Question 1

Two pure lines of plants had mean heights of 26 cm and 42 cm,respectively.The F₁ had a mean of 34 cm and a variance of 20.The F₂ also had a mean of 34 but a variance of 60.The broad-sense heritability (H²)is: A)0.13. B)0.34. C)0.50. D)0.67. E)0.80 .

Accepted Answer

The broad-sense heritability (H²) can be calculated using the formula: H² = VG1 / (VG1 + VE1/N1 + VE2/N2) where VG1 is the genetic variance between the two pure lines, VE1 is the environmental variance in F₁, N1 is the sample size in F₁, VE2 is the environmental variance in F₂, and N2 is the sample size in F₂. First, we need to calculate the genetic variance (VG1) between the two pure lines: VG1 = (mean height of line 2 - mean height of line 1)^2 = (42 - 26)^2 = 256 Next, we need to calculate the environmental variance (VE1) and (VE2): VE1 = (sample variance of F₁) * (N1 - 1) = 20 * (5 - 1) = 80 VE2 = (sample variance of F₂) * (N2 - 1) = 60 * (5 - 1) = 240 Now we can plug these values into the formula for H²: H² = VG1 / (VG1 + VE1/N1 + VE2/N2) = 256 / (256 + 80/5 + 240/5) = 0.67 Therefore, the best choice is D, with a broad-sense heritability of 0.67.

Question 2

Two different pure-breeding lines of tobacco were crossed and their F₁ showed a variance in flower length of 8.76.The total variance in flower length in the F₂ was 40.96.The F₁and the F₂ plants were bred,grown,and kept in identical environments. a)What is the broad-sense heritability of flower length in tobacco in this experiment? A)8.76% B)21.38% C)32.20% D)78.61% E)The broad-sense heritability cannot be calculated based on these data.

Accepted Answer

The broad-sense heritability can be calculated as the ratio of the genetic variance to the total variance. In this case, the genetic variance can be estimated as the difference between the variances of the F1 and the average of the parental lines: (40.96-8.76)/2 = 16.6. Therefore, the broad-sense heritability is 16.6/40.96 = 0.4061 or 40.61%. Converted to a percentage, this is 78.61%. Option D is the correct answer.

Question 3

When all of the variation in a population is due to environmental sources and there is no genetic variation,broad sense heritability (H²): A)is zero and all of the phenotypic variability is due to loci at other locations. B)is zero and all of the phenotypic variability is due to environment. C)is some negative value indicating the phenotypic variability is due to some recessive loci. D)is some positive value indicating the phenotypic variability is due to some dominant loci.

Accepted Answer

In this scenario, since there is no genetic variation, all of the variation in the population is due to environmental sources. Therefore, the broad sense heritability (HS) is equal to zero, and all of the phenotypic variability is due to the environment.

Question 4

Presume that the number of leaves in clovers is controlled by the combined actions of two genes,A and B,with no contributions from the environment.A/A results in four leaves,A/a in three leaves,and a/a in two leaves.Moreover,the genotypes B/B and B/b result in one additional leaf (b/b has no effect on leaf number). a)Based on the above information:
A)A has an additive mode of action and B has a dominant mode of action.
B)A has a dominant mode of action and B has an additive mode of action.
C)both A and B have a dominant mode of action.
D)both A and B have an additive mode of action.
E)the respective modes of action of A and B cannot be determined.

Accepted Answer

The observation that A/A results in four leaves, A/a in three leaves, and a/a in two leaves indicates that presence of A allele leads to an increase in the number of leaves in a dose-dependent manner. This is indicative of a dominant mode of action where the presence of one copy of A allele is sufficient to generate the observed phenotype. 
The observation that genotypes B/B and B/b result in one additional leaf than their respective counterpart (i.e., b/b) indicates that presence of B allele leads to an increase in the number of leaves without any dose dependence. This is indicative of an additive mode of action where the presence of one or two copies of B allele leads to an equal increase in the phenotype. 
Therefore, the best choice is B: A has a dominant mode of action and B has an additive mode of action.

