Suppose a study is being conducted to understand the potential heritability of artistic ability. Specifically, the study wants to recruit pairs of monozygotic and dizygotic twins raised in the same or in different environments. For each scenario, what is the theoretical genetic variance and environmental variance? 1. Monozygotic in the same environment
2. Monozygotic in a different environment
3. Dizygotic in the same environment
4. Dizygotic in a different environment
Answer Bank
0% genetic variance, 100% environmental variance
0% genetic variance, 0% environmental variance
50% genetic variance, 100% environmental variance
50% genetic variance, 0% environmental variance
Using the data collected from the study, what equation could be used to estimate the degree of phenotypic variability of artistic ability that is due to genetic factors?
H² = 2(rmz-rpz), where H² is broad-sense heritability
h2 VA/VP, where h’ is narrow-sense heritability
h2 = 2b, where ha is narrow-sense heritability
Vp = VA + VD + V1 + VE + VGE, where Vp is phenotypic variance

Genetic and environmental variance determine the phenotypic variability among individuals. 1) b / 2) a / 3) d / 4) c. The equation h² = VA/VP could be used to estimate the degree of phenotypic variability due to genetic factors.

1. Monozygotic in the same environment ⇒ 0% genetic variance, 0% environmental variance

Monozygotic twins have no genotypic variation, they share the genotypic charge, meaning there is no difference between them.

Since they grow in the exact same environment, they are exposed to the same factors and, hence, they express no difference.

2. Monozygotic in a different environment ⇒ 0% genetic variance, 100% environmental variance

Monozygotic twins have no genotypic variation, they share the genotypic charge, meaning there is no difference between them.

However, since they were exposed to different environmental factors, they express differences that are due to external elements or conditions.

3. Dizygotic in the same environment ⇒ 50% genetic variance, 0% environmental variance

Part of the genetic charge might be different because they developed from different eggs. They are not identical twins, so they express genotypic variation.

Environmental factors do not play a significant role because twins were equally exposed to the same conditions, so there is no varying environmental influence.

4. Dizygotic in a different environment ⇒ 50% genetic variance, 100% environmental variance

Part of the genetic charge might be different because they developed from different eggs. They are not identical twins, so they express genotypic variation.

Besides, as twins grew in different environments, which means they also express phenotypic differences that are due to external elements or conditions.

Heritability in the narrow sense -h²- refers to the proportion of the phenotypic variability that is influenced by genetic factors.

Heritability in the narrow sense is the measure of the genetic component to which additive genetic variance contributes.

The heritability might be used to determine how the population will respond to the selection done.

The equation that could be used to estimate the degree of phenotypic variability of artistic ability that is due to genetic factors is

                  h² = VA/VP ⇒ where h’ is narrow-sense heritability

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Explanation:

Monozygotic in the same environment

Genetic Variance will be zero and the environmental variance will be zero this is because a monozygotic twin shares the same gene 100% and because they stay in the same environment they are equally exposed to the same thing this makes environmental variance Zero.

Monozygotic in a different environment

Monozygotic in different environment will have an environmental variance 100% this is because they are expose to different environment leading to variation in what they know and 0% genetic variance

Dizygotic in the same environment

They have 0% environmental variation and 50% genetic variation this is because they are fraternal twins and not 100% genetically only 50% of their traits are identical

Dizygotic in a different environment

They will have 100% environmental variance because they are raised in different envrionment leading to variation in idea and what they know and 50% genetic variance because Dizygotic twin are not genetically identical.

The formula to calculate phenotypic variance due to genetic variation is

Hb2 = 2(rmz - rdz) this is in accordance to falconer's formula for twin heritability and its estimatebroad sense heritage.

Where hb2 is broad sense heritability

eukarya because is is like alche and b is bacteria and d is animals

explanation:

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