combinations of mRNA code can vary and be spliced into different combinations of exons which in turn translate into different amino acid combinations and different proteins
This is necessary to form the complexity of proteins we have with just four nucleotide bases
61
Know the general characteristics of the human genome – e.g., number of base pairs, number of genes, number of base pairs in genes
* Number of base pairs in human genome: 3.2 billion
* Number of genes in human genome: 21-25 thousand
* 1000 to 2 million base pairs per gene
62
Know what the three main surprises of the human genome project were
1. The majority of the human genome consists of non-coding regions (don’t code for proteins)
2. Humans have a small number of protein-coding genes (21-25 thousand compared with rice that may have 37 thousand)
3. We share 98% of our DNA with chimpanzees and 99.9% with other humans
63
Know the different kinds of genetic variation across individuals
1. Sequence: Different nucleotide at a given position in the genome
2. Structure
Insertions (small number of DNA bases that has been inserted)
Deletions (small number of DNA bases that has been deleted)
Copy number variants (>1000 bases that has been deleted or inserted)
Duplication: piece of DNA that is duplicated
3. Organization
Inversion: a piece of chromosome that is inverted
Translocation: pieces of non-homologous chromosomes (like chromosome 4 and 20) that exchange info so a piece of chromosome 4 is now on 20 and vice versa
64
Know what mutations are
Occurs when there is a mistake in copying DNA and result in polymorphisms that lead to different alleles of a gene
65
Know the characteristics of human chromosomes (e.g., number, sex vs. autosomes, parts of chromosomes) and those of other species (examples we discussed)
Humans have 23 chromosomes
22 autosomal chromosomes
2 sex chromosomes (an X and X or X and Y)
Each chromosome has a short p arm and and longer q arm that are joined at the centromere
Number of chromosomes varies widely with fruit flies having 4, dogs 39, carp 52
66
Know the difference between gametes and somatic cells
* Gametes: sex cells, like egg and sperm
* Somatic cells: all other cells
67
Know the difference between mitosis and meiosis
* Mitosis: This is the cell division process that occurs in somatic cells
* Meiosis: This is the cell division process that occurs in gametes
68
Know the definition of epigenetics
Changes in gene expression that are not due to alterations in the DNA sequence itself
69
1st type of epigenetic modification
X-inactivation: In females who have 2 X chromosomes, only one X is active in all daughter cells of somatic cells (also called Lyonization)
o Occurs early in development
o One X in each daughter cell randomly inactivates (can be either mother’s or father’s in each cell)
Ex: Tortoise shell/calico cats
o Tortoise shell: gene for fur color is on the X chromosome and one allele gives black fur, the other orange fur. Due to X-inactivation though female cats will have some somatic cells that have the orange fur color allele because it is on one X and the black fur color allele in other somatic cells that have the other X active
o Calico cats: same gene fur color thing that is going on with tortoise shell cats, but with an additional autosomal condition called piebalding that leads to unpigmented (white) fur.
70
2nd type of epigenetic modification
Imprinting: some genes are only expressed if they are inherited from the mother, others are only expressed if inherited from the father
Ex:
o Angelman syndrome occurs when there is a deletion on chromosome 15 that is inherited from the mother (severe mental disability, awkward gait, frequent inappropriate laughter)
o Prader-Willi syndrome occurs when there is a deletion on chromosome 15 that is inherited from the father (caused overeating/obesity, short stature, temper outbursts, depression)
71
Know what DNA methylation is
o A way to regulate gene expression
o Controls by altering access to a gene during transcription
o A gene that has a CpG island that is unmethylated allows access by transcription factors
o A gene that has a CpG island that is methylated is blocks access to the gene by transcription factors and thus does not get transcribed
72
Animal Model- Selection studies
o Selection studies: If a trait is heritable can breed selectively for it
o In lab experiments you typically select strains that have high levels of a trait of interest, strains that have low levels of a trait of interest, and maintain an unselected control line
o General results interpretations:
If lines between the two selectively bred strains (like high and low activity mice) continue to separate with each generation then it implies many genes contribute to a trait
If lines separate within a few generations and then don’t continue to separate then it implies there are only a couple of genes responsible for a trait
73
Animal Model- Inbred strain studies
Inbred strains: brothers and sisters are mated for at least 20 generations; this makes the animals in the inbred strains virtually clones of each other (like monozygotic twins)
o Results of inbred strain studies:
Because individuals in the inbred strain are genetically identical, differences within a strain are likely due to environmental influences
Mothering effects: Do two crosses of inbred strains—C57BL/6 mother x BALB/c father and the reverse. If differences in a phenotype are found then may indicate mothering effect
* Can further separate prenatal vs. postnatal mothering effects by cross fostering to see which mice exhibit the phenotype more
Gene by Environment interaction: environments of inbred strains can be manipulated to examine gene x environment interaction
* Ex: inbred strains can differ across differing environments like C57/BAL6 mice being more active than BALB/c mice in Colorado, but not in Florida
74
Know what a diallel design is
Compares several inbred strains and all possible F1 crosses between them
Ex: In crosses between various inbred strains results found that the F1 crosses tended to be the average between the scores of the parents
75
Be familiar with examples of animal models of human behavior or disorder discussed in class (e.g., anxiety, alcohol sensitivity, autism, schizophrenia).
o Anxiety
Examine heritability of fearfulness/anxiety in mice
* Choose mice to selectively breed that consists of mice with low anxiety= mice with high activity in open field, elevated plus mazes
* Choose another strain to selectively breed that consists of mice with high anxiety= mice with low activity in open field and elevated plus mazes
* Also maintain a control line that was not selectively bred
* Interpretation: if trait is heritable then the high anxiety mice should get more fearful (less active) over time, the mice with low anxiety should get less fearful (more active) over time, and the control line should stay relatively stable
o Alcohol sensitivity
Selected traits (short-sleep, long-sleep) mice to mimic alcohol sensitivity phenotype in humans.
