Genetics - MCAT Biological and Biochemical Foundations of Living Systems
Card 1 of 888
Cellular division is an essential part of the cell cycle. When a cell divides it passes genetic information to daughter cells. The amount of genetic information passed on to daughter cells depends on whether the cell undergoes mitosis or meiosis. Mitosis is the most common form of cell division. All somatic cells undergo mitosis, whereas only germ cells undergo meiosis. Meiosis is very important because it produces gametes (sperm and eggs) that are required for sexual reproduction. Human germ cells have 46 chromosomes (2n = 46) and undergo meiosis to produce four haploid daughter cells (gametes).
An individual containing three sex chromosomes (XXY) is called a polysomic individual. What is the reason for polysomy?
Cellular division is an essential part of the cell cycle. When a cell divides it passes genetic information to daughter cells. The amount of genetic information passed on to daughter cells depends on whether the cell undergoes mitosis or meiosis. Mitosis is the most common form of cell division. All somatic cells undergo mitosis, whereas only germ cells undergo meiosis. Meiosis is very important because it produces gametes (sperm and eggs) that are required for sexual reproduction. Human germ cells have 46 chromosomes (2n = 46) and undergo meiosis to produce four haploid daughter cells (gametes).
An individual containing three sex chromosomes (XXY) is called a polysomic individual. What is the reason for polysomy?
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The question states that the individual with three sex chromosomes has a condition called polysomy. Recall that a normal individual will only carry two copies of a chromosome.
This abnormality occurs when meiosis isn’t carried out properly. A daughter cell from meiosis can contain an extra chromosome if sister chromatids don’t separate properly during anaphase II (a process called nondisjunction). This extra chromosome can be carried over to the offspring, giving rise to a polysomic individual.
Fertilization of two sperm cells with two eggs gives rise to fraternal twins (non-identical twins) and splitting of a single fertilized egg into two or more eggs gives rise to monozygotic twins (identical twins); therefore, you can eliminate these two answer choices. After meiosis, females always possess multiple polar bodies. During meiosis in females, most of the cellular content is transferred to a single daughter cell: the egg. The remaining daughter cells contain the remnants and are called polar bodies. These polar bodies don’t participate in development and fertilization.
The question states that the individual with three sex chromosomes has a condition called polysomy. Recall that a normal individual will only carry two copies of a chromosome.
This abnormality occurs when meiosis isn’t carried out properly. A daughter cell from meiosis can contain an extra chromosome if sister chromatids don’t separate properly during anaphase II (a process called nondisjunction). This extra chromosome can be carried over to the offspring, giving rise to a polysomic individual.
Fertilization of two sperm cells with two eggs gives rise to fraternal twins (non-identical twins) and splitting of a single fertilized egg into two or more eggs gives rise to monozygotic twins (identical twins); therefore, you can eliminate these two answer choices. After meiosis, females always possess multiple polar bodies. During meiosis in females, most of the cellular content is transferred to a single daughter cell: the egg. The remaining daughter cells contain the remnants and are called polar bodies. These polar bodies don’t participate in development and fertilization.
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Which of the following mutations might lead to the formation of a recessive allele?
I. Silent
II. Frameshift
III. Nonsense
IV. Missense
Which of the following mutations might lead to the formation of a recessive allele?
I. Silent
II. Frameshift
III. Nonsense
IV. Missense
Tap to reveal answer
Recessive alleles are often created by mutated versions of functional genes that encode broken/nonfunctional proteins. The question is thus asking us which of the mutations is likely to result in a nonfunctional or abnormal protein.
Silent mutations are types of mutations that result in the insertion of the same amino acid due to the degeneracy of the genetic code and, therefore, will not cause any noticeable change in the protein. Essentially, even organisms with the mutation will still show the dominant allele.
Frameshift mutations change the reading frame used for translation and oftentimes result in premature stop codons. Nonsense mutations are mutations that specifically result in premature stop codons and generally result in nonfunctional proteins. Missense mutations lead to the insertion of a different amino acid at the normal site. This can lead to serious problems for the functionality of the protein, particularly if the mutation is present in the active site or another area that is highly conserved throughout evolution.
