Molecular Biology and Genetics - GRE Subject Test: Biochemistry, Cell, and Molecular Biology
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A population of insects exists in which black coloration is dominant to white. If there are 64 black insects and 36 white insects in the population, what is the recessive allele frequency?
A population of insects exists in which black coloration is dominant to white. If there are 64 black insects and 36 white insects in the population, what is the recessive allele frequency?
We can use the white insect population to figure out our recessive allele frequency if we use the Hardy-Weinberg equations:
From this equation we know that 
is actually the frequency of our homozygous recessive genotype. We can determine the value of this genotypic frequency based on the information in the question. Any white insects must be homozygous recessive.
Solve for the recessive allele frequency.
We can use the white insect population to figure out our recessive allele frequency if we use the Hardy-Weinberg equations:
From this equation we know that
Solve for the recessive allele frequency.
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A population of lizards is shown to have 36 members that are homozygous dominant, 48 members that are heterozygous, and 16 members that are homozygous recessive for a particular trait. The population displays Hardy-Weinberg equilibrium. What are the allele frequencies present in this population?
A population of lizards is shown to have 36 members that are homozygous dominant, 48 members that are heterozygous, and 16 members that are homozygous recessive for a particular trait. The population displays Hardy-Weinberg equilibrium. What are the allele frequencies present in this population?
To solve this question, we must use the Hardy-Weinberg equations:
There are two ways to solve this problem. The easier way is to use the second Hardy-Weinberg equation. We are told that the population is in Hardy-Weinberg equilibrium, so the observed genotype frequencies equal the expected genotype frequencies. Each term in the second Hardy-Weinberg equation can be used in coordination with the given phenotypic frequencies.
The dominant allele frequency is 0.6 and the recessive allele frequency is 0.4.
The second method involves using our population and the total number of alleles. Since our population totals to 100, we have a total of 200 alleles (two alleles per member in the population. Next, to get the frequencies we simply have to divide the total number of a single allele by the total number of alleles in the population. For the dominant allele frequency it would look like this:
We can then get the recessive allele frequency from the first Hardy-Weinberg equation:
Both methods result in the same answer.
To solve this question, we must use the Hardy-Weinberg equations:
There are two ways to solve this problem. The easier way is to use the second Hardy-Weinberg equation. We are told that the population is in Hardy-Weinberg equilibrium, so the observed genotype frequencies equal the expected genotype frequencies. Each term in the second Hardy-Weinberg equation can be used in coordination with the given phenotypic frequencies.
The dominant allele frequency is 0.6 and the recessive allele frequency is 0.4.
The second method involves using our population and the total number of alleles. Since our population totals to 100, we have a total of 200 alleles (two alleles per member in the population. Next, to get the frequencies we simply have to divide the total number of a single allele by the total number of alleles in the population. For the dominant allele frequency it would look like this:
We can then get the recessive allele frequency from the first Hardy-Weinberg equation:
Both methods result in the same answer.
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A population of beetles exists in which black coloration is dominant to white. The allele frequencies of the population were originally o.4 for the dominant allele and 0.6 for the recessive allele. A predator was introduced that selectively ate the white beetles. The new population consists of 36 homozygous dominant black beetles, 48 heterozygous beetles, and 16 white beetles. What are the new allele frequencies?
A population of beetles exists in which black coloration is dominant to white. The allele frequencies of the population were originally o.4 for the dominant allele and 0.6 for the recessive allele. A predator was introduced that selectively ate the white beetles. The new population consists of 36 homozygous dominant black beetles, 48 heterozygous beetles, and 16 white beetles. What are the new allele frequencies?
The original allele frequencies are actually superfluous information that we will not need to use in our calculation. We are given the populations of each genotype, so we can use the Hardy-Weinberg equations to solve for the allele frequencies.
In the second equation, 
gives the frequency of the homozygous dominant phenotype and 
gives the frequency of the homozygous recessive phenotype. Using the population ratios from the question we can solve for these values to find the allele frequencies.