Placed mice on their backs (which they don’t like) after low-doses of alcohol
Short-sleep mice were more active at low doses then long-sleep
This meant short-sleep mice were more like an alcoholic phenotype because the alcohol did not affect them as much
o Schizophrenia
Mice that were selected for schizophrenia trait were compared to those that didn’t have that trait
* Trait= startle response
Those without schizophrenia would have a startle response initially but then would get desensitized to it
Those with schizophrenia would have the same level of startle again and again
o Autism
With autism-like trait: mice not interested in other mice
Without autism-like trait: mice very interested in other mice
76
Know what induced mutations are and be able to give examples of induced mutations discussed in class.
o Induced mutation: using chemicals or other means to induce a mutation in a gene to see how it affects complex trait behavior
o Ex: sluggish (very slow) fruit flies
o Ex: “drop dead” fruit flies that fly/walk normal for a few days and then suddenly drop dead
77
What is gene targeting?
A process by which a gene is changed in a specific way to alter its function
Ex: knock-out
78
What is a knock-out?
Deleting key DNA sequences that prevent the gene from being transcribed
Has a limitation that the gene is inactivated throughout animal’s entire lifespan so the body begins to compensate
79
What is a conditional knock-out
Addresses limitation of “knock-out” by changing gene expression, not gene transcription. So can turn on and off a gene during the lifespan to better understand function
80
What is transgenics?
When mutated gene is transferred from one species to another
81
What is gene-silencing
Use double-stranded RNA to “knock down” expression of the gene that shares its sequence
82
What is RNA interference (RNAi) or small interfering RNAs (siRNA)?
Degrades complementary RNA transcripts and prevents a gene from being transcribed/translated
83
What is synteny homology
Parts of mouse chromosomes have the same genes in the same order as parts of human chromosomes and can be used to then identify gene or areas of interest for a particular trait
84
Know what markers are, be able to discuss an example of a marker, and how markers are useful for linkage
A marker: A piece of DNA that varies and can used to determine location of a particular gene
Ex: DNA polymorphisms
o A single base that differs between individuals
o Can be used to track chromosomal location of genes because we know where these markers occur and on which chromosome
85
Know the difference between the two QTL studies discussed in class: selection and recombinant inbred strain studies.
o QTL Studies (selection): Start out by breeding lines of mice that differ significantly in the behavior one is interested in, then see if there are any genetic differences between extreme behavioral groups (e.g., most fearful and least fearful) in the F2 generation
o QTL Studies (recombinant inbred strains): Start out by breeding lines of mice that differ genetically, then see if there are any behavioral differences between the strains
86
Be able to describe the results of Leonard Heston’s study on schizophrenia.
o Tried to challenge the previous assumption that schizophrenia was due to early parent interactions (where mother heaped lots of love at one moment and then was very distant in another)
o Method: Interview/track 47 adopted-away offspring of mothers who were hospitalized with schizophrenia and compare to control group of matched adoptees with no family history of mental illness
o Found that 5/47 adopted-away adoptees of schizophrenic mothers developed schizophrenia (similar to rates of children who were reared by schizophrenic mothers) and 0/47 of the adoptees with no family history of mental illness developed schizophrenia
o Conclusion: genetics play a large role in schizophrenia
87
Know the difference between the adoptees’ study method and the adoptees’ family method
o Adoptees’ study method: Heston’s study was an example of an adoptees’ study method where you start with affected parents and then look at incidence of disease in their adopted-away offspring
o Adoptees’ family method: Start with adoptees who themselves are affected and compares to adoptees who aren’t affected then looks at incidence of disorder in their respective adoptive and biological families
88
Know the methodological issues involved in adoption studies. Be able to discuss why each issue is a potential problem in adoption studies.
