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Bio 140 4th LE > Module 10 Quantitative Genetics > Flashcards

Module 10 Quantitative Genetics Flashcards

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1
Q

Influenced by alleles of 2 or more genes and the environment

A

Complex Traits

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2
Q

Also known as multifactorial traits

A

Complex Traits

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3
Q

Condition where in a single genotype has multiple phenotypes and vice versa, based on environment.

Occurs in complex traits

A

Complex Inheritance

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4
Q

Three categories of traits with complex inheritance

A

Continuous/Quantitative
Meristic Traits
Threshold Traits

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5
Q

These include traits that are measurable (i.e. Height, blood pressure, etc.)

A

Continuous/Quantitative Traits

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6
Q

Includes traits that are countable.

A

Meristic

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7
Q

Traits with 2 or less phenotypic classes with inheritance governed by multiple genes. “On or off”

A

Threshold Trait

(If genotypes exceed threshold, it is expressed hence “on or off”)

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8
Q

Sources of Phenotypic Variation

A

Genotypic Variation

Environmental Variation

Genotype-Environment Interaction

Genotype-Environment Association

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9
Q

Phenotypic variation unaffected by environment

A

Genotypic Variation

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10
Q

Variation wherein the same genotype exhibits different phenotypes

A

Environmental Variation (All Variation is environmental i.e. Body Weight)

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11
Q

Type of Variation wherein one genotype outperforms the other based on environment

A

Genotype-environment interaction

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12
Q

Type of variation where genotypes are not randomly distributed in all possible environments

A

Genotype-Environment Association

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13
Q

Differentiate Genotype-Environment interaction and association

A

In interaction, the environment directly boosts or hinders a specific genotype.

In association, there is no direct favoritism but a pattern is still observable. The distribution is non-random.

Ex. In Manila, android phones are more common. In Makati, iPhones are more common. Not necessarily caused by environment but the pattern is there.

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14
Q

How do you estimate the number of genes involved in a polygenic trait

A

Fraction of F2 as extreme as one parent = (1/4)^n

To find n,

n= log (P(F2))/log(1/4)

where P(F2) is the fraction of F2 with same genotype extreme as parent

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15
Q

Apply Polygenic trait gene number estimation:

In a plant, height varies from 6 cm to 36 cm. When 6- and 36 cm plants are crossed, all F1 plants are 21 cm. In the F2 generation, a continuous range of heights is observed. Most are around 21 cm, and 3 of 200 are as short as 6-cm P1 parent.

How many gene pairs are involved?

A

n= log (P(F2))/log(1/4)

P(F2) = 3/200

n = log(3/200)/log(1/4) = -1.8/-0.6 = 3

n = 3 genes

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16
Q

How do you estimate for allelic contribution?

A

a = D/2n; where
a=Contribution,
D=Difference between the parents,
n= number of gene pairs

17
Q

Apply allelic contribution:

In a plant, height varies from 6 cm to 36 cm. When 6- and 36 cm plants are crossed, all F1 plants are 21 cm. In the F2 generation, a continuous range of heights is observed. Most are around 21 cm, and 3 of 200 are as short as 6-cm P1 parent.

Assuming that there are 3 gene pairs involved, how much does each allele contribute?

A

a = D/2n

Given:
Parent 1 = 36, Parent 2 = 6, n=3

Solution:
D=36-6=30

a=30/(2*3)=30/6=5

5cm per allele

18
Q

In a plant, height varies from 6 cm to 36 cm. When 6- and 36 cm plants are crossed, all F1 plants are 21 cm. In the F2 generation, a continuous range of heights is observed. Most are around 21 cm, and 3 of 200 are as short as 6-cm P1 parent.

Assuming that there are 3 gene pairs and each allele contributes 5 cm to the height, what are the different genotype/s that can lead to 31 cm long plants?

A

Given: n=3, a=5 cm, min=6, max=36
Required: Genotype(s) for 31 cm
Solution:
Let the dominant genes be labelled as A, B, and C with their recessive counterparts a, b, and c.

We have the max, 36 cm, which we can assume as AABBCC and our minimum, 6 cm, as aabbcc.

Using the minimum, you can do
Y = min + Xa, where a is the allelic contribution, X is the number of contributory alleles and Y is the target.

