MCAT Biology : Biology
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Example Question #105 : Genetics
The Hardy-Weinberg principle is a tool used to predict the allele and disease frequencies in a given population. If we know the proportion of either the recessive or dominant allele, we can calculate the proportion of population with homozygous dominant, heterozygous, and recessive genotypes using Hardy-Weinberg principles. For example, let’s assume the frequency of the dominant allele for a particular gene is 
This principle also allows us to calculate the proportion of genotypes by squaring both sides of the above equation.
Where 
A researcher is analyzing a group of indigenous people on a remote island. He observes that a rare, autosomal recessive allele is found in 25% of the population. How many people on this island are unaffected by this disease? There are 100 people living on this island.
We need to use the Hardy-Weinberg principle to solve this problem.
Where 
Since the disease is recessive, both homozygous dominant and heterozygous individuals will be unaffected by this disease. Thus we need to find 
This means that about 94 people will be unaffected (out of 100 people).
Example Question #1171 : Mcat Biological Sciences
The Hardy-Weinberg principle is a tool used to predict the allele and disease frequencies in a given population. If we know the proportion of either the recessive or dominant allele, we can calculate the proportion of population with homozygous dominant, heterozygous, and recessive genotypes using Hardy-Weinberg principles. For example, let’s assume the frequency of the dominant allele for a particular gene is 
This principle also allows us to calculate the proportion of genotypes by squaring both sides of the above equation.
Where
Which of the following is not an assumption of Hardy-Weinberg principle?
Diploid organisms
Only sexual reproduction occurs in the population
Small population size
Random mating
Small population size
Hardy-Weinberg principle is a tool used to predict the frequency of alleles in a given population; however, it is only valid if the following assumptions are met.
1). no migration or transfer of individuals
2). large population
3). few or no mutations
4). random mating
5). no natural selection
If a population meets these five assumptions, then a it is said to be in Hardy-Weinberg equilibrium. Note that this rarely, if ever, occurs in nature.
Example Question #107 : Genetics
The Hardy-Weinberg principle is a tool used to predict the allele and disease frequencies in a given population. If we know the proportion of either the recessive or dominant allele, we can calculate the proportion of population with homozygous dominant, heterozygous, and recessive genotypes using Hardy-Weinberg principles. For example, let’s assume the frequency of the dominant allele for a particular gene is 
This principle also allows us to calculate the proportion of genotypes by squaring both sides of the above equation.
Where
Which of the following is true regarding allele and disease frequency?
Disease frequency is always higher
Allele frequency is always higher
Allele and disease frequencies depend on the population size
Allele and disease frequencies depend on the type of allele
Allele and disease frequencies depend on the type of allele
Allele frequency signifies how commonly a particular allele is found in a population whereas disease frequency signifies the prevalence of the disease. Allele frequency is usually designated as 
To determine which frequency would be higher, we need to first determine whether the disease is recessive or dominant. If the disease is recessive, then the allele frequency will be higher (because the disease frequency will be 
Example Question #108 : Genetics
The Hardy-Weinberg principle is a tool used to predict the allele and disease frequencies in a given population. If we know the proportion of either the recessive or dominant allele, we can calculate the proportion of population with homozygous dominant, heterozygous, and recessive genotypes using Hardy-Weinberg principles. For example, let’s assume the frequency of the dominant allele for a particular gene is 
This principle also allows us to calculate the proportion of genotypes by squaring both sides of the above equation.
Where
In a patient, it is observed that a gene on an autosomal chromosome is silenced whereas its homologous counterpart is active. The activated gene codes for a protein that causes a particular type of disease. It is found that 2 percent of the population have this activated gene. What is the incidence of this rare disease in the population?
We need to use Hardy-Weinberg principle for this question.
Where 
The question states that 2 percent of the population have this allele. The question also states that having the disease allele in just one chromosome causes the disease (heterozygous individuals); therefore, the disease must be dominant. So,
The proportion of affected individuals is:
This means that 4 percent of the population will have this disease.
Example Question #102 : Cell Biology, Molecular Biology, And Genetics
The concept of genomic imprinting is important in human genetics. In genomic imprinting, a certain region of DNA is only expressed by one of the two chromosomes that make up a typical homologous pair. In healthy individuals, genomic imprinting results in the silencing of genes in a certain section of the maternal chromosome 15. The DNA in this part of the chromosome is "turned off" by the addition of methyl groups to the DNA molecule. Healthy people will thus only have expression of this section of chromosome 15 from paternally-derived DNA.
The two classic human diseases that illustrate defects in genomic imprinting are Prader-Willi and Angelman Syndromes. In Prader-Willi Syndrome, the section of paternal chromosome 15 that is usually expressed is disrupted, such as by a chromosomal deletion. In Angelman Syndrome, maternal genes in this section are deleted, while paternal genes are silenced. Prader-Willi Syndrome is thus closely linked to paternal inheritance, while Angelman Syndrome is linked to maternal inheritance.