Recessive alleles are often created by mutated versions of functional genes that encode broken/nonfunctional proteins. The question is thus asking us which of the mutations is likely to result in a nonfunctional or abnormal protein.
Silent mutations are types of mutations that result in the insertion of the same amino acid due to the degeneracy of the genetic code and, therefore, will not cause any noticeable change in the protein. Essentially, even organisms with the mutation will still show the dominant allele.
Frameshift mutations change the reading frame used for translation and oftentimes result in premature stop codons. Nonsense mutations are mutations that specifically result in premature stop codons and generally result in nonfunctional proteins. Missense mutations lead to the insertion of a different amino acid at the normal site. This can lead to serious problems for the functionality of the protein, particularly if the mutation is present in the active site or another area that is highly conserved throughout evolution.
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Which of the following mutations will not change the amino acid sequence of the polypeptide created during translation?
Which of the following mutations will not change the amino acid sequence of the polypeptide created during translation?
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A silent mutation is a mutation that results in the inclusion of the same amino acid in the polypeptide product, despite a change in the DNA template sequence. Silent mutations generally involve the third nucleotide in the codon sequence, since this unit is often interchangeable due to the degeneracy of the coding sequence.
A missense mutation results in a different amino acid inclusion based on codon alteration. A nonsense mutation is the insertion of a new stop codon as a result of mutation, resulting in a truncated polypeptide and severely hindering functionality. An insertion or frameshift mutation can change the reading frame of the mRNA template, drastically changing the identity and primary sequence of the polypeptide product; these are the most dangerous type of mutation, as the product may be harmful as well as nonfunctional.
A silent mutation is a mutation that results in the inclusion of the same amino acid in the polypeptide product, despite a change in the DNA template sequence. Silent mutations generally involve the third nucleotide in the codon sequence, since this unit is often interchangeable due to the degeneracy of the coding sequence.
A missense mutation results in a different amino acid inclusion based on codon alteration. A nonsense mutation is the insertion of a new stop codon as a result of mutation, resulting in a truncated polypeptide and severely hindering functionality. An insertion or frameshift mutation can change the reading frame of the mRNA template, drastically changing the identity and primary sequence of the polypeptide product; these are the most dangerous type of mutation, as the product may be harmful as well as nonfunctional.
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Which of the following DNA changes is least likely to alter the translated amino acid sequence created by the sequence below?
5'-CGATGAACG-3'
Which of the following DNA changes is least likely to alter the translated amino acid sequence created by the sequence below?
5'-CGATGAACG-3'
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First, you want to narrow down the answers to those with the fewest changes (1 base change). Next, you needed to select the single base change on the third base of the codon. A silent mutation (a DNA mutation that doesn't result in an amino acid change) is most likely to occur when the third base of a codon is changed.
Original sequence: 5'-CGA TGA ACG-3'
Unaltered mRNA transcript: 3'-GCU ACU UGC-5'
Note that the mRNA sequence is antiparallel, changing the orientation of the codons. mRNA is read in the 5'-to-3' direction by the ribosome, meaning that the third base in the codon will be to the left of the 3-base sequence. The given RNA sequence will be read as CGU UCA UCG.
Correct answer: 5'-CGA CGA ACG-3'
Altered mRNA transcript: 3'-GCU GCU UGC-5'
The sequence will be read as CGU UCG UCG.
The correct answer involves a mutation in the third nitrogenous base of the second codon. Both UCA (original) and UCG (altered) code for serine.
First, you want to narrow down the answers to those with the fewest changes (1 base change). Next, you needed to select the single base change on the third base of the codon. A silent mutation (a DNA mutation that doesn't result in an amino acid change) is most likely to occur when the third base of a codon is changed.
Original sequence: 5'-CGA TGA ACG-3'
Unaltered mRNA transcript: 3'-GCU ACU UGC-5'
Note that the mRNA sequence is antiparallel, changing the orientation of the codons. mRNA is read in the 5'-to-3' direction by the ribosome, meaning that the third base in the codon will be to the left of the 3-base sequence. The given RNA sequence will be read as CGU UCA UCG.
Correct answer: 5'-CGA CGA ACG-3'
Altered mRNA transcript: 3'-GCU GCU UGC-5'
The sequence will be read as CGU UCG UCG.