31=6+5X-> x=5, 5 capital letters

Therefore, the genotypes for 31cm are AABBCc, AABbCC, and AaBBCC.

19
Q

Variance Method of Estimation of gene pairs involved for a polygenic trait can be estimated by what formula?

A

N = (D^2)/(8*(VPF2-VPF1)) where
N=Number of Genes,
D=Difference between parental means,
VPF2= Phenotypic Variance of the F2 population,
VPF1= Phenotypic Variance of the F1 population

20
Q

How to calculate the genotypic variance of the F2 generation (VGF2)? Why?

A

VGF2 =a^2*(N/2) = VPF2-VPF1

F1 Variance due to genotype = 0.
From parent to F1 generation, only affected by environmental factors.
N= alleles involved = 2n

F1 to F2 generation is environmental plus Genotypic variation.

Therefore VPF2-VPF1 = VGF2

21
Q

Assumptions of genotypic estimation

A

▪ Same environment
▪ The alleles of each gene are additive
▪ The genes contribute equally to the trait
▪ The genes are unlinked
▪ The original parental strains homozygous for alternative alleles
▪ No dominance
▪ No interaction

22
Q

The Flemish breed of rabbits has an average body weight of 3,000 grams. The Himalayan breed has a mean of 1,875 g. Matings between these two breeds produce an intermediate F1 with a standard deviation of 162 g. The variability of the F2 is greater, as indicated by a standard deviation of 230 g.

Estimate the number of pair of factors contributing to mature body weight in rabbits.

A

Given:
Mean 1 = 3000 g; Mean 2 = 1875 g
StDevF1 = 162 g; StDevF2 = 230 g

Required: N

Solution: N = (D^2)/(8*(VPF2-VPF1))
StDev^2=V, therefore

N= ((3000-1875)^2)/(8(230^2-162^2))
N= (1125^2)/(8(52900-26244))
N=1265625/213248=5.93

Therefore, N is Approximately 6.

23
Q

The Flemish breed of rabbits has an average body weight of 3,000 grams. The Himalayan breed has a mean of 1,875 g. Matings between these two breeds produce an intermediate F1 with a standard deviation of 162 g. The variability of the F2 is greater, as indicated by a standard deviation of 230 g.

Assuming that there are 6 gene pairs involved, estimate the average quantitative contribution of each active allele.

A

Given:
N=6
Mean 1 = 3000 g; Mean 2 = 1875 g

Required: a

Solution:
Still follows original formula for a

a=D/2n = (3000-1875)/(2*6)= 1125/12 = 93.75 g

Therefore, each allele contributes 93.75 g

24
Q

How to calculate for proportions of contributory alleles for n loci.

A

Use binomial expansion (C+c)^n, get the coefficient.

Ex. What is the proportion of genotypes with 2 contributory alleles in the offspring of a dihybrid cross.

Given: AaBb x AaBb
Required: P(2C)
Solution:
n=2; x=2; Dihybrid cross = 16 different outcomes; 2 Contributory

4C2(C^2)(c^2), get the coefficient,
4C2 = 4!/(2!*2!) = 12/2 = 6

Therefore, proportion is 6/16

25
Q

What is the effect of non-additivity or interactions between the gene pairs to the distribution?

A

It becomes asymmetrical as assumptions are not met.

26
Q

For example: The minimum weight of a fruit is 10 g and there are two pairs of genes involved with each gene contribution equal to 3 g.

What are the proportions of each different genotype and their expected measurements?

A

Given: Minimum = 10; a=3 g, n=4
Required: Weights and proportions
Solution: Binomial Expansion (C+c)^n
(C+c)^n = C^4 + (4C3)C^3*c + (4C2) C^2c^2 + (4C2)Cc^3 + c^4

4C4 (C^4) = 10 + (34) = 1/16 (22 g)
4C3 (C^3 c) = 10 + (33) = 4/16 (19 g)
4C2 (C^2 c^2))=10+(32) = 6/16 (16 g)
4C1 (Cc^3) = 10 + (31) = 4/16 (13 g)
4C0 (c^4) = 10 = 1/16 (10 g)

27
Q

For example: The minimum weight of a fruit is 10 g and there are two pairs of genes involved with each gene contribution equal to 3 g.
The following are the proportions per weight:

1/16 (22 g) 4/16 (19 g) 6/16 (16 g) 4/16 (13 g) 1/16 (10 g)

If gene dominance was exhibited, what is the new expected proportion and measured weights?

