Lecture 21 Flashcards

1
Q

vContrast polygenic traits and monogenic traits.

A

Polygenic traits: are dependent on contributions from multiple genes

Monogenic traits: traits dependen on the contribution of a single gene(eg mendels pea plants)

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2
Q

Is it possible for a monogenic trait to be quantitative? Explain your reasoning.

A

Quantitative traits: are measurable traits represented by a number and are typically polygeneic traits

however monogenic traits can be quantitative

e.g: Mendel with the tall and short flowers

while the trait is dependent on one gene the trait can still exist on a continuum of length

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3
Q

Is it possible for a quantitative trait to give rise to discreet variation? Explain your reasoning.

A

Quantative traits usually do not give rise to discreet variation however if there is a threshold expression of a gene that gives rise to a phenotype there will be discrete variation/

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4
Q

Define multi-factorial trait

A

There is an environmental influence as well as a genetic infleunce from multiple genes ( e.g height)

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5
Q

What is the mid-parent value for a trait? Why is the mid-parent value very often an underestimate
of offspring height in humans?

A

the mid-parent value is the average of the mother and father’s traits.

it is often an underestimate of offspring height in humans as it does not consider the environmental effect of dietary quality on height.

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6
Q

Distinguish between major and minor effect genes and how these ideas relate to the production of
different eye colors in humans.

A

major genes: genotype
two major genes with a large effect on eye color where the genotype at each gene determine if your eye color in the blue/green category or the brown eye category

minor genes: have small effect and in the case of eye color they alter the ey color very slightly

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7
Q

If a trait is determined by 5 additive genes, each with two alternative alleles, how many phenotypic categories would be expected in the absence of an environmental effect on phenotype.

A

11 phenotypic categories ( 0,1,2,3,4,5,6,7,8,9,10) gene dosages for the alternative alleles
2n+1 gene dosages(given two alleles per genen in population)

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8
Q

Suppose height were determined by three genes and there was no environmental effect, and each
gene had “tall” alleles and “short” alleles.

You observe and individual with a phenotype indicating the
presence of exactly 4 “tall” alleles. Can you infer their genotype? Is the same true if they had the
phenotype indicative of 6 “tall” alleles?

A

you would not be able to infer much about the genotype as there is no 1:1 correspondence between: phenotype and genotype.

However, if you had 6 tall alleles you would be able to tell that all genes are homozygous dominant.

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9
Q

Why is variation in F1s very low when parental lines that differ in a quantitative trait are crossed?

A

Edward East: took two pure breeding lines of tobacco plant with a short corolla and long corolla parental ( petals)
- corolla length is a quantitative trait
- there is also an environmental variation

crossing the two parents results in the f1s being an intermediate for the trait ( heterozygous) and mostly genetically uniform as they are directly receiving all alleletypes from each parent( with each parent being homozygous at almost all loci)

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10
Q

Why does variance increase when the F1s are crossed to one another

A

when you cross f1s you result in f2s with increased phenotypic variance due to the introduction of independent assortment and the increase in genes.

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11
Q

If the variance does not increase substantially in the F2s, is that indicative of many or few genes
underlying a trait? Explain using the ftle-Wright equation

A

n= D^2/8(s,2-s,1)
n=lower bound of the genes influencing. a trait

D+ difference in phenotype between parental

s1=variance in f1s
s2: variance in f2s

if f2 variance is not substantially larger than f1 variance the denominator would be smaller making the value n the or number of genes larger

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12
Q

Define genetic liability.

A

the innate likelihood of developing a disease based on a genotype

certain alleles increase genetic liability by passing a threshold for diseases/vs non disease–> contribute to. continuous variation of genetic liability

if you cross two heterozygotes each has three genes that contribute to a disease

you will have offspring with 0-6 alleles possibly contributing to the disease however in terms of genetic liability there can be a threshold where offspring need 5 or more liability alleles across the 3 loci to be affected

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13
Q

Give an example of when VG = 0. Give an example when VE = 0. Why are these cases useful?