The correct answer involves a mutation in the third nitrogenous base of the second codon. Both UCA (original) and UCG (altered) code for serine.
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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?
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?
Tap to reveal answer
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.
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.
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Prions are the suspected cause of a wide variety of neurodegenerative diseases in mammals. According to prevailing theory, prions are infectious particles made only of protein and found in high concentrations in the brains of infected animals. All mammals produce normal prion protein, PrPC, a transmembrane protein whose function remains unclear.
Infectious prions, PrPRes, induce conformational changes in the existing PrPC proteins according to the following reaction:
PrPC + PrPRes → PrPRes + PrPRes
The PrPRes is then suspected to accumulate in the nervous tissue of infected patients and cause disease. This model of transmission generates replicated proteins, but does so bypassing the standard model of the central dogma of molecular biology. Transcription and translation apparently do not play a role in this replication process.
This theory is a major departure from previously established biological dogma. A scientist decides to test the protein-only theory of prion propagation. He establishes his experiment as follows:
Homogenized brain matter of infected rabbits is injected into the brains of healthy rabbits, as per the following table:
Rabbit 1 and 2: injected with normal saline on days 1 and 2
The above trials serve as controls.
Rabbit 3 and 4: injected with homogenized brain matter on days 1 and 2
The above trials use unmodified brain matter.
Rabbit 5 and 6: injected with irradiated homogenized brain matter on days 1 and 2
The above trials use brain matter that has been irradiated to destroy nucleic acids in the homogenate.
Rabbit 7 and 8: injected with protein-free centrifuged homogenized brain matter on days 1 and 2
The above trials use brain matter that has been centrifuged to generate a protein-free homogenate and a protein-rich homogenate based on molecular weight.
Rabbit 9 and 10: injected with boiled homogenized brain matter on days 1 and 2
The above trials use brain matter that have been boiled to destroy any bacterial contaminants in the homogenate.
Some patients have a predisposition to developing neurodegenerative disease due to genetic mutations in the PrPC gene. If a father knows that he has the mutation, but never develops the disease, and the mother knows she is not a carrier, what is the probability that their son will develop the disease? Assume that this disease follows simple Mendelian inheritance patterns.
Prions are the suspected cause of a wide variety of neurodegenerative diseases in mammals. According to prevailing theory, prions are infectious particles made only of protein and found in high concentrations in the brains of infected animals. All mammals produce normal prion protein, PrPC, a transmembrane protein whose function remains unclear.
Infectious prions, PrPRes, induce conformational changes in the existing PrPC proteins according to the following reaction:
PrPC + PrPRes → PrPRes + PrPRes
The PrPRes is then suspected to accumulate in the nervous tissue of infected patients and cause disease. This model of transmission generates replicated proteins, but does so bypassing the standard model of the central dogma of molecular biology. Transcription and translation apparently do not play a role in this replication process.
This theory is a major departure from previously established biological dogma. A scientist decides to test the protein-only theory of prion propagation. He establishes his experiment as follows:
Homogenized brain matter of infected rabbits is injected into the brains of healthy rabbits, as per the following table:
Rabbit 1 and 2: injected with normal saline on days 1 and 2
The above trials serve as controls.
Rabbit 3 and 4: injected with homogenized brain matter on days 1 and 2
The above trials use unmodified brain matter.
Rabbit 5 and 6: injected with irradiated homogenized brain matter on days 1 and 2
The above trials use brain matter that has been irradiated to destroy nucleic acids in the homogenate.
Rabbit 7 and 8: injected with protein-free centrifuged homogenized brain matter on days 1 and 2
The above trials use brain matter that has been centrifuged to generate a protein-free homogenate and a protein-rich homogenate based on molecular weight.
Rabbit 9 and 10: injected with boiled homogenized brain matter on days 1 and 2
The above trials use brain matter that have been boiled to destroy any bacterial contaminants in the homogenate.
Some patients have a predisposition to developing neurodegenerative disease due to genetic mutations in the PrPC gene. If a father knows that he has the mutation, but never develops the disease, and the mother knows she is not a carrier, what is the probability that their son will develop the disease? Assume that this disease follows simple Mendelian inheritance patterns.
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This is a simple Mendelian recessive trait. It must be recessive if the father is a carrier but never affected.
