MCAT Biology › 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.
Chromosome 15 is an autosome. Which of the following is (are) true of all autosomes?
I. They contain histones
II. They determine chromosomal sex
III. They align on the metaphase plate during mitosis
I and III
I and II
I, II, and III
II and III
I only
Autosomes are the chromosomes that are not sex chromosomes. Any numbered chromosome (1 through 22) is an autosome, while the X and Y chromosomes (the 23rd pair) are the sex chromosomes. Statement II is only true of the X and Y chromosomes. Statements I and III are true of all chromosomes.
9% of a population of mice suffer from muscle spasms due to an autosomal recessive disease. What percentage of the population are carriers for the disease?
This question requires application of the two Hardy-Weinberg equations:
We know that the disease is autosomal recessive, which means that affected individuals must be homozygous. The frequency of the homozygous recessive genotype is given by the term in the second Hardy-Weinberg equation. We can use the given genotype frequency to solve for the recessive allele frequency.
Use the first Hardy-Weinberg equation to determine the dominant allele frequency.
The frequency of the heterozygous genotype (carriers) is given by the term of the second Hardy-Weinberg equation. Use the values of the allele frequencies to calculate this term.
Pea plants have two independently assorted genes that code for seed shape (round or wrinkled) and seed color (yellow or green), respectively. A researcher crosses two pea plants and observes that all F1 offspring have the same phenotype: round shape and yellow seeds. He then performs a test cross with an F1 offspring and observes four different phenotypes in a 1:1:1:1 ratio. Based on this information the researcher concludes the genotypes of the parents.
Which of the following is the correct pairing of recessive alleles?
Wrinkled and green
Round and green
Round and yellow
Wrinkled and yellow
The passage states that the F1 generation only had round, yellow seeds. Test crossing an F1 offspring lead to an equal ratio of four different phenotypes. When you test cross, you are crossing the F1 offspring with a homozygous recessive individual; therefore, the test cross individual had recessive seed shape and seed color.
???? x aabb
If round and yellow were recessive, then the F2 offspring would all be round and yellow (because you would cross round/yellow with round/yellow).
aabb x aabb (all offspring round and yellow)
If only round was recessive then you wouldn’t get any wrinkled seeds in F2 generation; similarly, if only yellow was recessive then you wouldn’t get any green seeds.
Aabb x aabb (all offspring have one recessive trait, bb)
If round and yellow are both dominant, then means that wrinkled and green are recessive. If the F1 offspring is heterozygous for both traits, then we can see the observed ratios from the test cross.
AaBb x aabb
1 AaBb (round/yellow), 1 aaBb (wrinkled/yellow), 1 Aabb (round/green), 1 aabb (wrinkled/green)
We can conclude that green and wrinkled must be recessive to yellow and round.
Passage:
This has seemed a fatal objection to the chromosome view, but it may not be so, as Spillman has argued, so long as it has not yet been shown that all of the dominant characters may be present at the same time. But even admitting this possible way of eluding the objection, the other point raised above concerning the absence of groupings of characters in Mendelian inheritance seems a fatal objection to the chromosome theory, so long as that theory attempts to locate each character in a special chromosome. We shall have occasion to return to this point later.
In recent years most workers in Mendelian inheritance have adopted a new method of formulating their theory. Characters that Mendelize are no longer allelomorphic to each other, but each character has for its pair the absence of that character. This is the presence and absence theory. We can apply this hypothesis to the chromosome theory. For examples, let us assume a new variety or race arises by the loss of a character from that chromosome that has heretofore carried it. The chromosome still remains in existence, since it may carry many other characters besides the one that was lost, and it becomes in the hybrid the mate of the one still retaining that character. If now separation occurs, two classes of germ-cells result, one with and the other without the character; and the observed numerical proportions follow. There is nothing in this assumption that meets with any greater difficulty on the chromosome separation hypothesis than on the earlier view of paired allelomorphs, but it meets with the same difficulties, and as an assumption is neither more nor less in accord with the postulated mechanism.
Excerpt from Morgan, T. H. 1910. Chromosomes and heredity. The American Naturalist, 44:449–496.
