Studied by 7 people
In humans, one allele of the APOE gene, called APOE–ε2, can result in a high tolerance of cholesterol. Cholesterol is a vital substance for humans but may lead to heart disease in an older adult with a history of high cholesterol diets. High cholesterol diets are becoming more prevalent in the United States. Currently only about 2% of humans carry the APOE–ε2 allele.
Which of the following states a valid null hypothesis about the future distribution of APOE alleles in future generations in the United States?
The variant protects an individual from a condition that is only common among humans beyond reproductive age, so the frequency of the allele will likely not change much in the future because it is not influenced by natural selection.
Students observed the distribution of different color phenotypes in northern ravine salamanders (Plethodon electromorphus) before and after a spring flood. The data are shown in Table 1.
Table 1. Observed number of salamanders by phenotype before and after a spring flood
Salamander Distribution before the Flood (n) | Salamander Distribution after the Flood (n) | |
Black phenotype | 13 | 7 |
Dark-brown phenotype | 25 | 14 |
Light-brown phenotype | 6 | 3 |
Which of the following is an appropriate null hypothesis regarding the phenotypic frequencies of this population of salamanders before and after the flood?
The proportions of black, dark-brown, and light-brown salamanders before and after the flood are not significantly different, and any observed differences are due to random chance.
The California newt, Taricha torosa, lives in the coastal areas around Los Angeles. Which of the following is a valid null hypothesis relating fitness to survival of a bottleneck event in a coastal area where a small, isolated population of California newts resides?
Surviving the bottleneck event will be random, so any change in the allelic frequencies of the salamander population is not attributed to fitness.
A gene that influences coat color in domestic cats is located on the X chromosome. A female cat that is heterozygous for the gene (XBXO) has a calico-colored coat. In a genetics experiment, researchers mate a calico-colored female cat (XBXO) with an orange-colored male cat (XOY) to produce an F1 generation. The researchers record observations for the cats in the F1 generation and plan to use the data to perform a chi-square goodness-of-fit test for a model of X-linked inheritance. The data for the chi-square goodness-of-fit test are presented in Table 1.
Table 1. Data for the chi-square goodness-of-fit test
PhenotypeGenotypeObservedExpected | |||
Calico-colored female | XBXO | 15 | 10 |
Orange-colored female | XOXO | 6 | 10 |
Black-colored male | XBY | 11 | 10 |
Orange-colored male | XOY | 8 | 10 |
The researchers calculate a chi-square value of 4.6 and choose a significance level of p=0.05. Which of the following statements best completes the chi-square goodness-of-fit test?
The null hypothesis cannot be rejected because the chi-square value is less than the critical value.
In order to determine the effects of age on the accumulation of mitochondrial mutations, mitochondrial DNA samples from young mice (3 months) and old mice (30 months) were observed for mutations.
Table 1. Number of mitochondrial mutations in 106 base pairs
MouseYoungOld | ||
A | 1 | |
B | 1 | |
C | 3 | |
D | 2 | |
E | 4 | |
F | 10 | |
G | 15 | |
H | 12 | |
I | 9 | |
J | 11 |
Which of the following is a correct analysis of this data set?
There is an increase in the mean number of mutations for the two age groups of 9.2 mutations per 106 base pairs. This is more critical in female mammals since mitochondria are maternally inherited.
An experiment was performed to determine the mode of inheritance of two mouse genes, one for fur color and one for fur length. It is known that black fur (B) is dominant over white fur (b) and that long fur (L) is dominant over short fur (l). To determine how the genes are inherited, a cross was performed between two true-breeding mice, one with long black fur and one with short white fur. Their progeny, the F1 generation, all had long black fur. Five F1 male-female pairs were then crossed with one another. The F2 generation phenotypes for each cross are shown in Table·1.
Table 1. Number of F2 generation phenotypes for five crosses
PhenotypeCross 1Cross 2Cross 3Cross 4Cross 5 | |||||
Long black fur | 6 | 5 | 5 | 6 | 7 |
Long white fur | 1 | 1 | 0 | 2 | 2 |
Short black fur | 0 | 2 | 1 | 1 | 1 |
Short white fur | 2 | 3 | 3 | 2 | 2 |
Which of the following is the mean number per cross of F2 generation offspring that are the result of crossing over?
2.2
Pigeons demonstrate ZW sex determination, such that a ZZ genotype produces a male and a ZW genotype produces a female. The gene for feather color is located on the Z chromosome, and the red allele is dominant over the brown allele. Three crosses between brown male pigeons and red female pigeons were performed, and the results are shown below.
Table 1. Offspring from three separate crosses of a brown male pigeon and a red female pigeon
Number of Offspring | |||
Phenotype | Cross 1 | Cross 2 | Cross 3 |
Red | 11 | 9 | 7 |
Brown | 9 | 11 | 13 |
9
In fruit flies, sepia eyes and ebony body are traits that display autosomal recessive patterns of inheritance. To investigate whether the traits are genetically linked, students cross wild-type flies with a line of flies that have sepia eyes and ebony bodies. The students observe that all the flies in the F1 generation have normal eyes and normal bodies. The students allow the flies in the F1 generation to mate and produce an F2 generation. The students then record observations for the flies in the F2 generation and use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the chi-square goodness-of-fit test is presented in Table 1.