This is a simple Mendelian recessive trait. It must be recessive if the father is a carrier but never affected.
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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?
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?
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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.
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.
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A botanist sees that when he breeds a plants with blue flowers with a plant with red flowers, the resulting generation are plants that all have a 1 : 1 ratio of blue : red flowers. He knows that the two parents are homozygous for the trait of color. What phenomenon most likely explains the 1 : 1 ratio in the filial generation?
A botanist sees that when he breeds a plants with blue flowers with a plant with red flowers, the resulting generation are plants that all have a 1 : 1 ratio of blue : red flowers. He knows that the two parents are homozygous for the trait of color. What phenomenon most likely explains the 1 : 1 ratio in the filial generation?
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Codominance occurs when each allele for a certain gene isn't completely dominant over the other, so both are expressed. We see both fully blue and fully red flowers on each plant, so codominance fits our situation in this question. Incomplete dominance would be the correct answer if the filial plants had flowers that were a combination of blue and red, so they would be purple. Expressivity is the term for when a trait is expressed to different degrees in a population, like some plants having flowers that are more blue than others. Penetrance is when only a certain percentage of the population expresses the trait at all.
Codominance occurs when each allele for a certain gene isn't completely dominant over the other, so both are expressed. We see both fully blue and fully red flowers on each plant, so codominance fits our situation in this question. Incomplete dominance would be the correct answer if the filial plants had flowers that were a combination of blue and red, so they would be purple. Expressivity is the term for when a trait is expressed to different degrees in a population, like some plants having flowers that are more blue than others. Penetrance is when only a certain percentage of the population expresses the trait at all.
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In a dihybrid cross, what fraction of offspring will display both recessive phenotypes?
In a dihybrid cross, what fraction of offspring will display both recessive phenotypes?
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The correct answer is 1/16. A dihybrid cross results in a phenotypic ratio of 9:3:3:1. The 1 in the ratio is indicative of the double recessive phenotype, with a homozygous recessive genotype for both traits. Thus, only 1/16 of the offspring will display both recessive phenotypes.
The correct answer is 1/16. A dihybrid cross results in a phenotypic ratio of 9:3:3:1. The 1 in the ratio is indicative of the double recessive phenotype, with a homozygous recessive genotype for both traits. Thus, only 1/16 of the offspring will display both recessive phenotypes.
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The pattern of inheritance of Syndrome V has been documented in a family. According to the pedigree shown below, which mode of inheritance cannot explain the pattern?
The pattern of inheritance of Syndrome V has been documented in a family. According to the pedigree shown below, which mode of inheritance cannot explain the pattern?
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This problem asks you to use concepts of inheritance and Mendelian genetics. The best approach to this problem is to rule out possiblities rather than to find the actual mode of inheritance, as the latter can be a much more difficult and time-consuming process. First off, we know that Y-linked inheritance could not explain this pattern because we see that in generation 1 (G1), the male is affected. If he is affected, all of his sons (who inherit his Y chromosome) would also be affected. There is one son in G2 who is not. Similarly, dominant X-linked inheritance could not explain this pattern; recall that the daughters inherit two copies of the X chromosome, and one is always inactivated. Were the trait X-linked dominant, then the girls of generation 3 (G3) would be affected, having received a copy of the affected gene from their father. Revisiting all other options, we see that any of the remaining inheritance patterns could possibly explain what we see.
This problem asks you to use concepts of inheritance and Mendelian genetics. The best approach to this problem is to rule out possiblities rather than to find the actual mode of inheritance, as the latter can be a much more difficult and time-consuming process. First off, we know that Y-linked inheritance could not explain this pattern because we see that in generation 1 (G1), the male is affected. If he is affected, all of his sons (who inherit his Y chromosome) would also be affected. There is one son in G2 who is not. Similarly, dominant X-linked inheritance could not explain this pattern; recall that the daughters inherit two copies of the X chromosome, and one is always inactivated. Were the trait X-linked dominant, then the girls of generation 3 (G3) would be affected, having received a copy of the affected gene from their father. Revisiting all other options, we see that any of the remaining inheritance patterns could possibly explain what we see.
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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?
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?
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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.
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.
