BIO101 F’98



BIOL276 Name _______________________

Dr. Young

Quantitative and Population Genetics Problems Set

Many of the questions/problems in the text are very useful for practice. Without ignoring the other questions/problems, make sure that you can work through the following:

Chapter 21, Questions 12, 14, 16, 18 (answers are in the text)

Chapter 22, Questions 6, 7, 8 (answers to 6 and 8 are in the text)

Formula

Variance [pic] Standard deviation [pic]

Covariance [pic] Correlation coefficient [pic]

Phenotypic variance VP = VG + VE + VGE

VG = VA + VD + VI VP = VA + VD + VI + VE + VGE

Broad-sense heritability [pic] Narrow-sense heritability [pic]

Response to selection R = h2 x S where S is the selection differential

[pic]

1. Two pairs of independently segregating genes with two alleles each, A/a and B/b, determine plant height additively in a population. The homozygote AA BB is 50 cm tall, and the homozygote aa bb is 30 cm tall.

a. What is your prediction of the F1 height in a cross between two homozygous stocks?

Since the cross is AABB x aabb, the F1 genotype will be AaBb. Since capital alleles additively determine height, and individuals with four capital alleles have a height of 50 cm, while individuals with no capital alleles have a height of 30 cm, each capital allele appears to confer (50 – 30)/4 = 5 cm of height over the 30 cm base. AaBb individuals with two capital alleles should have an intermediate height of 40 cm.

b. What genotypes in the F2 will show a height of 40 cm after an F1 x F1 cross?

Any individuals with two capital alleles will show a height of 40 cm. Thus, AaBb, AAbb, and aaBB individuals will be 40 cm high.

c. What will be the F2 frequency of the 40 cm plants?

In the F2, 1/16 of the progeny are AAbb, 4/16 are AaBb, and 1/16 are aaBB. Thus, 6/16 = 3/8 of the progeny will be 40 cm high.

d. What assumptions have you made in answering these questions?

In answering this question, we have assumed that the A and B loci assort independently and that each locus and each allele contribute equally to the phenotype.

2. A quantitative geneticist determines the following variance components for leaf width in a population of wildflowers growing along a roadside in Kentucky:

Additive genetic variance (VA) = 4.2

Dominance genetic variance (VD) = 1.6

Genic interaction variance (VI) = 0.3

Environmental variance (VE) = 2.7

Genetic-environmental variance (VGE) = 0.0

a. Calculate the broad-sense heritability and the narrow-sense heritability for leaf width in this population of wildflowers.

To calculate the either heritability, we need to calculate VP = VG + VE + VGE. VG = VA + VD + VI = 4.2 + 1.6 + 0.3 = 6.1. Therefore VP = 6.1 + 2.7 + 0.0 = 8.8.

Broad-sense heritability H2 = VG/VP = 6.1/8.8 = 0.69.

Narrow-sense heritability h2 =VA/VP = 4.2/8.8 = 0.48.

b. What do these calculated heritabilities indicate about the genetic nature of leaf width variation in this plant?

Broad-sense heritability tells us how much of phenotypic variance is due to genetic variance. Narrow-sense heritability tells us how much of phenotypic variance is due to additive genetic variance.

3. Ten male Harvard students were weighed in 1916. Their weights are given below. Calculate the mean, variance, and standard deviation for these weights.

Weight (kg) of Harvard students (class of 1920): 51, 69, 69, 57, 61, 57, 75, 105, 69, 63

|xi ||xi – mean| |(xi – mean)2 |

|51 |16.6 |275.56 |

|69 |1.4 |1.96 |

|69 |1.4 |1.96 |

|57 |10.6 |112.36 |

|61 |6.6 |43.56 |

|57 |10.6 |112.36 |

|75 |7.4 |54.76 |

|105 |37.4 |1398.76 |

|69 |1.4 |1.96 |

|63 |4.6 |21.16 |

|Sum = 676; Mean = 67.6 | |∑ = 2024.4 |

Variance = 2024.4/9 = 224.9

Standard deviation = s = 15

4. The heights of mothers and daughters are given in the following table.

|Height of mother (inches) |Height of daughter (inches) |

|64 |66 |

|65 |66 |

|66 |68 |

|64 |65 |

|63 |65 |

|63 |62 |

|59 |62 |

|62 |64 |

|61 |63 |

|60 |62 |

Calculate the correlation coefficient for the heights of the mothers and daughters.

|x |y |

|Dominance genetic variance (VD) |0.3 |

|Genic interaction variance (VI) |0.1 |

|Environmental variance (VE) |0.4 |

|Genic-environmental interaction variance (VGE) |0.0 |

a. What is the narrow-sense heritability of tail length?

Narrow-sense heritability is VA/VP = 0.5/1.3 = 0.38.

b. What is the broad-sense heritability of tail length?

Broad-sense heritability is VG/VP = (VA + VD + VI))/VP = 0.9/1.3 = 0.69.

6. A characteristic has a narrow-sense heritability of 0.6.

a. If the dominance variance (VD) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Explain.

The narrow-sense heritability will decrease. Narrow-sense heritability is VA/VP. Increasing the VD will increase the total phenotypic variance VP. If VA remains unchanged, then the proportion VA/VP will become smaller.

b. What will happen to the broad-sense heritability? Explain.