Figure 1 shows the chromosome 15 homologous pair for a child with Prader-Willi Syndrome. The parental chromosomes are also shown. The genes on the mother’s chromosomes are silenced normally, as represented by the black boxes. At once, there is also a chromosomal deletion on one of the paternal chromosomes. The result is that the child does not have any genes expressed that are normally found on that region of this chromosome.
A scientist discovers another genetic disease that has similar symptoms to Prader-Willi Syndrome. He discovers that this disease is recessive, and caused not by changes to chromosome 15, but by a point mutation on chromosome 3. He calculates that 
The Hardy-Weinberg equation for this situation is as follows, where 
We want to solve for 
Example Question #1151 : Biology
A rare recessive mutation causes rabbits that are normally white to be pink. If one out of every 625 rabbits is pink, what percentage of the population is heterozygous?
We can use the Hardy-Weinberg equilibrium formulas to calculate the allele frequencies.
We know that the frequency of homozygous recessive (pink) rabbits is 
Now that we know the value of 
The frequency of heterozygotes is equal to 
Example Question #1 : Dna, Rna, And Proteins
A culture of human tissue is being grown in a lab to study mitosis. A solution containing radioactively labelled cytosines was added to the culture in the middle of prophase, and then growth was halted at the end of telophase. Where would the scientists see radioactively labelled DNA?
In the nuclei of every cell
In the cells produced at the end of telophase—only the daughter cells
In the mother cells only—not in the cells produced at the end of telophase
Only in the nuclei of half of the cells
No where
No where
DNA is replicated in S phase. Prophase is a part of mitosis, or M phase. Since all of the DNA that would be present at the end of telophase had already been synthesized in S phase, none of the radioactively labelled cytosines would be incorporated into the DNA of any cells in the culture.
Example Question #1 : Dna Replication And Repair
Which answer choice correctly matches the molecule with its function in DNA replication?
DNase—adds nucleotides to new strands
Single-stranded binding proteins—untangles supercoils
Polymerase—adds nucleotides to new strands
Primase—unzips DNA
Topoisomerase—untangles supercoils
Single-stranded binding proteins—prevents reannealing of DNA during replication
Topoisomerase—prevents reannealing of DNA during replication
Single-stranded binding proteins—untangles supercoils
Polymerase—adds RNA primers prior to replication
Primase—adds nucleotides to new strands
Topoisomerase—untangles supercoils
Single-stranded binding proteins—prevents reannealing of DNA during replication
Topoisomerase functions to untangle the supercoiling of DNA, which is when DNA overwinds into itself. This mechanism facilitates the unwinding action of helicase during replication. Single-stranded binding proteins bind to the two unzipped DNA strands to prevent them from prematurely coming back together into a whole molecule; otherwise replication would be interrupted.
The other proteins discussed serve the following functions.
Polymerase—adds nucleotides to new strands
Primase—adds RNA primers prior to replication
DNase—cleaves and degrades DNA molecules
Example Question #1 : Dna, Rna, And Proteins
Which of the following is the first to act during DNA replication?
Helicase
DNA polymerase
DNA ligase
Primase
Helicase
Helicase is the first component of the DNA replication machinery to act during replication. It works by "unzipping" the double-stranded DNA so that replication can subsequently occur. Following the work of helicase, primase creates a primer to which the DNA polymerase will subsequently add deoxynucleotides and elongate the strand. DNA ligase acts at the end of replication by joining together the Okazaki fragments of the lagging strand.
Example Question #2 : Dna Replication And Repair
The Meselson-Stahl experiment provided the necessary evidence to discover the mechanism by which DNA replicates. They accomplished this discovery by first culturing DNA with the heavy 15N nitrogen isotope. They then allowed the "heavy" DNA to replicate with DNA grown in normal 14N nitrogen. The density of each generation of replicated DNA was recorded by marking its position in a test tube after centrifugation. The position of each generation was then compared to the positions of pure 15N DNA and pure 14N DNA.
Suppose that the first generation after replication revealed two bands after being centrifuged: one at the pure 14N mark, and one at the pure 15N mark. Which method of replication would this observation support?
Dispersive replication
Semiconservative replication
Conservative replication
Another generation would be needed in order to find a viable mechanism
Conservative replication
Conservative replication proposes that both strands of DNA act as the template, but do not separate during replication. If the heavy strands were to stay together, we would expect to see a "heavy" set of DNA at the 15N mark as well as a "normal" set of DNA at the 14N mark.
Both semiconservative and dispersive replication would predict a singular band of DNA in between the two marks.
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