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A woman is a carrier for a sex-linked disorder. She marries a man whose father had the disorder, and whose mother did not. The man is unaffected. If they have a child, what is the probability that the child is also a carrier?
A woman is a carrier for a sex-linked disorder. She marries a man whose father had the disorder, and whose mother did not. The man is unaffected. If they have a child, what is the probability that the child is also a carrier?
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In order for the woman to be a carrier, the disorder must be X-linked recessive, and the woman must be heterozygous for the allele. The information about the man's parents becomes irrelevant as soon as we know that he does not have the disorder. Males only carry one copy of the X-chromosome; thus, for the man to be unaffected, he must carry the dominant, unaffected allele. Now we know that the woman's genotype is 
and the man's is 
, where 
represented the affected allele. The Punnett square for this couple is drawn below.
Males cannot be carriers for X-linked traits, as they only carry one copy of the X-chromosome; thus, we are looking only at daughters (50%). Of the daughters, 50% will be heterozygous for the trait (). There is a one in four chance the child will be a female who is a carrier, or 25%.
In order for the woman to be a carrier, the disorder must be X-linked recessive, and the woman must be heterozygous for the allele. The information about the man's parents becomes irrelevant as soon as we know that he does not have the disorder. Males only carry one copy of the X-chromosome; thus, for the man to be unaffected, he must carry the dominant, unaffected allele. Now we know that the woman's genotype is
Males cannot be carriers for X-linked traits, as they only carry one copy of the X-chromosome; thus, we are looking only at daughters (50%). Of the daughters, 50% will be heterozygous for the trait (
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A mother is homozygous for blood type A and a father has blood type AB. Which of the following could describe their child's blood type?
A mother is homozygous for blood type A and a father has blood type AB. Which of the following could describe their child's blood type?
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We are told that the mother is homozygous for blood type A (AA) and the father has blood type AB (AB). Using a simple punnett squre we can find that 50% of the offspring would be blood type A and 50% would be blood type AB.
We are told that the mother is homozygous for blood type A (AA) and the father has blood type AB (AB). Using a simple punnett squre we can find that 50% of the offspring would be blood type A and 50% would be blood type AB.
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Sickle cell anemia is an autosomal recessive trait. If a mother, who is a carrier of the trait, mates with a man who is also a carrier of the trait, what is the possibility that a boy born to the couple will have the disease?
Sickle cell anemia is an autosomal recessive trait. If a mother, who is a carrier of the trait, mates with a man who is also a carrier of the trait, what is the possibility that a boy born to the couple will have the disease?
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There is a 25% chance that a child, regardless of sex, will have the disease. Sickle cell anemia is an autosomal recessive trait, so female and male offsprings are affected equally. Two carriers that mate will have a 25% chance of having a child with the disease, 50% chance of having a child who is also a carrier, and 25% chance of having a child who is normal, meaning he/she has two normal alleles. This can be seen by drawing a Punnett square for two heterozygous parents.
There is a 25% chance that a child, regardless of sex, will have the disease. Sickle cell anemia is an autosomal recessive trait, so female and male offsprings are affected equally. Two carriers that mate will have a 25% chance of having a child with the disease, 50% chance of having a child who is also a carrier, and 25% chance of having a child who is normal, meaning he/she has two normal alleles. This can be seen by drawing a Punnett square for two heterozygous parents.
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Red-green colorblindness is an X-linked recessive disorder. Jacob's paternal grandfather and father are both colorblind, but his mother has two normal alleles. What is the probability that Jacob is red-green colorblind?
Red-green colorblindness is an X-linked recessive disorder. Jacob's paternal grandfather and father are both colorblind, but his mother has two normal alleles. What is the probability that Jacob is red-green colorblind?
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To answer this question, you must have a basic knowledge of sex-linked disease genetics. Men are much more likely to suffer from X-linked disorders; unlike women, they have only one X chromosome, and therefore do not have a second copy that can compensate for the affected one. A man cannot, however, inherit a defective X chromosome from his father, because fathers pass on their Y chromosome to their sons. So, Jacob must inherit one of his mother's healthy X chromosomes, and there is no chance that he will be colorblind.