A

vp=vg+ve
In a clonal or inbred population, the inviduals are genetically uniform therefore VG=0 and all of the variance is driven by environment (never in a natural population), this allows us to measure the variance caused by environmental conditions

In a tightly control lab experiment ( Ve is zero) allowing vg to be measured.

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14
Q

Which component of VG is narrow sense heritability concerned with?

A

Broad sense Heritability = Vg/vp

narrow sense = Va/vp =h2

narrow sense heritability= va/vp–> important for determining how responsive things are to selection and allows us to select for given traits within a population that are desirable

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15
Q

If a trait has low heritability, does that mean the trait is not determined by genes? Explain.

A

Heritability = The proportion of
phenotypic variation attributable to
genetic variation

High heritability of a trait= a lot of the phenotypic variance is driven by differences in genotype

low heritability: phenotypic variances has more influence from environmental variation as opposed to genetic variation

It does NOT mean it’s NOT determined by genes there is just more environmental factors influencing the phenotypic variation.

e.g: deviations from having two eyes indifferent species is due to environment

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16
Q

Explain the method for using controlled crosses to measure broad sense heritability.

A

when vg cannot be partitioned you cross genetically uniform parentals

parentals: V(g) is =0 as the genetic variance is tightly controlled and they are genetically uniform

In F1s, Vg=0
If Vg=0, then v(p) = v(e)

f2:
Vp= Ve+Vg–> Vg=Vp-Ve

By using the previously measured environmental variances(Ve) found in parentals and f1, and subtracting from the total phenotypic variance in the f2 progeny we would get the Vg of the f2 offspring

we would then plug in the Vg and vp of the f2s into the equation H2= Vg/Vp to determine broad sense heritability

17
Q

Explain the logic of Falconer’s formula and the expectation we would have for this formula if a trait is highly heritable.

A

Monozygotic twins are genetically identical and result from zygote splitting into two daughter cells after fertilization

dizygotic twins: obtain different combinations of genes from parents and are the result of two separate fertilization events

obtain the individual’s correlation in phenotype for monozygotic twins and dizygotic twins

in monozygotic twins, you would expect the phenotype to be tightly correlated

In dizygotic twins, there would be some correlation but not as strong

You then do a (correlation of MZ- correlation of Dz twins)

If a trait is highly heritable the broad sense heritbaility would be larger

18
Q

What are two reasons twin studies overestimate heritability?

A
  1. MZ twins experience more similar environments as they develop in the same placenta than DZ twins indicating that VE is smaller in MZ twins than in DZ twins.

As a result, VG is often overestimated in MZ twins because shared environments of embryos is not considered.

  1. Ve is less in MZ than DZ, more similar treatment/social environment as the grow older
19
Q

If R = 56 and S = 244 for a trait, calculate narrow sense heritability

A

R= response to selection

S= slection differential

Narrow sense=h^2= R/S

r= population mean-offspring mean

s= mating/selected population mean- population mean

56/244

20
Q

What is the expectation for R if S = 100 and h2 = 0 and why?

A

R««100 as Small/large–> 0

21
Q

What is the purpose of QTL mapping?

A

QTL mapping is the process by which chromosomal regions containing quantitative trait loci or “regions that contain genes that contribute to the variation in quantitive traits “ are identified

22
Q

Suppose you want to perform a QTL mapping study in Drosophila concerning the genes that
affect bristle number. How would you go about doing that? Assume that any parental lines you
choose have easily genotyped marker loci throughout their genome.

A

Have two isogenic parental lines with different phenotypes of interest (high bristle number +low bristle number)

you must have two different known genetic markers( e.g: SNPs ) at which these parental lines differ

cross to generate f1 ( heterozygous for the phenotypic allele as well as heterozygous for the different genetic markers)

cross the f1s to generate f2 with increase in variance due to recombinations
- these offspring are the QTL mapping populations

determine if the phenotype for individuals who a re homozygous for one marker is different than the phenotype for those who are homozygous for the different marker

if the phenotype is linked to homozygosity for a specific marker, you can then infer there is a quantitative loci gene that is linked to the marker

you could also find the tail end of the distribution with higher bristle number and pcr and sequence.