Mendel and Inheritance Patterns
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MCAT Biology › Mendel and Inheritance Patterns
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-
Explanation
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.
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
Explanation
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.
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
Explanation
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.
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
Explanation
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.
Consider the pedigree of a recessive trait. Is the trait autosomal or sex-linked?
X-linked
Autosomal
The mode of inheritance cannot be determined
Y-linked
Explanation
This trait is X-linked, which can be determined by considering individual 4. This individual has a 50% chance of receiving an affected maternal chromosome (she is a carrier) and a 0% chance of receiving an affected paternal chromosome (he is unaffected), and yet he has the defect. We know that the son inherits the Y-chromosome from his father, and a singular X-chromosome from his mother; thus, we can assume he inherits the affected chromosome from his mother. He has no dominant X-chromosome to veil the trail, thus he expresses the trait despite it being recessive. This pattern indicates that the trait resides on the X-chromosome.
In peas, the gene for yellow color (C) is dominant to the gene for green color (c). To determine the genotype of an unknown yellow pea, what kind of pea should you cross with it?
Homozygous recessive (cc) or known heterozygous (Cc)
Homozygous dominant
Any genotype
It is impossible to determine
Another unknown green pea
Explanation
We are unsure if our yellow pea is homozygous dominant (CC) or heterozygous (Cc). Crossing it with a homozygous recessive (cc) green pea will yield only yellow peas if it is homozygous dominant, or a mix of green and yellow if it is heterozygous. We could also cross it with a known heterozygote, as we would see the same pattern as if we had crossed with a homozygous recessive.
Duchenne Muscular Dystrophy is an X-linked recessive genetic disorder, resulting in the loss of the dystrophin protein. In healthy muscle, dystrophin localizes to the sarcolemma and helps anchor the muscle fiber to the basal lamina. The loss of this protein results in progressive muscle weakness, and eventually death.
In the muscle fibers, the effects of the disease can be exacerbated by auto-immune interference. Weakness of the sarcolemma leads to damage and tears in the membrane. The body’s immune system recognizes the damage and attempts to repair it. However, since the damage exists as a chronic condition, leukocytes begin to present the damaged protein fragments as antigens, stimulating a targeted attack on the damaged parts of the muscle fiber. The attack causes inflammation, fibrosis, and necrosis, further weakening the muscle.
Studies have shown that despite the severe pathology of the muscle fibers, the innervation of the muscle is unaffected.
A young girl is diagnosed with Duchenne Muscular Dystrophy, and her mother is pregnant with a baby boy. Which of the following must be true?
Her father had the disease
Her mother had the disease
Her father's mother was a carrier
Her father's father had the disease
Her brother will have the disease
Explanation
The passage tells us that the disease gene is X-linked and recessive. If the young girl has the disease, then she must have a double recessive genotype, meaning that both of her parents have at least one recessive gene. Her mother must either be a carrier, or have the disease, but we cannot conclude one over the other. Her father must have the disease, since he only has one copy of the X-chromosome. He received his Y chromosome from his father, so we cannot make any judgments about the father's father. He received his X chromosome from his mother, meaning that she must have been a carrier or had the disease. As far as the unborn brother is concerned, he has a 50% chance of getting the disease if the mother is a carrier, and 100% if she has the disease.
The only thing we can say with 100% certainty is that the father has the disease.
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 true regarding a recessive allele?
In a heterozygous offspring the dominant allele is expressed and the recessive allele is suppressed
In a heterozygous offspring both the dominant and recessive alleles are expressed, but dominant allele is expressed more frequently
In a homozygous recessive offspring both recessive alleles are on the same chromosome
In a heterozygous offspring the dominant and the recessive alleles are on the same chromosome
Explanation
A Mendelian gene generally has two types of alleles: dominant and recessive. An individual always contains two sets of chromosomes (homologous chromosomes) that contain the gene. If both chromosomes contain a dominant allele, then the dominant trait is expressed. If both contain the recessive allele, then the recessive trait is expressed.
If an individual carries both alleles (a heterozygous individual), then only the dominant trait is observed. This occurs because the recessive allele is silenced and is not expressed in the presence of a dominant allele. If the recessive allele is expressed in conjunction with the dominant allele, then it is called incomplete dominance. In normal heterozygous genes, however, the recessive allele is not expressed.
Each chromosome carries only one copy of each gene, and cannot accommodate two alleles for a single trait.
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.
The parents of the child in Figure 1 decide to have a second child. This child is diagnosed at birth with both cystic fibrosis and Prader-Willi. Cystic fibrosis is an autosomal recessive disorder inherited on chromosome 7. The mother and father are both healthy. What was the probability of the child being born with both Prader-Willi and cystic fibrosis?
Explanation
To be diagnosed with both cystic fibrosis and Prader-Willi, this child would have inherited either of the maternal alleles and the deleted paternal allele (50% chance of this happening). If the mother's chromosome 15 genotype is said to be AA and the father is Aa, where a is the chromosomal deletion, we can see that the possible combinations are: AA, AA, Aa, Aa. There is a 50% chance of inheriting the chromosome with the deletion.
He would have also had to inherit both dysfunctional copies of chromosome 7. Since both parents were healthy, they must be heterozygous carriers. The child has a 25% chance of inheriting one damaged chromosome from each parent. If each parent has genotype Cc, we can see that the possible offspring are: CC, Cc, Cc, cc. There is a 25% chance of inheriting both recessive alleles for cystic fibrosis.
Combining these probabilities tells us the chance of inheriting both disorders.
Suppose two individuals with the genotypes PPQqRr and ppQqrr mate. What is the probability that their offspring will display the genotype PpQQRr?
1/8
1/2
1/4
1/16
1/32
Explanation
While this question looks complicated it can be broken down into three individual Punnett squares, and the probability of each of the three individual allele genotypes can be multiplied to give the desired answer.
Starting with P: Parent 1 is homozygous dominant (PP) and parent 2 is homozygous recessive (pp), therefore 100% of their offspring will be heterozygous (Pp), or 1/1.
For Q: Both parents are heterozygous (Qq); therefore 1/4 of their offspring will be QQ, 1/2 will be Qq, and 1/4 will be qq.
For R: Parent 1 is heterozygous (Rr) and parent 2 is homozygous recessive (rr); therefore 1/2 their offspring will be Rr and 1/2 will be rr.
Because the question asks the probability that the offspring is PpQQRr, we can multiply 1/1 * 1/4 * 1/2 = 1/8.