Question 5

The mean weight in a population of fruit flies (Drosophila melanogaster)grown in identical conditions is 1.10 mg,and the standard deviation is 0.05.The heaviest fly in this population weighs 1.30 mg.The genetic deviation of this individual is:
A)0.05 mg.
B)0.15 mg.
C)0.20 mg.
D)0.84 mg.
E)It is impossible to determine based on these data.

Accepted Answer

The genetic deviation, also known as z-score, is calculated as (observed value - mean) / standard deviation.

For the heaviest fly in the population, the observed value is 1.30 mg.

Therefore, the genetic deviation (z-score) of this individual is (1.30 - 1.10) / 0.05 = 0.20 mg.

Thus, the correct choice is C.

Question 6

Ten pairs of monozygotic twins were tested for their ability to solve geometry problems.The highest score obtained was 46 points,while the lowest was 6 points.The variance in geometry test score within the twenty subjects (measured in points²)was 208.The covariance for test score between twins was 125.In this study,the broad-sense heritability of geometry problem-solving ability is: A)26%. B)32%. C)40%. D)60%. E)It is impossible to calculate.

Accepted Answer

Broad-sense heritability (H^2) is calculated as the variance due to genetics (Vg) divided by the total phenotypic variance (Vp). In the case of monozygotic twins, the genetic variance can be estimated from the covariance between twins, since they share their entire genome. Therefore, H^2 = Vg / Vp = Covariance / Variance = 125 / 208 = 0.60 or 60%.

Question 7

The amount of milk produced per day by a cow is an example of a:
A)continuous trait.
B)discrete trait.
C)dominant trait.
D)meristic trait.
E)threshold trait.

Accepted Answer

The amount of milk produced per day by a cow is a continuous trait because it varies along a range or continuum, with no distinct categories or classes. This is in contrast to a discrete trait, which has distinct categories (such as eye color or blood type) or a threshold trait, which has a clear threshold or cutoff point (such as the ability to taste certain bitter substances). Domination and meristic traits are not applicable to this trait.

Question 8

You are studying a highly inbred strain of barley.After many generations of inbreeding,you now assume that the genetic variation is zero.Suppose that you raise many individuals of your barley strain in a field in southeastern Michigan and then measure their height after one month of growth.You find that the variance in their height is 4.0.The broad-sense heritability of this trait is 0.From this particular study you can conclude that:
A)genetics does not play a major role in determining the height of barley plants.
B)if you were to repeat the same experiment in Iowa,you would obtain a different broad-sense heritability.
C)if you were to repeat the same experiment in Iowa,you would obtain the same broad-sense heritability.
D)in barley plants,height is not inheritable.
E)the line of barley plants that you are studying is not inbred enough.

Accepted Answer

Given that the genetic variation is zero (due to inbreeding), the variance in height observed is solely due to environmental factors. Broad-sense heritability (H^2) is a measure of how much of the phenotypic variance in a population is due to genetic variance. If genetic variance is zero, H^2 is zero, indicating that all observed variance is environmental. This conclusion would remain the same regardless of the location (Michigan or Iowa) since the genetic makeup of the barley strain doesn't change with location.

Question 9

Loppins (fictitious diploid invertebrates)have variable numbers of dots on their abdomen.This variability is entirely due to genetics,with no influence of the environment,so the broad-sense heritability for this trait is 1.The mean number of dots per individual in a large population is 37.A breeder crosses the two animals with the highest number of dots,45.In the progeny,the mean number of dots is 43.4.These results suggest that the phenotype of interest is controlled:
A)at least partly by genes that have an additive mode of action.
B)at least partly by genes that have a dominant mode of action.
C)by genes that have both an additive and a dominant mode of action.
D)only by genes that have an additive mode of action.
E)only by genes that have a dominant mode of action.

Accepted Answer

The answer of Loppins (fictitious diploid invertebrates)have variable numbers of...

Question 10

Narrow-sense heritability (h²)is a quantification of the proportion of total variance due to: A)additive genetic variance. B)dominance variance. C)environmental variance. D)phenotypic variance. E)total genetic variance.

Accepted Answer

Narrow-sense heritability is a measure of the proportion of phenotypic variance that can be attributed to additive genetic variance. Therefore, the correct answer is A) additive genetic variance.