A

Given: a=3g, Dominance exhibited, n=4

Required: Proportion and weights

Solution: Since dominance exhibited, CC = Cc = 3g, 2 will contribute same as one. Therefore, the only phenotypes are C_ and cc

C_C_: 16g, 9/16
C_cc or ccC_: 13g, 5/16
cccc: 1/16 10g

28
Q

Estimation of threshold genes:
A certain affliction may be controlled by three pairs of genes and presence of less than three contributory genes would lead to the manifestation of the disease. If two normal but fully heterozygous individuals were to have children, what are the probabilities of them producing afflicted and normal children.

A

Given: N=3->n=6, n<3 = Afflicted
Required: Proportion of Normal and Afflicted
Solution: Since threshold, “on or off”, 2 phenotypes, 2^6= 64 outcomes

Using binomial expansion, since the threshold for affliction is 3, 1st to 4th term are normal and 5th to 7th are afflicted. Get the coefficients and sum.

1st to 4th: 1+6+15+20=42/64=21/32
5th to 7th:15+6+1=22/64=11/32

Normal: 21/32; Afflicted: 11/32

29
Q

This is the proportion of population’s phenotypic variation and is attributable to genetic factors

A

Heritability

30
Q

Phenotypic variance is the sum of?

A

Genetic Variation, Environmental variation, and Genetic-Environmental interaction variation.

These are variance components,
VP=VG+VE+VGE

VG can be further differentiated to
VA+VD+VI

Additive+Dominance+Interaction Variances

31
Q

Two types of heritability and their formulas

A

Broad-sense: H^2=VG/VP
Narrow Sense: H^2=VA/VP

32
Q

Given Parent, F1, and F2 Generations, how do we find VP, VE, and VG

A

VF1=VE
VF2=VG+VE
VP=VF2
VG=VF2-VF1

33
Q

Regression coefficient (b) is calculated by? What about correlation coefficient (r)?

A

Depending on the relationship

offspring-each parent:h^2=2b
offspring-midparent:h^2=b
Full siblings: h^2=2b
Half siblings:h^2=4b

Correlation coefficient (r) is same as b, isolate.

offspring-each parent:(h^2)/2=r
offspring-midparent:h^2=r
Full siblings: (h^2)/2=r
Half siblings:(h^2)/4=r

34
Q

If all the variation between offspring and one parent (e.g. their sires) is genetic, then r should be equal to 0.5; if r=0.2, then h2 is?

A

offspring-each parent: (h^2)/2=r
h^2=2r=2(0.2)=0.4

35
Q

How do you calculate heritability in response to selection

A

Using means:
Mean of the whole population M
Mean phenotype of parents MP
Mean phenotype of progeny Mp

h^2 = (Mp-M)/(MP-M)

36
Q

The average yearly milk production of a herd of cows is 18,000 lb. The average milk production of the individuals selected to be parents of the next generation is 20,000 lb. The average milk production of the offspring generation is 18,440 lb. Estimate the heritability of milk production in this population.

A

Given: 3 means, 18000, 20000, 18440
Required: h^2
Solution:
Average of Herd = M = 18000
Selected Parents = MP = 20000
Offspring = Mp = 18400

h^2 = (18440-18000)/(20000-18000)
h^2 = 0.22

37
Q

Most quantitative traits are not highly heritable, if given an h^2 value, what are the 3 classifications of heritability?

A

High Heritability: h^2>0.5
Medium Heritability: 0.2<h^2<0.5
Low Heritability: h^2<0.2

38
Q

regions of DNA which is associated with a particular phenotypic trait

A

Quantitative Trait Loci

39
Q

How often do you think of the Roman Empire

A

Do NOT confuse with the “hOLy” Roman Empire >:0

This is just a meme/joke to cut the tension. gl on the exam.

Forgot I even added this lmao

Bio 140 4th LE (6 decks)

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