1. decrease in frequency: adoption studies are harder because adoptions have become less frequent
2. representativeness: biological and adoptive families may not be representative of general population (ex: biological family may be lower SES or younger in age then the respective adoptive families which tend to have higher SES and be older)
3. prenatal environment: correlations in a trait between biological parents and an adopted-away child may reflect prenatal environment rather than genetic similarity
4. selective placement: in past has been evidence that adoptive children are placed in adoptive families that are more similar to their biological families (ex: children of bright biological parents get placed with adoptive families that are also bright)
89
Be able to discuss the two ways to test for prenatal effects in adoption studies.
o Compare adoptee’s biological half siblings related through mother and biological half-siblings related through father
o Can compare correlations of a trait to birth mothers and birth fathers and see if there are differences
90
Know the difference between MZ twin pairs and DZ twin pairs. Be able to describe what is meant that MZ twin pairs share 100% genetic similarity and DZ twin pairs share 50% genetic similarity
o Monzygotic twin pairs: share 100% of genetic information because they come from a single fertilized egg that separates
o Dizygotic twin pairs: share 50% of genetic information because they come from two separate fertilized eggs
91
Know the concept of IBD and the # of alleles shared IBD at a particular locus (and across the genome) for parent-offspring pairs, MZ pairs, DZ pairs, and full sibling pairs.
o IBD for parent-offspring: generally one allele IBD
o IBD for MZ pairs: always 2 alleles IBD
o IBD for DZ pairs/full siblings: can be 0 alleles (25% of time), 1 allele IBD (50%) of time, or 2 alleles IBD (25%) of time
92
Know the methods of zygosity determination
o If twins differ in any DNA marker they must be DZ twins
o Can ask parents how similar their twins are across physical traits- this is about 90% accurate compared to using DNA markers
93
Know the methodological issues involved in twin studies. Be able to discuss why each issue is a potential problem in twin studies
1. Equal environments assumption: assumes that environmentally caused similarity is roughly the same for both types of twins reared in the same family
May not be the case because MZ twins may be dressed alike, put in same classes, etc. compared to DZ twins.
Errors would result in inflated heritability estimates
Has been tested and generally traits follow the equal environments assumption
2. Generalizability: twins may not be representative of population for several reasons
Prenatal environment: more hostile for twins who are sharing resources
Postnatal environment: parents are more taxed having to take care of two kids rather than one
3. Assumption of no epistasis
Assume that coefficient for genetic relatedness of MZ twins= 1
Assume that coefficient for genetic relatedness of DZ twins is 0.5 for additive traits, .25 for dominant traits, and varies for epistatic traits
Then we will over inflate genetic estimates if it turns out MZ=1 and DZ= 0.25, rather than 0.5 as assumed
94
Know and be able to describe at least two ways to test the equal environments assumption
1. Examine in twins who were mislabeled as MZ and are really DZ and see if environment is actually more similar
2. Ask parents directly if they treat their MZ twins similarly or differently
95
Understand the concept of effect size and statistical significance
o Effect size: the extent to which individual differences for the trait in the population can be accounted for by genetic differences among individuals
Ex: effect size for untreated PKU on cognitive ability is huge on the individual, but because it occurs in 1/10,000 individuals the affect of PKU on cognitive ability in the entire population is small
o Statistical significance: the reliability of the effect
96
Understand what h2, c2, and e2 mean
o h2= heritability: the proportion of phenotypic variance that can be accounted for by genetic differences among individuals
o c2= magnitude of shared environmental traits: influences that are shared by family members and that tend to make members more similar on a particular trait
o e2= magnitude of nonshared environmental traits/includes measurement of error: influences that make members dissimilar on a particular trait
97
Understand the difference between shared environmental influences and nonshared environmental influences
o Shared environment for family members: by definition 100% shared environment
o Nonshared environment for family members: by definition 0% nonshared environment for family members
98
Know the h2, c2, and e2 equations and how to calculate given a correlation between family members
h2 + c2 + e2 = 1 (NOTE WHEN CALCULATING h2 c2 e2 you are not setting equation equal to 1, you are setting it equal to the r value you are given!!!)
h2 = 2(rMZ – rDZ)
c2 = rMZ – h2
e2 = 1 – rMZ
99
Understand the difference between narrow sense heritability and broad sense heritability
o Narrow sense heritability: only the proportion of phenotypic variance explained by additive genetic effects
o Broad sense heritability: refers to all sources of genetic variance, whether the genes operate in an additive manner or not; includes dominance and epistasis
Note: If the MZ correlation is more than twice the DZ correlation it is evidence for dominance or epistasis
100
Know why it is not possible to test both c2 and d2 in a study including only MZ twins reared together and DZ twins reared together
Cannot examine c2 and d2 together with just MZ and DZ twin information because both c2 and d2 depend on the same information (correlation between MZ and DZ)
o Evidence for d2 if DZ correlation is less than half MZ correlation BUT
o Also evidence for c2 if DZ correlation is greater than half MZ correlation
101
Know the reasons for the lack of studies testing models including epistatic interactions in human studies?