To answer this question, you must have a basic knowledge of sex-linked disease genetics. Men are much more likely to suffer from X-linked disorders; unlike women, they have only one X chromosome, and therefore do not have a second copy that can compensate for the affected one. A man cannot, however, inherit a defective X chromosome from his father, because fathers pass on their Y chromosome to their sons. So, Jacob must inherit one of his mother's healthy X chromosomes, and there is no chance that he will be colorblind.
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Suppose two individuals with the genotypes PPQqRr and ppQqrr mate. What is the probability that their offspring will display the genotype PpQQRr?
Suppose two individuals with the genotypes PPQqRr and ppQqrr mate. What is the probability that their offspring will display the genotype PpQQRr?
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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.
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.
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The gene for muscular dystrophy is X-linked. A female carrier and an unaffected male have one daughter together. The daughter has a son with an unaffected male. What is the probability that the son will not be affected?
The gene for muscular dystrophy is X-linked. A female carrier and an unaffected male have one daughter together. The daughter has a son with an unaffected male. What is the probability that the son will not be affected?
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First, we know that muscular dystrophy must be recessive, as the grandmother in the question is a carrier.
Below is the Punnett Square for the grandparents, where the mother is a carrier and father is unaffected. X represents an unaffected allele, while X' represents the allele for dystrophy. The probability their daughter is a carrier is 50%, and the probability she is unaffected is 50%.
If the daughter is a carrier, the Punnett square between her and an unaffected male (the father) would be identical to that of the grandparents. In this case, the chance of the son being affected is 50%.
If the daughter is not a carrier, then her son cannot be affected.
In order for the son to be affected, two events must take place: the daughter must be a carrier (50% chance), and the son must inherit the affected allele (50% chance).
There is a 25% chance that the son will be affected; thus, there is a 75% chance that he will be unaffected.
| | X’ | X | |
| ----- | --- | -- |
| X | X’X | XX |
| Y | X’Y | XY |
First, we know that muscular dystrophy must be recessive, as the grandmother in the question is a carrier.
Below is the Punnett Square for the grandparents, where the mother is a carrier and father is unaffected. X represents an unaffected allele, while X' represents the allele for dystrophy. The probability their daughter is a carrier is 50%, and the probability she is unaffected is 50%.
If the daughter is a carrier, the Punnett square between her and an unaffected male (the father) would be identical to that of the grandparents. In this case, the chance of the son being affected is 50%.
If the daughter is not a carrier, then her son cannot be affected.
In order for the son to be affected, two events must take place: the daughter must be a carrier (50% chance), and the son must inherit the affected allele (50% chance).
There is a 25% chance that the son will be affected; thus, there is a 75% chance that he will be unaffected.
| | X’ | X | | | ----- | --- | -- | | X | X’X | XX | | Y | X’Y | XY |
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Consider the pedigree of a recessive trait. Is the trait autosomal or sex-linked?
Consider the pedigree of a recessive trait. Is the trait autosomal or sex-linked?
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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.
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.
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Consider the pedigree. What is the probability of individual 22 being affected?
Consider the pedigree. What is the probability of individual 22 being affected?
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Individual 22 is male, and the trait it X-linked recessive. We know he will inherit the Y-chromosome from the unknown father, and a singular X-chromosome from the affected mother. Because the mother is affected, we know she must have two affected X-chromosomes. No matter which chromosome is passed to individual 22, he will inherit the trait.
Individual 22 is male, and the trait it X-linked recessive. We know he will inherit the Y-chromosome from the unknown father, and a singular X-chromosome from the affected mother. Because the mother is affected, we know she must have two affected X-chromosomes. No matter which chromosome is passed to individual 22, he will inherit the trait.
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Consider the pedigree. What is the probabilty of individual 18 being affected?
Consider the pedigree. What is the probabilty of individual 18 being affected?
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We can tell from the pedigree that the trait is X-linked recessive. Individual 18 is female; her mother is a carrier, and her father is affected. She will receive one affected paternal X-chromosome (100%) and has a 50% chance of received an affected maternal copy. The probability of her being affected is given by the calculation.
We can tell from the pedigree that the trait is X-linked recessive. Individual 18 is female; her mother is a carrier, and her father is affected. She will receive one affected paternal X-chromosome (100%) and has a 50% chance of received an affected maternal copy. The probability of her being affected is given by the calculation.
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