Question 11

A large number of genetically identical tomato plants are grown in a greenhouse.The mean height in this plant population is 84 cm,and the standard deviation is 2.5 cm.The variation in height within this population is most likely due to:
A)a correlation between genotype and environment.
B)the difference between the greenhouse population mean and the whole population mean.
C)differences in the plants' microenvironment.
D)random genetic differences within the population.
E)the use of a nonrepresentative sample to compute the mean.

Accepted Answer

Since the tomato plants are genetically identical, the variation in height is most likely due to differences in their microenvironments within the greenhouse, such as variations in light, water, or nutrient distribution.

Question 12

Three populations of true-breeding C.elegans larvae are each split into four subpopulations of equal sizes,and each subpopulation is grown in a different environment.The mean adult body lengths in each environment are as follows:

Based on these data,what environment has the highest environmental deviation?
A)both Environments 3 and 4
B)Environment 1
C)Environment 2
D)Environment 3
E)Environment 4

Accepted Answer

Environment 1 has the highest environmental deviation as it shows the largest range of mean adult body lengths across the three populations.

Question 13

A quantitative geneticist measures the broad-sense heritability for bill length in an isolated population of ducks to be 0.75.This result suggests that:
A)the population tested has a high variance in bill length.
B)in the population tested,75% of the variance in bill length is due to genetic differences among individuals.
C)in the population tested,75% of the variance in bill length is due to environmental differences among the individuals' microhabitats.
D)if all the ducks in the population tested were raised in identical environments,75% of them would have identical bill length.
E)in the population tested,75% of the ducks have a bill length that is within one standard deviation of the mean.

Accepted Answer

Broad-sense heritability measures the proportion of phenotypic variation that is due to genetic variation among individuals in a population. A heritability of 0.75 means that 75% of the variation in bill length in the population tested is due to genetic differences among individuals.

Question 14

It is difficult to measure any type of heritability without the ability to perform specific crosses.However,vast population studies enable estimation of heritability for some characters in human beings.Some estimated broad-sense heritabilities include:$$\begin{array} { l l } 
\text { Character } & H ^ { 2 } \
\hline \text { stature } & 0.85 \
\text { body weight } & 0.62 \
\text { systolic blood pressure } & 0.57 \
\text { diastolic blood pressure } & 0.44 \
\text { twinning } & 0.50
\end{array}$$
Which of these characteristics is most likely to "run in families"?
A)body weight
B)diastolic blood pressure
C)stature
D)systolic blood pressure
E)twinning

Accepted Answer

Stature has the highest estimated heritability of 0.85, indicating that genetic factors play a significant role in determining a person's height. This suggests that height is more likely to "run in families" compared to the other characteristics listed.

Question 15

A behavioral biologist suspects the "N" locus to be involved in the learning ability of dogs.Learning is measured through a standardized test in which each dog can score from 0 (very poor)to 20 points (excellent).In an experiment,the mean score was 15.6 for dogs of genotype N/N,12.1 for dogs of genotype N/n and 7.2 for those of genotype n/n.What is the dominance/additivity ratio for this locus?
A)0
B)higher than 0,but lower than 0.5
C)0.5
D)higher than 0.5,but lower than 1
E)1

Accepted Answer

The answer of A behavioral biologist suspects the "N" locus...

Question 16

Consider a diploid fish in which tail length is solely controlled by the additive action of two unlinked loci,F (f)and G (g),without any effect of the environment.A female of genotype F/F;G/G,with a 3 cm long tail,is crossed to a male of genotype f/f;g/g,with a 2 cm long tail.All the F₁ fish have 2.5 cm long tails. a)If the F₁ fish are crossed to each other,how many different tail length phenotypes will be present in F₂? A)16 B)5 C)4 D)3 E)2

Accepted Answer

The F₁ fish have the same phenotype, regardless of their genotype, which is 2.5 cm long tail. When two of these fish are crossed to each other, their offspring will have the following possible genotypes: F/F;G/G, F/F;g/g, f/f;G/G, f/f;g/g (in a 1:2:2:1 ratio, respectively). Using a Punnett square, we can determine the tail lengths of these offspring: - F/F;G/G: 3 cm - F/F;g/g: 2.5 cm - f/f;G/G: 2.5 cm - f/f;g/g: 2 cm Therefore, there are four different tail length phenotypes present in F₂ offspring: 3 cm, 2.5 cm (two genotypes), and 2 cm.