Too complex.. Would have to examine relationship between all genetic loci involved in a trait which is hard to do given there are so many loci involved in complex traits
102
Be able to describe at least two ways that the extended twin family designs are superior to the classical twin design
o Can give information about the etiology of shared environmental influences
o Can give information about assortative mating(which generally inflates our estimation of A)
103
Know what heritability does and does not mean and what it does and does not imply
o Heritability: refers to the genetic contribution to individual differences (variance in a population), not to the phenotype of a single individual
Ex: 90% height heritability does not mean you grew to 90% of your height because of heritable aspects and 10% to environment, it means that within a population the differences in height can be 90% explained by heritability
o Heritability is also for a particular population at a particular time
o Heritability estimates can change over time
Ex: Heritability of reading comprehension increased from 20th to 21st century as the environmental influences became less variable
o High heritability of a trait does not mean we should only try to improve performance for those with high levels of that trait
104
Be able to describe the pattern of results consistent with sibling competition/rater effects vs. sibling imitation/cooperation effects
o Sibling competition/rater effects: both dominance/non-additivity and sibling contrast effects will lead to very low DZ correlations so may seem like higher dominance/non-additive effects when in reality there is a competition effect; However, variances change with MZ variances being lower than DZ variances when contrast effect
o Sibling imitation/cooperation effects: imitation effects are hard to distinguish from those of the shared environment and leads to higher DZ correlations; Variance increases most for most closely related individuals (higher variance in MZ twins)
105
Discuss the conclusions suggested when there are sex differences in twin correlations
o Sex differences in twin correlations point towards sex effects in either genetics or environment
o If the correlation for opposite-sex twin pairs is significantly less than that for same-sex DZ twin pairs – it may be concluded that different genetic or shared environmental factors are operating in males and females
106
Discuss at least two questions that can be addressed using a Cholesky bivariate model?
o Are there genetic/environmental influences specific to the second variable?
o To what degree is the covariation between the two variables due to common genetic, shared environmental, and nonshared environmental influences?
107
Interpret the table from the Slutske et al. (2002) paper
o Behavioral undercontrol accounted for about 88% of the genetic covariation between alcohol dependence and conduct disorder
108
What is the difference between an independent pathway model and a common pathway model?
o Independent pathway model: a single genetic factor, a single shared environmental factor or dominance factor, and a single nonshared environmental factor explain the covariation among the four measures
o Common pathway model: a restricted submodel of the independent pathway model; represented as a latent phenotype, which is partitioned into genetic, shared environmental/dominance, and nonshared environmental influences
109
How are the common factor model and the simplex model similar? How are they different?
Both take developmental trajectories into account (how things change with age)
o Simplex model – measures taken closer in time will be more highly correlated than measures further apart in time
o Common factor model – assumes no specific correlation pattern as a function of time
110
What were the major conclusions of the review by Kendler and Baker (2007) regarding genetic influences on measures of the environment?
o Traditional models of psychiatric epidemiology – often assume that relationship between individuals and their environment is unidirectional, from environment to person
o However, a person's genetics can influence their environment (ex: self-select for different environments)
111
A) Define the three types of gene-environment correlation, B) describe an example of each type of gene-environment correlation, and C) name the possible source(s) of environmental influence for each type of gene-environment correlation.
1. Passive GE correlation: children passively inherit from their parents family environments that are correlated with their genetic propensities (for passive GE you are always thinking about the parents---parents are providing genes and environment
a. Example: musically gifted children--- parents provide genes and likely the environment to help them flourish
b. Source of environmental influence: parents and siblings
2. Evocative GE correlation: when individuals evoke reactions from other people on the basis of their genetic propensities; talking about how a child’s genetic propensities influence their environment
a. Example: Musically gifted children may be singled out for special opportunities at school
b. Source of environmental influence: anybody
3. Active GE correlation: individuals select, modify, construct, or reconstruct experiences that are correlated with their genetic propensities
a. Example: adoptees become more genetically similar to their biological parents over time when they have the opportunity to choose their own environment, rather than the environment chosen by adoptive parents
b. Source of environmental influence: anybody/anything
112
Discuss the three methods used to detect genotype-environment correlation. Describe each method, and discuss which type of gene-environment correlation can be detected using this method.
1. compare correlations between environmental measures and traits in nonadoptive and adoptive families
a. correlation between measure of family environment and a psychological trait of children could be environmental in origin, or genetic factors might also contribute to the correlation
b. this indirect genetic path between family environment and children’s traits is not present, because adoptive parents are not genetically related to their adopted children
c. Genetic contribution to the covariation between family environment and children’s traits is implied if the correlation is greater in nonadoptive families than in adoptive families
d. reflects passive genotype-environment correlation, because children in nonadoptive families passively inherit from their parents both genes and environment that are correlated with the trait
2. involves correlations between biological parents’ traits and adoptive families’ environment; addresses evocative and active genotype-environment correlation
3. bivariate genetic analysis of the correlation between an environmental measure and a trait
a. estimates the extent to which genetic effects on one measure overlap with genetic effects on another measure
b. Evocative or active genotype-environment correlation – implied if genetic effects on an environmental measure overlap with genetic effects on a trait measure
113
Neiderhiser et al. (2004) used a child-based design and a parent-based design to examine the genetics of parenting. Describe the difference between a child-based design and a parent-based design. What is the limitation of using just a child-based design?
o Environment vs. outcome= child-based design; children’s genes are the unit of measurement; these are indirect effects of environment (we don’t know the genes that cause maltreatment)
o Another type of design: parent-based; we are trying to get more information on parents’ genes by looking at MZ and DZ parents, not children. Idea being that if there is passive rGE, MZ parents will be more similar in their parenting than DZ parents
114
Explain why passive GE correlation is evidenced the presence of shared environmental influences in a child-based design and genetic influences in a parent-based design.