The broad-sense heritability VG/VP will increase. VG is the sum of VA + VD + VI. VP is the sum of VA + VD + VI + VE. Increasing the numerator and denominator of the fraction by the same arithmetic increment will result in a larger fraction, if the fraction is less than 1, as must be the case for VG/VP.

c. If the environmental variance (VE) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Explain.

The narrow sense heritability VA/VP will decrease because the total phenotypic variance VP will increase if VE increases.

d. What will happen to the broad-sense heritability? Explain.

The broad-sense heritability VG/VP will decrease because VP = VG + VE will increase.

7. A graduate student, Lorena, is studying a population of bluebonnets along a roadside. The plants in this population are genetically variable. Lorena counts the seeds produced by 100 plants and measures the mean and variance of seed number. The variance is 20. Selecting one plant, Lorena takes cuttings from it and cultivates these cuttings in the greenhouse, eventually producing many genetically identical clones of the same plant. She then transplants these clones into the roadside population, allows them to grow for one year, and then counts the number of seeds produced by each of the cloned plants. The graduate student finds that the variance of seed number among these cloned plants is 5. From the phenotypic variance of the genetically variable and genetically identical plants, she calculates the broad-sense heritability.

What is the broad-sense heritability of seed number for the roadside population of bluebonnets?

In the genetically identical population, VG = 0, and VP = VE = 5. In the original population, VG = VP – VE = 20 – 5 = 15. The broad-sense heritability is then VG/VP = 15/20 = 0.75.

8. Joe is breeding cockroaches in his dorm room. He finds that the average wing length in his population of cockroaches is 4 cm. He picks six cockroaches that have the largest wings; the average wing length among these selected cockroaches is 10 cm. Joe interbreeds these selected cockroaches. From previous studies, he knows that the narrow-sense heritability for wing length in his population of cockroaches is 0.6.

a. Calculate the selection differential and expected response to selection for wing length in these cockroaches.

Use the equation: R = h2 x S, where S is the selection differential. In this case, S = 10 cm – 4 cm = 6 cm, and we are given that the narrow-sense heritability h2 is 0.6. Therefore, the response to selection R = (0.6) (6 cm) = 3.6 cm.

b. What should be the average wing length of the progeny of the selected cockroaches?

The average wing length of the progeny should be the mean wing length of the population plus R: 4 cm + 3.6 cm = 7.6 cm.

9. Three characteristics in beef cattle – body weight, fat content, and tenderness – are measured and the following variance components are estimated:

| |Body weight |Fat content |Tenderness |

|VA |22 |45 |12 |

|VD |10 |25 |5 |

|VI |3 |8 |2 |

|VE |42 |64 |8 |

|VGE |0 |0 |1 |

In this population, which characteristic would respond best to selection? Explain your reasoning.

Tenderness would respond best because it has the highest narrow-sense heritability. The response to selection is given by R = h2 x S, where the narrow-sense heritability h2 is equal to VA/VP.

10. The narrow-sense heritability of wing length in a population of Drosophila melanogaster is 0.8. The narrow-sense heritability for head width in the same population is 0.9. The genetic correlation between wing length and head width is -0.86. If a geneticist selects for increased wing length in these flies, what will happen to head width?

The head width will decrease. These two traits have a high negative genetic correlation. Therefore, selection for one trait will affect the other trait inversely.

11. Pigs have been domesticated from wild boars. Would you expect to find higher heritability for weight among domestic pigs or wild boars? Explain your answer.

Wild boars will probably have higher heritability than domestic pigs. Domestic pigs, because of many generations of breeding and selection, are likely to have less variance, and higher degree of homozygosity, for genes that affect commercial traits such as weight.

12. Full color (D) in domestic cats is dominant over dilute color (d). Of 325 cats observed, 194 have full color and 131 have dilute color.

a. If these cats are in Hardy-Weinberg equilibrium for the dilution locus, what is the frequency of the dilute allele?

f (d) = f (dd) = q2 = 131/325 = 0.403; q = 0.635.

b. How many of the 194 cats with full color are likely to be heterozygous?

If q = f (d) = 0.635, then p = 1 – q = 0.365.

f (Dd) = 2pq = 2 (0.365) (0.635) = 0.464; (0.464) (325) = 151 heterozygous cats.

13. Tay-Sachs disease is an autosomal recessive disorder. Among Ashkenazi Jews, the frequency of Tay-Sachs disease is 1 in 3,600. If the Ashkenazi population is mating randomly for the Tay-Sachs gene, what proportion of the population consists of heterozygous carriers of the Tay-Sachs allele?

q2 = 1/3600; q = 0.017. If q = 0.017, then p = 1 – 0.017 = 0.983. The frequency of heterozygous carriers = 2pq = 2 (0.983) (0.017) = 0.033; approximately 1 in 30 are carriers.

14. In the plant Lotus corniculatus, cyanogenic glycoside protects the plants against insect pests and even grazing by cattle. This glycoside is due to a simple dominant allele. A population of L. corniculatus consists of 77 plants that possess cyanogenic glycoside and 56 that lack the compound. What is the frequency of the dominant allele that results in the presence of cyanogenic glycoside in this population?

If this population is in Hardy-Weinberg equilibrium for this gene, then q2 = 56/(77 + 56) = 56/133 = 0.42; q = 0.65. Then, p = 1 – q = 1 – 0.65 = 0.35.

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