Asssume that in a particular studied species of rodents, coat color is an autosomally-inherited trait, with black fur being the dominant phenotype and white fur being the recessive phenotype. In an experiment, a mouse with black fur mates with a mouse with a mouse with white fur. The resultant 26 offspring mice all have grey fur color. Which of the following genetic principles is best demonstrated by this result?
Incomplete Dominance
Codominance
X-linkage
Nondisjunction
Spontaneous mutation
The genetic principle best demonstrated here is incomplete dominance. Incomplete dominance means that if a given trait has two alleles, neither allele is individually dominant to the other allele, and therefore, if an organism is heterozygous for that trait, it will display a phenotype that is intermediate between that coded for by each individual alleles.
In this case, the parent mice are most likely BB (black fur) and bb (white fur), therefore homozygotes for fur color. As a result, any offspring that they have, will have genotype Bb. If this trait was inherited in such a manner that black fur was completely dominant to white fur, or vice versa, the mice would have either one of those fur colors in the progeny. Because the resulting fur color is grey, which is intermediate between black and white, this is evidence that the genetic principle at play here is incomplete dominance.
Co-dominance is not the correct answer because a heterozygote for a co-dominant trait would demonstrate both phenotypes simultaneously (e.g. black and white fur color), rather than a blending of the individual phenotypes (e.g. grey fur color) as in these mice.
X-linkage of fur color would not describe the presence of an intermediate phenotype, nor would nondisjunction.
Spontaneous mutation could, in theory, result in a phenotype different from that of the parent mice. However, given that the resultant phenotype for the progeny mice was intermediate to that of the parents' fur colors, and the fact that all progeny demonstrated this phenotype, rather than just one or two mice, this is a more compelling demonstration of incomplete dominance.
A scientist is working with a new species of insect and is specifically observing the inheritance of two traits: eye color and antennae shape. Eye colors come in red (dominant) and white (recessive), and antennae come in long shapes (dominant) and short shapes (recessive). He performs a dihybrid cross between two insects heterozygous for both traits and observes a ratio of 3:1 (red eyes and long antennae: white eyes short antennae). Which of the following explanations most likely explains the observed ratio?
The genes for eye color and antennae length are linked
The recessive phenotypes are lethal
The insects have incredibly high rates of recombination
Application of Mendel's law of independent assortment
If Mendel's law of independent assortment applied to this case, we would expect to see a normal 9:3:3:1 phenotypic ratio of offspring. Also, we know that the recessive phenotypes are not lethal because we are told that approximately 25% of the progeny were homozygous recessive for both traits. The most likely explanation is that the genes are linked (located very closely on the same chromosome) and, thus, segregate together. Even if the insects had high rates of recombination it would not affect these genes that are, according to the offspring ratios, completely linked.
A man with type A– blood and a woman with type AB+ blood have a child. Which blood type is impossible for that child to have?
O-
B+
A+
AB+
A-
Blood type is inherited as a codominant trait and relies on alleles for blood antibodies as well as Rh (Rhesus) factor. The father's blood type is A–, so he has no Rh factor and must be either AA or AO. The mother must be AB with an Rh factor.
Father possibilities: A-A- or A-O-
Mother possibilities: A-B+ or A+B- or A+B+
Based on these possibilities, we cannot conclude if the child will be positive or negative for Rh factor; however, since the mother has no allele for O blood type, we can conclude that the child cannot have O type blood. The child could receive AA, AO, BO, or AB.
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.
Chromosome 15 is an autosome. Which of the following is (are) true of all autosomes?
I. They contain histones
II. They determine chromosomal sex
III. They align on the metaphase plate during mitosis
I and III
I and II
I, II, and III
II and III
I only
Autosomes are the chromosomes that are not sex chromosomes. Any numbered chromosome (1 through 22) is an autosome, while the X and Y chromosomes (the 23rd pair) are the sex chromosomes. Statement II is only true of the X and Y chromosomes. Statements I and III are true of all chromosomes.
9% of a population of mice suffer from muscle spasms due to an autosomal recessive disease. What percentage of the population are carriers for the disease?
This question requires application of the two Hardy-Weinberg equations:
We know that the disease is autosomal recessive, which means that affected individuals must be homozygous. The frequency of the homozygous recessive genotype is given by the term in the second Hardy-Weinberg equation. We can use the given genotype frequency to solve for the recessive allele frequency.
Use the first Hardy-Weinberg equation to determine the dominant allele frequency.