Explain why nonpassive GE correlation is evidenced by the presence of genetic influences in a child-based design and shared and/or nonshared environmental influences in a parent-based design.
o Passive GE correlation:
The parent’s genes for being difficult is leading to both negative parenting environment and child’s behavior
If there is passive rGE then there would be shared environmental influences in a child-based design. The key to passive rGE is that the parent’s genes are influencing the environment. This means that parent’s are going to influence the child’s environment in the same way, no matter whether the child is easy or difficult to parent. It is independent of characteristics of the child. Can’t distinguish what is shared environment (c2) and what is passive rGE
Passive rGE in a parent-based design would show up as genetic influences. Parents who are more genetically similar are going to parent more similarly
o Nonpassive GE correlation
If evidence for non-passive rGE would show up as genetic influences in a child-based design
Greater genetic similarity of siblings – leads to greater similarity in parenting received
Little or no evidence of genetic influences on parenting in a parent-based design, and just shared/nonshared environmental influences
115
Define genotype-environment interaction and discuss at least two examples of early findings of GE interaction discussed in class
Definition: The effect of an environmental variable on phenotype depends on genotype -OR-
The effect of a genotype on phenotype depends on the environment
Example of early findings 1: Bohman 1996 study that found criminal behavior of adopted children was highest when both the biological and adoptive parents also had a criminal background; About 10% (13% to 24%) higher in this group compared to adoptees where neither had a criminal background
Example of early findings 2: Heath (1989 and 1998) found that heritability of alcoholism and depression was higher in unmarried women compared to married women.
116
Describe the four mechanisms by which social context may moderate gene expression and provide an example of each
1.) contextual triggering
2.) social context as compensation
3.) social context as social control
4.) social context as enhancement
1.) contextual triggering: You need both the specific environmental context and the specific geneotype for a specific phenotype; Ex: Kendler co-twin study where the group who was at highest risk for developing depression was the MZ twins where one twin was affected (genetic risk) and the other twin had a stressful life event (environmental risk)
2.) social context as compensation: in the absence of environmental risk, all groups defined by differing levels of genetic susceptibility exhibited the same propensity a particular phenotype; Ex: Kendler co-twin study where twin groups who did not have stressful life events saw almost equal levels of MDD between affected and unaffected MZ twin groups
3.) social context as social control : In settings marked by high levels of social control – heritability decreases
In contexts marked by low levels of social control – heritability increases; Ex: Dunne study that showed heritability of age at first intercourse among Australian youth was lower in individuals born between 1922 and 1952; heritability of age at first intercourse increased in individuals in the later-born group (1952-1965) as there was more availability of birth control
4.) social context as enhancement: Among individuals without a genetic diathesis – social context can also interact with genes to facilitate higher levels of developmental functioning; Ex: Heath 1985 study that showed as educational reforms were put into place in Norway (making it easier to continue education) then heritability of educational attainment increased from 0.41 in 1915-1939 cohort to 0.74 in 1940-1949 cohort
117
Dick et al. (2001) suggest that past results indicate that “protective” environments like marriage or low stress levels can reduce the impact of genetic predispositions to various clinical problems. Describe the results of Rowe et al. (1999), which goes against this hypothesis, and their explanation of the results.
Study of aggression in children with the environmental effect being family warmth. They found that in "protective" environments (like those with more family warmth) a genetic predisposition was required for aggression (so did not reduce impact, but in fact was opposite effect)
Whereas in environments that weren't desirable (like low family warmth levels) then the environmental influences actually became more influential in development of aggression.
118
Dick et al. (2001) examined a series of socioregional variables hypothesized to be more proximal to the interaction effect found in their previous study examining the urban/rural distinction. What were their conclusions regarding the evidence of GE interaction with these variables?
Variables examined/the variable's GE interaction
1.) Percent of young adults; As percentage of young adults increased – heritability increased and magnitude of shared environmental influences decreased
2.) % of migration into and out of different regions; As % of migration increased – heritability increased and magnitude of shared environmental influences decreased
3.) per capita expenditure on alcohol in each region relative to the mean amount spent in the entire country; no significant moderating effect on unique environmental factors (E); moderating effect of genetic influences did not reach significance (A); but significant moderation of shared environmental influences (C)
OVERALL CONCLUSIONS: Communities characterized by more young adult role models and greater social mobility – allow for increased expression of genetic dispositions
119
Discuss how Dick et al. (2001) tested an alternative explanation of their results, and the conclusions.
One alternative explanation provided to explain difference in rural/urban AUD rates was that it's possible there is an GxE interaction where families who are more likely to drink are selecting urban environments where alcohol is more available; Looked at families in urban vs. rural environments and found no difference in drinking rates meaning that it is not likely that one's genetic propensity to drink is what is causing them to choose an urban vs. rural environment.
120
Turkheimer et al. (2005) suggest that the measured C design cannot provide estimates of environmental effects unbiased by genetic factors. Discuss Turkheimer’s arguments supporting this statement. (Include in this discussion Turkheimer et al.’s findings on the VA 30,000 dataset).
Most studies began concluding that environmental factors shared by siblings do not influence children at all because
in many instances variables are only measured at the level of the family, and therefore can only vary between families, not within them.