The problem could also be solved by summing the total number of alleles, and dividing the total of each individual allele by this number. Keep in mind that each individual carries two alleles.
The original allele frequencies are actually superfluous information that we will not need to use in our calculation. We are given the populations of each genotype, so we can use the Hardy-Weinberg equations to solve for the allele frequencies.
In the second equation,
The problem could also be solved by summing the total number of alleles, and dividing the total of each individual allele by this number. Keep in mind that each individual carries two alleles.
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Which of the following choices are likely to change the allele frequencies of the indicated populations?
I. A geographic barrier isolating a small subset of a larger population
II. The introduction of a predator that only preys upon the homozygous dominant members of the population
III. A population that displays completely random mating
Which of the following choices are likely to change the allele frequencies of the indicated populations?
I. A geographic barrier isolating a small subset of a larger population
II. The introduction of a predator that only preys upon the homozygous dominant members of the population
III. A population that displays completely random mating
Allele frequencies are the measure of an allele in relation to the total number of alleles in the given population. Introducing a predator that only preys upon homozygous dominant members will cause the number of dominant alleles to drop significantly and will, therefore, change allele frequencies. This would be an example of the bottleneck effect. Isolating a small subset of a population is going to change allele frequencies because that small subset is not likely to accurately represent the original population. This is an example of the founder effect.
Random mating is actually a factor that helps maintain allele frequencies, and is a requirement for Hardy-Weinberg equilibrium.
Allele frequencies are the measure of an allele in relation to the total number of alleles in the given population. Introducing a predator that only preys upon homozygous dominant members will cause the number of dominant alleles to drop significantly and will, therefore, change allele frequencies. This would be an example of the bottleneck effect. Isolating a small subset of a population is going to change allele frequencies because that small subset is not likely to accurately represent the original population. This is an example of the founder effect.
Random mating is actually a factor that helps maintain allele frequencies, and is a requirement for Hardy-Weinberg equilibrium.
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Which of the following is not a condition for Hardy-Weinberg equilibrium?
Which of the following is not a condition for Hardy-Weinberg equilibrium?
Of the choices, the only one that is not a Hardy-Weinberg assumption is that natural selection is occurring on the population. In fact, the exact opposite is a Hardy-Weinberg assumption. If natural selection is occurring on a population, over a large period of time, it is likely to have an effect on allele frequencies within the population.
All other answers are requirements in order for Hardy-Weinberg equilibrium to be in effect: large population size, random mating, and negligible mutation frequencies.
Of the choices, the only one that is not a Hardy-Weinberg assumption is that natural selection is occurring on the population. In fact, the exact opposite is a Hardy-Weinberg assumption. If natural selection is occurring on a population, over a large period of time, it is likely to have an effect on allele frequencies within the population.
All other answers are requirements in order for Hardy-Weinberg equilibrium to be in effect: large population size, random mating, and negligible mutation frequencies.
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Which of the following conditions are not necessary for a population to be in Hardy-Weinberg equilibrium?
Which of the following conditions are not necessary for a population to be in Hardy-Weinberg equilibrium?
The Hardy-Weinberg equilibrium states that the frequency of alleles at a locus remains constant from generation to generation. In order for this to be the case, natural selection cannot affect the alleles under consideration. All other answer choices describe conditions that do need to be met for Hardy-Weinberg equilibrium to be displayed. Note that the conditions for Hardy-Weinberg equilibrium are not met in nature.
The Hardy-Weinberg equilibrium states that the frequency of alleles at a locus remains constant from generation to generation. In order for this to be the case, natural selection cannot affect the alleles under consideration. All other answer choices describe conditions that do need to be met for Hardy-Weinberg equilibrium to be displayed. Note that the conditions for Hardy-Weinberg equilibrium are not met in nature.
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An isolated population consists of 10 males and 10 females. Two individuals are carriers of the recessive blue eye allele. Assuming all Hardy-Weinberg conditions are met. What is the frequency of the blue eye phenotype in the population?
