Module 10 Quantitative Genetics

Question 1

Influenced by alleles of 2 or more genes and the environment

Answer 1

Complex Traits

Question 2

Also known as multifactorial traits

Answer 2

Complex Traits

Question 3

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

Occurs in complex traits

Answer 3

Complex Inheritance

Question 4

Three categories of traits with complex inheritance

Answer 4

Continuous/Quantitative
Meristic Traits
Threshold Traits

Question 5

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

Answer 5

Continuous/Quantitative Traits

Question 6

Includes traits that are countable.

Answer 6

Meristic

Question 7

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

Answer 7

Threshold Trait

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

Question 8

Sources of Phenotypic Variation

Answer 8

Genotypic Variation

Environmental Variation

Genotype-Environment Interaction

Genotype-Environment Association

Question 9

Phenotypic variation unaffected by environment

Answer 9

Genotypic Variation

Question 10

Variation wherein the same genotype exhibits different phenotypes

Answer 10

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

Question 11

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

Answer 11

Genotype-environment interaction

Question 12

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

Answer 12

Genotype-Environment Association

Question 13

Differentiate Genotype-Environment interaction and association

Answer 13

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.

Question 14

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

Answer 14

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

Question 15

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?

Answer 15

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

Question 16

How do you estimate for allelic contribution?

Answer 16

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

Question 17

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?

Answer 17

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

Question 18

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?

Answer 18

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.

Question 19

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

Answer 19

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

Question 20

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

Answer 20

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

Question 21

Assumptions of genotypic estimation

Answer 21

▪ 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

Question 22

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.

Answer 22

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.

Question 23

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.

Answer 23

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

Question 24

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

Answer 24

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

Question 25

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

Answer 25

It becomes asymmetrical as assumptions are not met.

Question 26

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?

Answer 26

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)

Question 27

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?

Answer 27

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

Question 28

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.

Answer 28

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

Question 29

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

Answer 29

Heritability

Question 30

Phenotypic variance is the sum of?

Answer 30

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

Question 31

Two types of heritability and their formulas

Answer 31

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

Question 32

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

Answer 32

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

Question 33

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

Answer 33

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

Question 34

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?

Answer 34

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

Question 35

How do you calculate heritability in response to selection

Answer 35

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

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

Question 36

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.

Answer 36

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

Question 37

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

Answer 37

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

Question 38

regions of DNA which is associated with a particular phenotypic trait

Answer 38

Quantitative Trait Loci

Question 39

How often do you think of the Roman Empire

Answer 39

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