The frequency of the heterozygous genotype (carriers) is given by the term of the second Hardy-Weinberg equation. Use the values of the allele frequencies to calculate this term.
Passage:
This has seemed a fatal objection to the chromosome view, but it may not be so, as Spillman has argued, so long as it has not yet been shown that all of the dominant characters may be present at the same time. But even admitting this possible way of eluding the objection, the other point raised above concerning the absence of groupings of characters in Mendelian inheritance seems a fatal objection to the chromosome theory, so long as that theory attempts to locate each character in a special chromosome. We shall have occasion to return to this point later.
In recent years most workers in Mendelian inheritance have adopted a new method of formulating their theory. Characters that Mendelize are no longer allelomorphic to each other, but each character has for its pair the absence of that character. This is the presence and absence theory. We can apply this hypothesis to the chromosome theory. For examples, let us assume a new variety or race arises by the loss of a character from that chromosome that has heretofore carried it. The chromosome still remains in existence, since it may carry many other characters besides the one that was lost, and it becomes in the hybrid the mate of the one still retaining that character. If now separation occurs, two classes of germ-cells result, one with and the other without the character; and the observed numerical proportions follow. There is nothing in this assumption that meets with any greater difficulty on the chromosome separation hypothesis than on the earlier view of paired allelomorphs, but it meets with the same difficulties, and as an assumption is neither more nor less in accord with the postulated mechanism.
Excerpt from Morgan, T. H. 1910. Chromosomes and heredity. The American Naturalist, 44:449–496.
Asssume that in a particular studied species of rodents, coat color is an autosomally-inherited trait, with black fur being the dominant phenotype and white fur being the recessive phenotype. In an experiment, a mouse with black fur mates with a mouse with a mouse with white fur. The resultant 26 offspring mice all have grey fur color. Which of the following genetic principles is best demonstrated by this result?
Incomplete Dominance
Codominance
X-linkage
Nondisjunction
Spontaneous mutation
The genetic principle best demonstrated here is incomplete dominance. Incomplete dominance means that if a given trait has two alleles, neither allele is individually dominant to the other allele, and therefore, if an organism is heterozygous for that trait, it will display a phenotype that is intermediate between that coded for by each individual alleles.
In this case, the parent mice are most likely BB (black fur) and bb (white fur), therefore homozygotes for fur color. As a result, any offspring that they have, will have genotype Bb. If this trait was inherited in such a manner that black fur was completely dominant to white fur, or vice versa, the mice would have either one of those fur colors in the progeny. Because the resulting fur color is grey, which is intermediate between black and white, this is evidence that the genetic principle at play here is incomplete dominance.
Co-dominance is not the correct answer because a heterozygote for a co-dominant trait would demonstrate both phenotypes simultaneously (e.g. black and white fur color), rather than a blending of the individual phenotypes (e.g. grey fur color) as in these mice.
X-linkage of fur color would not describe the presence of an intermediate phenotype, nor would nondisjunction.
Spontaneous mutation could, in theory, result in a phenotype different from that of the parent mice. However, given that the resultant phenotype for the progeny mice was intermediate to that of the parents' fur colors, and the fact that all progeny demonstrated this phenotype, rather than just one or two mice, this is a more compelling demonstration of incomplete dominance.
A scientist is working with a new species of insect and is specifically observing the inheritance of two traits: eye color and antennae shape. Eye colors come in red (dominant) and white (recessive), and antennae come in long shapes (dominant) and short shapes (recessive). He performs a dihybrid cross between two insects heterozygous for both traits and observes a ratio of 3:1 (red eyes and long antennae: white eyes short antennae). Which of the following explanations most likely explains the observed ratio?
The genes for eye color and antennae length are linked
The recessive phenotypes are lethal
The insects have incredibly high rates of recombination
Application of Mendel's law of independent assortment
If Mendel's law of independent assortment applied to this case, we would expect to see a normal 9:3:3:1 phenotypic ratio of offspring. Also, we know that the recessive phenotypes are not lethal because we are told that approximately 25% of the progeny were homozygous recessive for both traits. The most likely explanation is that the genes are linked (located very closely on the same chromosome) and, thus, segregate together. Even if the insects had high rates of recombination it would not affect these genes that are, according to the offspring ratios, completely linked.