An isolated population consists of 10 males and 10 females. Two individuals are carriers of the recessive blue eye allele. Assuming all Hardy-Weinberg conditions are met. What is the frequency of the blue eye phenotype in the population?
Use the two Hardy-Weinberg equations:
Above, 
is the frequency of the dominant allele, and 
is the frequency of the recessive allele in the isolated population.
Since there are 20 people in total on the island, that means that there are 40 alleles for eye color. 2 of the 40 are for the blue allele:
We are looking for the blue eye phenotype, which can only result from two recessive alleles.
Use the two Hardy-Weinberg equations:
Above,
Since there are 20 people in total on the island, that means that there are 40 alleles for eye color. 2 of the 40 are for the blue allele:
We are looking for the blue eye phenotype, which can only result from two recessive alleles.
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Within his rat population, a scientist is trying to generate twice as many recessive homozygotes as heterozygotes. What allelic frequency would accomplish this?
Within his rat population, a scientist is trying to generate twice as many recessive homozygotes as heterozygotes. What allelic frequency would accomplish this?
Use the Hardy-Weinberg equations:
The equation he will need to set up is the following:
Solve for 
and substitute the first equation into the equation above.
Simplify.
Lastly, find 
.
Use the Hardy-Weinberg equations:
The equation he will need to set up is the following:
Solve for
Simplify.
Lastly, find
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Assuming Hardy-Weinberg equilibrium conditions, what are the heterozygote (Bb) and homozygote recessive (bb) genotypes for a gene if the homozygote dominant (BB) genotype is 0.45?
Assuming Hardy-Weinberg equilibrium conditions, what are the heterozygote (Bb) and homozygote recessive (bb) genotypes for a gene if the homozygote dominant (BB) genotype is 0.45?
The correct answer is Bb = 0.44 and bb = 0.11.
Since we know BB = 0.45 and the equations for allele frequencies when Hardy-Weinberg equilibrium conditions are met:
and
We solve for B first:
Now we can solve for the homozygote recessive.
Lastly, solve for the heterozygote.
The correct answer is Bb = 0.44 and bb = 0.11.
Since we know BB = 0.45 and the equations for allele frequencies when Hardy-Weinberg equilibrium conditions are met:
and
We solve for B first:
Now we can solve for the homozygote recessive.
Lastly, solve for the heterozygote.
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Which of the following is not an assumption of the Hardy-Weinberg equilibrium?
Which of the following is not an assumption of the Hardy-Weinberg equilibrium?
Non-random mating is not an assumption of the Hardy-Weinberg equilibrium, in fact, in order to make predictions about the next generation, random mating must be assumed. Additionally, no new mutations, no gene flow, no genetic drift, and no natural selection must also occur. If any of these phenomenon are present in a population, we can not estimate allele frequencies in subsequent generations due to chance, rather selective pressures may favor one allele over another allele.
Non-random mating is not an assumption of the Hardy-Weinberg equilibrium, in fact, in order to make predictions about the next generation, random mating must be assumed. Additionally, no new mutations, no gene flow, no genetic drift, and no natural selection must also occur. If any of these phenomenon are present in a population, we can not estimate allele frequencies in subsequent generations due to chance, rather selective pressures may favor one allele over another allele.
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What are the phenotypic ratios for a given population for which the proportion of the dominant allele is 0.55 and that of the recessive allele is 0.45?
What are the phenotypic ratios for a given population for which the proportion of the dominant allele is 0.55 and that of the recessive allele is 0.45?
To solve this problem, assume Hardy-Weinberg equilibrium and use the associated equations to solve:
is dominant allele and 
is recessive allele
To find the phenotype ratios:
homozygous dominant
heterozygous
homozygous recessive
To solve this problem, assume Hardy-Weinberg equilibrium and use the associated equations to solve:
To find the phenotype ratios:
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Which of the following is not a tenet of Hardy-Weinberg equilibrium?
Which of the following is not a tenet of Hardy-Weinberg equilibrium?