Table 1. Setup for the students’ chi-square goodness-of-fit test
PhenotypeObservedExpected | ||
Normal eyes, normal body | 231 | 279 |
Normal eyes, ebony body | 86 | 93 |
Sepia eyes, normal body | 97 | 93 |
Sepia eyes, ebony body | 82 | 31 |
The students calculate a chi-squared value of 92.86 and compare it with a critical value of 7.82. Which of the following best completes the chi-square goodness-of-fit test?
Responses
The null hypothesis can be rejected, and the students should conclude that the data may have resulted from genetic linkage.
A student carries out a genetics experiment with fruit flies to investigate the inheritance pattern of the white eye trait. The student crosses a homozygous white-eyed female with a wild-type male and records observations about the flies in the F1 generation. The student plans to use the F1 data to perform a chi-square goodness-of-fit test for a model based on an X-linked recessive pattern of inheritance. The student will use one degree of freedom and a significance level of p=0.05. The setup for the student’s chi-square goodness-of-fit test is presented in Table 1.
Table 1. Setup for the student’s chi-square goodness-of-fit test
Phenotype | Observed | Expected |
Red-eyed female | 53 | 50 |
White-eyed male | 47 | 50 |
The student calculates a chi-square value of 0.36. Which of the following statements best completes the student’s chi-square goodness-of-fit test?
The critical value is 3.84, and the student cannot reject the null hypothesis.
In eastern gray squirrels, Sciurus carolinensis, the allele for black fur (B) is dominant to the allele for gray fur (b). In a particular population of gray squirrels, 64% have black fur and 36% have gray fur. A researcher calculated the allelic frequencies to be B=0.4 and b=0.6. Five years later, the researcher returned to the location and determined the allelic frequencies within the squirrel population to be B=0.6 and b=0.4.
Which of the following could best explain the increase in the frequency of the B allele in the population after five years?
The frequency of the B allele increased due to the selective pressures of the environment.
Gaucher disease type 1 (GD1) is a recessive genetic disease that affects 1 in 900 individuals in a particular population. GD1 is caused by a mutation in the enzyme glucocerebrosidase.
Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of the wild-type (nonmutant) allele for the enzyme glucocerebrosidase.
Responses
0.967
In a long‑term study of a population of gray squirrels, researchers observed that most of the squirrels had gray fur. However, there were some individuals in the population with white fur (albino). Genetic analysis revealed that the albino condition is expressed by individuals who are homozygous recessive (gg). Over a ten-year period, the average frequency of albino squirrels in the population was 0.18. If the population is assumed to be in Hardy-Weinberg equilibrium, the average frequency of the dominant allele is closest to which of the following?
Responses
0.58
In a large, isolated population of an insect species, a specific gene locus has one dominant allele (A) and one recessive allele (a). The genotype frequencies of the gene were collected for ten generations, as shown in Table 1.
Table 1. Genotypic frequencies over ten generations in an insect species
AA | Aa | aa | |
Generation 1 | 0.47 | 0.43 | 0.10 |
Generation 5 | 0.32 | 0.60 | 0.08 |
Generation 10 | 0.20 | 0.75 | 0.05 |
Which of the following could best account for the change in genotypic frequencies over the ten generations?
The population is not exhibiting random mating between individuals.
Which of the following statements best explains how a condition of Hardy-Weinberg equilibrium results in a population that exhibits stable allele frequencies (i.e., a nonevolving population) ?
Without migration or mutation, new alleles cannot be introduced to the population.
Humans vary in their ability to taste the bitter chemical compound phenylthiocarbamide (PTC). The taster phenotype is dominant to the nontaster phenotype. Researchers tested the PTC taster phenotype of individuals from an isolated population. In the sample, 780 individuals were able to taste PTC and 70 individuals were not able to taste PTC.
Assuming the population is in Hardy-Weinberg equilibrium, what is the frequency of the non-taster allele?
0.29
An African violet grower observes that genetically identical African violet plants growing near the walls of the greenhouse have white flowers, that plants growing farther away from the walls have pale blue flowers, and that plants growing nearest the center of the greenhouse have dark blue flowers.
Which of the following best explains the differences in flower color of the African violets in the greenhouse?
An enzyme responsible for flower color does not fold correctly in cooler temperatures, and the greenhouse is warmest in the center.
A scientist studying phenotypic variation in a species of butterfly observed that genetically identical caterpillars grown in similar cages but exposed to different colored lights developed into butterflies with differences in wing color and body size, as shown in Table 1.
Table 1. Effect of Exposing Identical Caterpillars to Specific Colors of Light
Phenotype of Adult ButterflyCaterpillars Exposed to Red LightCaterpillars Exposed to Blue Light | ||
Wing color | Darker | Lighter |
Body size | Smaller | Larger |
Which of the following best explains the cause of the phenotypic variation observed in the butterflies?
There was differential gene expression of wing color and body size in response to the colors of light the caterpillars were exposed to.
Australian dragon lizards have a ZW sex-determination system. The male genotype is homogametic (ZZ), and the female genotype is heterogametic (ZW). However, all eggs incubated at temperatures above 32°C tend to develop into females.
Which of the following best explains how the development of phenotypic female Australian dragon lizards with a ZZ genotype occurs when incubation temperatures are above 32°C?
Incubation temperatures above 32°C inhibit the genes on the Z chromosome that produce proteins necessary for male development.