Ex: Flawed logic that parental SES is an obligatory shared environmental variable between twins (the parent SES is the same for both twins at the same time) then it is a purely environmental variable, but we know that there are genetic factors that also contribute to the parental SES and thus the parental SES has a genetic component as well as a shared environmental component
Ex with twin data example – VA 30,000 dataset: Looking at educational attainment in twins with the obligatory shared environmental factor of parental divorce. Found that the estimate does not control for genetic or environmental effects on either divorce or educational attainment.
121
Purcell and Koenen (2005) also suggest that environmental mediation models are logically flawed when examining obligatory-shared environments.
Discuss Purcell and Koenen’s arguments supporting this statement. (Include in this discussion Purcell and Koenen’s proof-of-principle simulation results).
Arguments supporting:
A.) Because a twin study of the variable ‘parental SES’ would clearly yield a heritability estimate of zero
due to the fact that for an obligatory-shared environmental variable: rMZ = rDZ = 1.0 (so all variance would be attributed to the shared environment) which we know is not true, parental SES is in part genetic.
B.) Even if we were to consider parental SES to be fully genetic we can't measure that in a child-based twin design because in this design we are assuming heritablity of parental genotypes to be the same between DZ and MZ twins
Proof-of-principle simulation results:
Did two simulations
1) one where they looked at family-wide environment as a function of parental phenotypes (and expected there to be environmental mediation about 5% of the time)
2) one where the loked at family-wide environment as a function of parental genotypes (where they expected no environmental mediation)
They found that in the 1st simulation there was environmental mediation 96% of the time and in the 2nd it was present 97% of the time
122
Describe the logic of the discordant-twin design
Asking if differences between twin 1 and twin 2 are due to the effect of an exposure variable ; trying to get at causation
If we see no differences at the individual level, between discordant MZ twins, or between discordant DZ twins then it implies that the effect of the exposure entirely determines the phenotype.
If we see no differences between discordant MZ twins, some between discordant DZ twins, and and most at the individual level then that implies phenotype is entirely determined by genetic factors.
If we see some differences between discordant MZs, more between discordant DZs, and most between individuals then it implies there are genetic effects and effects of exposure on phenotype.
123
Describe the two factors that increases/decreases power in the phenotypic causation model
Decrease: measurement error
Increase: when there are differences in causes of variation in one trait versus another (e.g., trait 1 has low heritability and trait 2 has high heritability)
124
Describe the logic of the Children of Twins (CoT) design
If the outcome variable is genetic in cause then the children of the twin parent 1 and children of twin parent 2 will be affected
If the outcome variable is environmental in cause then only the children of the twin parent who also is also affected will exhibit the outcome (ex: twin parent 1 uses MJ and so does their child, but twin parent 2 doesn't use MJ and thus their child doesn't either)
125
Describe how Burt et al. (2009) examined the association between divorce and delinquency, and the major conclusions of this study.
Aimed to better explain the relationship between divorce and deliquency. Wanted to better rule out passive rGE by looking at differences in deliquency between both adopted and non-adopted youth and youth who differed on whether their parents got divorced prior to them or during their lifetime
Evidence for genetic effects (and the possibly passive rGE in disguise):
If biological children had same rates of delinquency if divorce happened before they were born and after they were born; and if bio children had higher rates than adopted children
Evidence for environmental effects:
If bio children had higher deliquency if divorce happened during their lifetime, would also not see a difference between adopted and non-adopted youth
Findings: indicate evidence for enviornmental effects. Deliquency was higher in all groups if divorce happened during their lifetime
Post-hoc analyses showed that remarriage moderated this affect such that individuals whose parents got divorced and then remarried showed less deliquency likely because there was more parent monitoring
126
Which of the following designs are appropriate for studying obligatory shared influences? (discordant twin design/co-twin control, phenotypic causation model, children of twins, adoption study)
Children of twins because the other one's necessitate that the twins differ on a variable, not that they have that variable in common
127
How much genetic difference is there when you compare individuals of the same race in the variable part of human DNA?
A lot (95.5% of the total differences in variable human DNA)
128
Relative to the amount of genetic differences that exist between individuals of the same race, how much extra genetic difference exists between individuals of different racial groups?
A little (4.5% of the total differences in variable human DNA)
129
Describe at least two distinguishing features of linkage vs. association
Linkage purpose: To test if a particular gene associated with a disease is at a particular locus
Association: To test if a particular allele at a locus is associated with a disorder
1.)Linkage: A few hundred markers could be used to scan the whole genome
Association: Many thousands of DNA markers needed for a genome scan
2.) Linkage: Can only detect QTLs of large effect sizes
Association: Statistical power to detect QTLs of small effect size
130
Describe how case control studies are conducted
They look for allelic frequency differences between affected individuals (cases) and unaffected individuals (controls)
Has a problem if the controls and cases are not matched by ethnicity because there are differences in allele frequencies between ethnic groups due to population stratification
131
Describe what population stratification is and at least two methods used to address population stratification
Population stratification occurs when people are of different ethnicities because over time different ethnic groups developed different allele frequencies
Can be addressed by:
1.) controlling for it using things like principal components analysis or
2.) using family-based design where you are looking at individuals who are of the same ethnicity
132
Describe how the transmission disequilibrium test is conducted
Uses trios consisting of an affected individual and their biological parents
Idea is that the affected child had to have inherited the alleles from their parents and rests on the idea that if affected individuals are inheriting "affected/case" alleles more often than chance it is evidence for association
133
Define linkage disequilibrium and what an r2 of 0 or 1 means.