The Hardy-Weinberg equilibrium does not account for genetic drift. The Hardy-Weinberg law states that genetic frequencies will remain constant in a population from generation to generation in the absence of evolutionary influences. Therefore, there is no migration, natural selection, nonrandom mating, or small populations in a Hardy-Weinberg population.
The Hardy-Weinberg equilibrium does not account for genetic drift. The Hardy-Weinberg law states that genetic frequencies will remain constant in a population from generation to generation in the absence of evolutionary influences. Therefore, there is no migration, natural selection, nonrandom mating, or small populations in a Hardy-Weinberg population.
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In a species of ant, black coloration is dominant to white. A scientist is operating under the assumption that this gene follows basic Mendelian principles; however, after crossing two heterozygotes he obtained a ratio of 2:1 of dominant to recessive offspring. Which of the following could explain this result?
In a species of ant, black coloration is dominant to white. A scientist is operating under the assumption that this gene follows basic Mendelian principles; however, after crossing two heterozygotes he obtained a ratio of 2:1 of dominant to recessive offspring. Which of the following could explain this result?
The only answer that properly explains a 2:1 ratio is if the homozygous dominant phenotype is lethal. In our punnett square, this would give a two-thirds chance for a heterozygous offspring and a one-third chance for a recessive offspring. We know the recessive phenotype is not lethal because homozygous recessive offspring were produced.
Parents: Aa x Aa
Offspring genotypes: AA, Aa, Aa, aa
Offspring phenotypes: lethal, dominant, dominant, recessive (only three live offspring produced)
Offspring ratio: 2 dominant to 1 recessive
There is also no evidence that codominance or incomplete dominance is present. If black and white spotted offspring or gray offspring were produced then these theories would have merit.
The only answer that properly explains a 2:1 ratio is if the homozygous dominant phenotype is lethal. In our punnett square, this would give a two-thirds chance for a heterozygous offspring and a one-third chance for a recessive offspring. We know the recessive phenotype is not lethal because homozygous recessive offspring were produced.
Parents: Aa x Aa
Offspring genotypes: AA, Aa, Aa, aa
Offspring phenotypes: lethal, dominant, dominant, recessive (only three live offspring produced)
Offspring ratio: 2 dominant to 1 recessive
There is also no evidence that codominance or incomplete dominance is present. If black and white spotted offspring or gray offspring were produced then these theories would have merit.
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A scientist has been working with a new species of plant. He has found that there are two separate genes, which segregate according to standard Mendelian genetics, that are capable of producing the same phenotype. A single dominant allele from either gene confers red coloration of the plant's flowers. Without any dominant alleles the flowers are white. If he crosses two plants heterozygous for both traits, what will be the resulting phenotypic ratios of the offspring?
A scientist has been working with a new species of plant. He has found that there are two separate genes, which segregate according to standard Mendelian genetics, that are capable of producing the same phenotype. A single dominant allele from either gene confers red coloration of the plant's flowers. Without any dominant alleles the flowers are white. If he crosses two plants heterozygous for both traits, what will be the resulting phenotypic ratios of the offspring?
This problem requires a standard dihybrid cross. The crossed genotypes are AaBb x AaBb. This results in a phenotypic ratio of 9 dominant for both traits, 3 dominant for a single trait, 3 dominant for the other trait, and 1 recessive for both traits. In this cross, it will result in: 9 AxBx, 3 Axbb, 3 aaBx, and 1 aabb.
Since we know that the genes are both capable of making the red coloration we actually need to add together all of the choices that contain at least a single dominant allele. Essentially, AxBx, Axbb, and aaBx all show the exact same phenotype. This leaves us with a 15:1 ratio of red to white flowers.
This problem requires a standard dihybrid cross. The crossed genotypes are AaBb x AaBb. This results in a phenotypic ratio of 9 dominant for both traits, 3 dominant for a single trait, 3 dominant for the other trait, and 1 recessive for both traits. In this cross, it will result in: 9 AxBx, 3 Axbb, 3 aaBx, and 1 aabb.