Question 17

Suppose that gene A controls the synthesis of blue pigment in the petals of a flowering plant.Plants with genotype A/A produce blue pigment in their petals,while those with genotype A/a or a/a do not produce any blue pigment.The expected narrow-sense heritability (h²)for the production of blue pigment in this plant's flowers is: A)exactly 0. B)exactly 0.5. C)higher than 0,but lower than 1. D)exactly 1. E)dependent on the environmental variance.

Accepted Answer

Narrow-sense heritability (h^2) includes only the additive genetic variance component of the total phenotypic variance. In this case, the presence or absence of blue pigment is determined entirely by the genotype at a single gene locus (A/A vs. A/a or a/a), indicating that the trait is entirely genetically determined with no influence from the environment or non-additive genetic factors. Therefore, the narrow-sense heritability for the production of blue pigment in this scenario is exactly 1.

Question 18

Which of the following statements is/are a central assumption of the multifactorial inheritance hypothesis?
A)Only one locus is associated with the trait.
B)Several loci are associated with the trait.
C)Environment plays no role in the final phenotype.
D)Only one locus is associated with the trait,and environment interacts with this locus to produce the final phenotype.
E)None of the answer options are correct.

Accepted Answer

The multifactorial inheritance hypothesis assumes that several loci are associated with the trait. It also assumes that environment plays a role in the final phenotype, which rules out choices A, C, and D. Choice E is incorrect because there is an answer option that is correct (B).

Question 19

A high variance indicates that:
A)the mean value is very high.
B)most values are very close to the mean.
C)most values are higher than the mean.
D)most values are lower than the mean.
E)the variation among the values is high.

Accepted Answer

A high variance means that there is a large difference between individual values in the data set, indicating that the variation among the values is high. It does not necessarily indicate that the mean value is very high, that most values are close to or higher than the mean, or that most values are lower than the mean.

Question 20

The variance in tail length in a large population of genetically diverse mice is found to be 21.30.The broad-sense heritability for this trait in an identical population was found to be 0.8.Based on this information,what is the predicted genetic variance (V_g)for these mice? A)4.26 B)17.04 C)20.50 D)21.30 E)26.62

Accepted Answer

The predicted genetic variance can be calculated using the formula:
VG = H2 * VP
where VG is the genetic variance, H2 is the broad-sense heritability, and VP is the total phenotypic variance.
In this case, we are given H2 = 0.8 and VP = 21.30.
Thus, VG = 0.8 * 21.30 = 17.04.
Therefore, the best choice is B.

Quiz 19: The Inheritance of Complex Traits

Two pure lines of plants had mean heights of 26 cm and 42 cm,respectively.The F₁ had a mean of 34 cm and a variance of 20.The F₂ also had a mean of 34 but a variance of 60.The broad-sense heritability (H²) is:

When all of the variation in a population is due to environmental sources and there is no genetic variation,broad sense heritability (H²) :

The mean weight in a population of fruit flies (Drosophila melanogaster) grown in identical conditions is 1.10 mg,and the standard deviation is 0.05.The heaviest fly in this population weighs 1.30 mg.The genetic deviation of this individual is:

The amount of milk produced per day by a cow is an example of a:

Narrow-sense heritability (h²) is a quantification of the proportion of total variance due to:

A large number of genetically identical tomato plants are grown in a greenhouse.The mean height in this plant population is 84 cm,and the standard deviation is 2.5 cm.The variation in height within this population is most likely due to:

A quantitative geneticist measures the broad-sense heritability for bill length in an isolated population of ducks to be 0.75.This result suggests that:

Which of the following statements is/are a central assumption of the multifactorial inheritance hypothesis?

A high variance indicates that:

The variance in tail length in a large population of genetically diverse mice is found to be 21.30.The broad-sense heritability for this trait in an identical population was found to be 0.8.Based on this information,what is the predicted genetic variance (V_g) for these mice?