Linkage disequilibrium refers to the non-random association of loci (loci that are very close to each other on a chromosome are more likely to assort together during recombination)
r^2 of 1= complete linkage disequlibrium (genotype at one locus will always correspond with a particular genotype at another locus)
r^2 of 0= complete equilibrium (genotype at one locus provides no information about the genotype at another locus)
134
Describe at least two problems that have led to lack of replications in candidate gene studies in the literature
1.) low statistical power, were expecting genes with larger effect sizes than actually exist for most traits
2.) Differences in MAF between populations leading to issues when population stratification is not controlled for (too many false positives)
135
Describe at least one of the problems noted by Duncan and Keller regarding candidate GxE studies in the literature in detail.
There was evidence for publication bias: As replication attempts found positive, vs negative, vs. pure negative (as it got more negative) the sample size had increased; This means that people who were showing a true negative finding were being necessitated to have much larger sample sizes than those who were replicating positive findings. So it was harder to publish negative findings (needed large sample sizes)
136
Describe what the genome-wide association study is
Association between SNPs across the genome and phenotype of interest; For association studies you need a SNP that is much closer to the actual SNP so you need more SNPs than you do in linkage studies
137
Describe two possible outcomes when there is a significant association with a SNP in a GWAS
* The SNP is directly associated with the trait
* The SNP is in high LD with a SNP that is directly associated
138
Describe how population stratification is addressed in GWAS
Generally there is use of PCA that compares genome- wide allele frequencies to those of HapMap ethnic groups and then the principal components are treated as covariates in a model
139
Describe how multiple testing is addressed in GWAS
Bonferroni correction: Divide the alpha of 0.05 by the # of statistical tests performed (ex: if one used 1 million SNPs then 0.05/1million= 5x 10^-8
140
Describe how meta-analyses of multiple GWAS can be accomplished when different studies use different SNP marker sets
Imputation: exploit known LD patterns and haplotype frequencies from HapMap or 1000 Genomes project to estimate genotypes for SNPs not directly genotyped in the study
141
Discuss the main results of the Schizophrenia Psychiatric GWAS Consortium’s mega-analysis (2011)
Found that 7 loci for schizophrenia reached genome-wide significance including 5 newly identified loci
142
Describe what Yang et al.’s GCTA (genome-wide complex trait analysis) does
GCTA is the software that uses the statistical method GREML (genomic based restricted maximum likelihood) estimations which aim to address the issue of highly stringent p-values that are used when doing multiple testing correction.
It fits all SNPs simultaneously instead of testing for a relationship one-by-one and effects are treated as random and the variance explained is using all SNPs together. It is used to estimate the relatedness between individuals and then exclude the close relatives from analyses.
143
What were Yang et al.’s major conclusions regarding height in humans?
When estimating variance using all SNPs Yang found that 45% of the variance in human height could be explained. This was in contrast to methods that were previously just using SNPs that met genome-wide significance (only 5% of variance was explained with these methods)
144
Discuss what LDSC (LD Score Regression) does, and when it can be useful
Allows for the estimation of SNP heritability when only summary statistics are available. This is in contrast to GCTA which requires an individual's actual genotype.
145
Discuss at least two factors that will impact the validity of LDSC results
1.) Match between LD scores from reference population and target population used for GWAS needed
2.)Underestimates heritability if the trait is not polygenic
146
Describe the difference between gene analysis and gene-set analysis
In gene analysis: seeing if all markers are related to a gene and if that is related to the phenotype of interest
gene-set analysis : individual genes are aggregated to groups of genes sharing certain biological, functional, or other characteristics
147
Describe why the competitive gene-set analysis is preferable over the self-contained gene-set analysis
Self-contained – Whether the genes in a gene-set are jointly associated with the phenotype of interest
Competitive – Whether the genes in a gene-set are more strongly associated with the phenotype of interest than other genes; Competitive is preferred because it is more stringent and reduces the rate of type 1 error
148
Describe what stratified LDSC (partitioned heritability) accomplishes
Partitions heritability from GWAS summary statistics while accounting for linked markers
149
Summarize Doust et al.’s gene analysis, gene-set analysis, and stratified LDSC (partitioned heritability) results
Was looking at heritability of dyslexia in a case-control design where there were many cases (51,000)
Gene analysis results: Found 173 associated genes after bonferroni correction
Gene-set analysis: Broad analysis looking at all known pathways didn't find any biologically relevant pathways after B. Correction. However, when doing a focused gene-set analysis and just 3 tests found that FOXP2 was associated
LDSC: Found enrichment in genes found in the frontal cortex and ACC
150
Describe the major steps of polygenic risk score analysis
1.) Identify a discovery sample (ex: PGC) and utilize GWAS summary statistics for trait of interest
2.) Identify a target sample (like a collected set of individuals in your own study) and generate GWAS summary statistics for trait of interest
3.) Ensure there is no overlap in individuals in discovery and target and then see what SNPs are overlapped in both samples
4.)Use something like LDSC to remove SNPs that are correlated (in high LD)