Since we know that the genes are both capable of making the red coloration we actually need to add together all of the choices that contain at least a single dominant allele. Essentially, AxBx, Axbb, and aaBx all show the exact same phenotype. This leaves us with a 15:1 ratio of red to white flowers.
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A child is curious to know what his blood type is, but he only knows his parents' blood types. If his mother had blood type A and his father had blood type AB, what are the potential blood types the child might have?
A child is curious to know what his blood type is, but he only knows his parents' blood types. If his mother had blood type A and his father had blood type AB, what are the potential blood types the child might have?
We are not given the mother's full genotype in the question; she could reasonably carry two A alleles, or an A allele and a recessive O allele. We know that the father must carry one copy of the A allele and one copy of the B allele.
Two punnett squares can answer this question, corresponding to the two possible maternal genotypes: one crossing AA x AB and the other crossing AO x AB. From the first cross there is a 50% chance of blood type A versus 50% chance of blood type AB (half AA and half AB). The second cross shows that there is a potential chance of 50% for type A, 25% for type AB, and 25% for type B (one AA, one OA, one AB, and one OB).
Based on these possibilities, the child could have blood type A, B, or AB. The child cannot have blood type O.
We are not given the mother's full genotype in the question; she could reasonably carry two A alleles, or an A allele and a recessive O allele. We know that the father must carry one copy of the A allele and one copy of the B allele.
Two punnett squares can answer this question, corresponding to the two possible maternal genotypes: one crossing AA x AB and the other crossing AO x AB. From the first cross there is a 50% chance of blood type A versus 50% chance of blood type AB (half AA and half AB). The second cross shows that there is a potential chance of 50% for type A, 25% for type AB, and 25% for type B (one AA, one OA, one AB, and one OB).
Based on these possibilities, the child could have blood type A, B, or AB. The child cannot have blood type O.
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A breeder performs a standard dihybrid cross between two plants that are heterozygous for both traits in question. How many unique genotypes could be present in the resulting offspring?
A breeder performs a standard dihybrid cross between two plants that are heterozygous for both traits in question. How many unique genotypes could be present in the resulting offspring?
There are nine distinct genotypes present after a standard dihybrid cross. This question can easily be answered by setting up a Punnett square (AaBb x AaBb) and counting the number of unique genotypes present after doing the cross. The numbers also conveniently work out that however many offspring display the dominant phenotype is equal to the number to of genotypes present (this is true for monohybrid and trihybrid crosses as well).
There are nine distinct genotypes present after a standard dihybrid cross. This question can easily be answered by setting up a Punnett square (AaBb x AaBb) and counting the number of unique genotypes present after doing the cross. The numbers also conveniently work out that however many offspring display the dominant phenotype is equal to the number to of genotypes present (this is true for monohybrid and trihybrid crosses as well).
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A scientist is performing a monohybrid homozygous cross: tall plants crossed with short plants. What fraction of the F2 generation are homozygous tall?
A scientist is performing a monohybrid homozygous cross: tall plants crossed with short plants. What fraction of the F2 generation are homozygous tall?
A monohybrid cross between two homozygous plants would involve a parental generation that looked like this: SS (tall) x ss (short). The F1 generation would produce only heterozygous tall plants (Ss). The F2 generation would produce offspring from the following cross: Ss x Ss. A punnett square would reveal that the F2 generation would have 25% homozygous tall (SS), 50% heterozygous tall (Ss), and 25% homozygous short plants. Note that we do not need information regarding which trait is dominant in this case, and we would still get the correct answer if we took short as the dominant phenotype.
A monohybrid cross between two homozygous plants would involve a parental generation that looked like this: SS (tall) x ss (short). The F1 generation would produce only heterozygous tall plants (Ss). The F2 generation would produce offspring from the following cross: Ss x Ss. A punnett square would reveal that the F2 generation would have 25% homozygous tall (SS), 50% heterozygous tall (Ss), and 25% homozygous short plants. Note that we do not need information regarding which trait is dominant in this case, and we would still get the correct answer if we took short as the dominant phenotype.
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In apple trees, the allele for white blossoms is dominant over the allele for pink blossoms. Two trees heterozygous for this gene are crossed. What is the phenotype ratio of the offspring?
In apple trees, the allele for white blossoms is dominant over the allele for pink blossoms. Two trees heterozygous for this gene are crossed. What is the phenotype ratio of the offspring?
The crossing of two heterozygous parents will yield 1 homozyougs dominant offspring, 1 homozygous recessive offspring and 2 heterozygous offspring. Since white blossoms is the dominant allele, the heterozygous offspring will be white leading to a phenotypic ratio of 3 white : 1 pink blossom.
The crossing of two heterozygous parents will yield 1 homozyougs dominant offspring, 1 homozygous recessive offspring and 2 heterozygous offspring. Since white blossoms is the dominant allele, the heterozygous offspring will be white leading to a phenotypic ratio of 3 white : 1 pink blossom.
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Assume complete dominance inheritance for the following question.
A pure-breeding red flower is mated with a pure-breeding white flower. All offspring are red in color; this is the F1 generation. Two of these offspring flowers are then mated with one another, and have F2 offspring.
Which of the following is true of the F2 offspring?
Assume complete dominance inheritance for the following question.
A pure-breeding red flower is mated with a pure-breeding white flower. All offspring are red in color; this is the F1 generation. Two of these offspring flowers are then mated with one another, and have F2 offspring.
Which of the following is true of the F2 offspring?
Since we had pure-breeding parents (also known as homozygotes for their respective colors), we can safely say the F1 offspring are heterozygotes and have a red allele and a white allele. When crossing these offspring with one another, we will expect to get a 3:1 ratio of red to white flowers. Not all flowers will be red, but 75% of the flowers will be.
Shown below is the punnett square that reflects this conclusion. (Note that "A" represents a red allele and "a" represents a white allele):
A a
A AA (red) Aa (red)
a Aa (red) aa (white)
Since we had pure-breeding parents (also known as homozygotes for their respective colors), we can safely say the F1 offspring are heterozygotes and have a red allele and a white allele. When crossing these offspring with one another, we will expect to get a 3:1 ratio of red to white flowers. Not all flowers will be red, but 75% of the flowers will be.
Shown below is the punnett square that reflects this conclusion. (Note that "A" represents a red allele and "a" represents a white allele):
A a
A AA (red) Aa (red)
a Aa (red) aa (white)
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Which of the following are examples of codominance?
I. A person with blood type AB
II. A flower that displays a red and white spotted phenotype (both colors are attributed to the same gene; homozygosity for either color makes a flower that is completely red or white)
III. A flower that displays a pink phenotype (a homozygous dominant flower is red and a homozygous recessive flower is white)
IV. An organism whose heterozygous phenotype is identical to the homozygous dominant phenotype
Which of the following are examples of codominance?
I. A person with blood type AB
II. A flower that displays a red and white spotted phenotype (both colors are attributed to the same gene; homozygosity for either color makes a flower that is completely red or white)
III. A flower that displays a pink phenotype (a homozygous dominant flower is red and a homozygous recessive flower is white)
IV. An organism whose heterozygous phenotype is identical to the homozygous dominant phenotype
Codominance occurs when both phenotypes are displayed equally and independently in the phenotype (without blending). This is the case with blood type and the red and white spotted flower. A person with blood type AB expresses proteins that will recognize both type A and type B. The red and white spotted flower equally expresses the two color phenotypes.
The pink flower is an example of incomplete dominance (blended phenotype). Option IV describes a normal dominant-recessive hierarchy, where only one copy of the dominant allele is needed to display the dominant phenotype.
Codominance occurs when both phenotypes are displayed equally and independently in the phenotype (without blending). This is the case with blood type and the red and white spotted flower. A person with blood type AB expresses proteins that will recognize both type A and type B. The red and white spotted flower equally expresses the two color phenotypes.
The pink flower is an example of incomplete dominance (blended phenotype). Option IV describes a normal dominant-recessive hierarchy, where only one copy of the dominant allele is needed to display the dominant phenotype.
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