5.)Limit this list further by SNPs that meet your significance threshold
6.)Generate a PGS for your individuals in your target sample
7.) Regress this PGS over the phenotype of interest for each individual
151
Describe what the polygenic score is
1.) Get a count of # of risk alleles present for a person
2.) Weight these based on the results from the discovery sample using something like a log odds ratio (to account for effect size of the risk allele)
152
Describe how sample size is related to the p value threshold that maximizes predictive ability
In large sample sizes a lower p-value allows one to maximize predictive ability (lower rate of type 1 error)
In smaller/under-powered sample sizes the variance explained is higher when all SNPs are included
153
Describe the major findings from GWAS studies of educational attainment in 2013 vs. 2018 vs. 2022
Each year saw an increase in sample size and correspondingly found an increase in significant loci/% variance explained. However, not enough information to show dominance effects
154
Discuss at least two problems with using polygenic scores to select embryos discussed by Turley et al. (2021)
The probability that parents have exactly two viable embryos, one in the top and one in the bottom quintile of polygenic scores, is less than 3% so the difference between education attainment is unlikely to be realized within families
155
Describe the logic of bivariate GREML and cross-trait LDSC
1.) Genetic variation estimated when case-case pairs and control-control
pairs more similar across the genome than case-control pairs
2.) Positive correlation when cases of one disorder show higher genetic
similarity to cases of the other disorder than their own controls
3.) Negative correlation possible if cases of one disorder show less
genetic similarity to cases of the other disorder than their own controls
4.) Zero correlation: If genome-wide relation between cases of one disorder is the same with cases as with the controls of another disorder
156
Describe what a QSNP is
Identifies SNPs that communicate risk for an individual phenotype and is thus a test of heterogeneity such that it assumes that the SNP does not act solely through a common factor
157
Discuss two major benefits of genomic SEM
1.) Calculate more predictive polygenic scores (Consistently
outperform those from univariate GWAS)
2.)Allows one to identify loci that cause divergence between traits
158
Know how genetic nurture influences heritability using GREML vs. from twin studies
1.) Genetic nurture effects on GREML heritability estimates: in GREML there is no distinction between direct effects and genetic nurture so genetic nurture may bias heritability estimates
2.) Genetic nurture effects from twin studies: Because genetic nurture is a type of passive rGE this gets counted in the c2 estimates in twin studies and thus gets cancelled out when doing the heritability estimates
159
Name three potential sources of influences other than direct genetic effects in GWAS associations from samples of unrelated individuals
1.) assortative mating
2.)genetic nurture
3.) population stratification
160
Discuss how the direct and population effect of the polygenic index (PGI) was examined in Okbay et al. (2022)
1.) Direct effect: measured by regressing the individual's phenotype on the individual's polygenic index (PGI) while controlling for both parents' PGI
2.)Population effect: regressing the individual's phenotype on the individual's PGI WITHOUT controlling for parents' PGI
161
Discuss the main findings of the C-Reactive Protein Coronary Heart Disease Genetics Consortium (CCGC; 2011) study and its main conclusions
What was the study: CRP may be associated with CHD because it may be a causal factor or CRP and CHD may have common factors that affect both of them. This study used MR to attempt to randomize participants (although you can't do that directly like assigning people to have higher or lower CRP, because of Mendel's law of random assortment can treat whether someone inherits an allele as random)
Findings: Genetic variations in CRP gene is associated with CRP levels, however, the presence of a risk allele is not associated with increased risk for CHD (log odds ratio near 1). Thus CRP does not cause CHD.
162
Discuss the three assumptions for instrumental variables in MR
1.) Relevance assumption – Instrumental variables are associated with the risk factor of interest
2.) Independence assumption – Instrumental variables share no common cause with the outcome (ex of violation: ancestry may affect ALDH2 and blood pressure separately)
3.)Exclusion restriction assumption – Instrumental variables do not affect the outcome except through the risk factor (ex of violation: horizontal pleiotropy like ALDH2 may affect alcohol consumption and tobacco consumption leading to increase in blood pressure through both)
163
Discuss how paternal age, de novo mutations, and psychiatric disorders are associated
De novo mutations are mutations that occur during cell division and that were not inherited from the parents. With increasing paternal age there is an increase in de novo mutations. De novo mutations have been associated with psychiatric disorders like schizophrenia and autism
164
Discuss the differences between h2, h2SNP, and h2WGS for height and BMI
h2: this is the heritability estimate from family-based studies and for height it is about 0.70- 0.85, for BMI it is around 0.4-0.6
h2SNP: this is the heritability estimate from GWAS studies using common SNPs and is 0.48 for height and 0.24 for BMI
h2WGS: this is the heritability estimate when sequencing the entire genome (so allows for inclusion of rare variants; it is 0.70 for height (close to family based studies) and 0.29 for BMI (lower than family-based studies)
165
Define copy number variants and discuss one example of a CNV that influences a psychiatric disorder
Definition: A large DNA segment (1kB or larger) that has a variable number of copies
Example: Deletion of 3Mb section to DNA on 22q11.2 which leads to Velo-cardiofacial syndrome, but 20-30% of individuals with this disorder develop schizophrenia like symptoms
