Concept explainers
Videos
To determine: The possible genotypes of children born if a individual who is heterozygous for a recessive mutation in enzyme 1 and enzyme 2 marries an individual having the same genotype.
Introduction:
Explanation of Solution
It is given that both the individuals are heterozygous for a recessive mutation of enzymes 1 and 2. To find the possible genotypes of their children, genotypes of parents can be assumed as:
R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.
Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.
Genotype of the children from these gametes can be determined as follows:
| RD | Rd | rD | rd | |
| RD | RRDD | RRDd | RrDD | RrDd |
| Rd | RRDd | RRdd | RrDd | Rrdd |
| rD | RrDD | RrDd | rrDD | rrDd |
| rd | RrDd | Rrdd | rrDd | rrdd |
GenotypesRRDD, RRDd, RrDD, RrDd, RRDd, RrDd, RrDD, RrDd, and RrDd will not show any mutation in any of the enzymes.
Genotypes rrDD, rrDd, and rrDd will show mutation in enzyme 1 only.
Genotypes RRdd, Rrdd, and Rrdd will show mutation in enzyme 2 only.
Genotype rrdd will show mutation in both the enzymes 1 and 2.
To determine: The activity ofenzyme 1 and enzyme 2 for all the genotypes that are produced in children if an individual who is heterozygous for a recessive mutation in both, enzyme 1 and enzyme 2, marries an individual having the same genotype, assuming that there is 0% activity for mutant alleles and 50% activity in normal alleles.
Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.
Explanation of Solution
The possible genotypes of children born from parents who are heterozygous for a recessive mutation in both the enzymes are determined as follows:
R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.
Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.
Genotype of the children from these gametes can be determined as follows:
| RD | Rd | rD | rd | |
| RD | RRDD | RRDd | RrDD | RrDd |
| Rd | RRDd | RRdd | RrDd | Rrdd |
| rD | RrDD | RrDd | rrDD | rrDd |
| rd | RrDd | Rrdd | rrDd | rrdd |
Genotypes RRDD, RRDd, RrDD, RrDd, RRDd, RrDd, RrDD, RrDd, and RrDd will not show any mutation in any of the enzymes. Hence, both the enzymes will show 50% activity in these genotypes
Genotypes rrDD, rrDd, and rrDd will show mutation in enzyme 1 only. Hence, in children with these genotypes, enzyme 1 will show 0% activity and enzyme 2 will show 50 % activity.
Genotypes RRdd, Rrdd, and Rrdd will show mutation in enzyme 2 only. Hence, in children with these genotypes, enzyme 1 will show 50% activity and enzyme 2 will show 0% activity.
Genotype rrdd will show mutation in both the enzymes 1 and 2. Hence, in children with these genotypes, both, enzyme 1 and enzyme 2 will show 0% activity.
To determine: Whether compound C will be made or not in each genotype of children born, if an individual who is heterozygous for a recessive mutation in both, enzyme 1 and enzyme 2, marries an individual having the same genotype, and if compound C not produced, the compounds that will be in excess.
Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.
Explanation of Solution
Genotype RRDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts.
Genotype RRDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts.
Genotype RrDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype RRDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype RRdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess production of compound B. Compound A will be present in normal amounts.
Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype Rrdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess production of compound B. Compound A will be present in normal amounts.
Genotype RrDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype rrDD will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.
Genotype rrDd will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.
Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts
Genotype Rrdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess of compound B. Compound A will be present in normal amounts.
Genotype rrDd will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.
Genotype rrdd will show mutation in both, enzyme 1 and enzyme 2. Hence no compound will be metabolized and compound C will not be produced.
Want to see more full solutions like this?
Chapter 10 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
- Leber’s Hereditary Optic Neuropathy (LHON) is a disease that involves degeneration of neural cells in the retina and results in loss of central vision. The disease is caused by mutations in any one of three genes in the mitochondrial genome that encode proteins involved in oxidative phosphorylation. In a genetic counseling clinic, a woman and her husband seek advice on the potential that any of their children would be afflicted with LHON. The husband's mother and father, both exhibit symptoms of the disease, but the woman does not. What is a reasonable advising statement to make?arrow_forwardWhat are some possible social and ethical issues associated with mitochondrial replacement therapy (MRT)? Present arguments for and against approving MRT for prevention of the births of children with mitochondrial disorders.arrow_forwardPhenylketonuria (PKU) is an autosomal recessive disease that results from a defect in an enzyme that normally metabolizes the amino acid phenylalanine; when this enzyme is defective, high levels of phenylalanine cause brain damage, in the past, most children with PKU became intellectually disabled. Fortunately, intellectual disability can be prevented in these children by carefully controlling the amount of phenylalanine in the diet. The diet is usually applied during childhood when the brain development is taking place. As a result of this treatment, many people with PKU now reach reproductive age. Children born to women with PKU (who are no longer on a phenylalanine-restricted diet) frequently have low birth weight, developmental abnormalities, and intellectual disabilities. However, children of men with PKU do not have these problems. Describe the results depicted in Fg1. and us it to provide an explanation for these observations What type of genetic effect is…arrow_forward
- Phenylketonuria (PKU) is an autosomal recessive disease that results from a defect in an enzyme that normally metabolizes the amino acid phenylalanine; when this enzyme is defective, high levels of phenylalanine cause brain damage, in the past, most children with PKU became intellectually disabled. Fortunately, intellectual disability can be prevented in these children by carefully controlling the amount of phenylalanine in the diet. The diet is usually applied during childhood when the brain development is taking place. As a result of this treatment, many people with PKU now reach reproductive age. Children born to women with PKU (who are no longer on a phenylalanine-restricted diet) frequently have low birth weight, developmental abnormalities, and intellectual disabilities. However, children of men with PKU do not have these problems. QUESTION: Describe the results depicted in Fig Use the first question to provide an explanation for these observations. What type of genetic effect…arrow_forwardLeber Congenital Amaurosis (LCA) causes progressive vision loss due to defects in the gene that encodes RPE65 isomerase. Affected individuals are homozygous recessive for mutant alleles of the RPE65 gene. You are trying to determine the molecular nature of the mutations in three individuals with LCA. For ease of analysis, you may assume that each individual is homozygous for the same mutant allele (though the three individuals have different mutations than each other). You use the polymerase chain reaction to amplify DNA from each patient and you determine the sequence of the DNA and compare it to unaffected individuals. You identify the following differences. Note that the non-template strand of DNA is given and the changes are highlighted using red boldface. You can assume that the sequences are in the first reading frame (eg. the first three nucleotides of each sequence is a codon). The coding region of the gene is 1602 bp and the position of the sequences shown below is…arrow_forwardFeather color in parakeets is produced by the blending of pigments produced from two biosynthetic pathways shown below. Four independently assorting genes (AA, BB, CC, and DD) produce enzymes that catalyze separate steps of the pathways. For the questions below, use an uppercase letter to indicate a dominant allele producing full enzymatic activity and a lowercase letter to indicate a recessive allele producing no functional enzyme. Feather colors produced by mixing pigments are green (yellow + blue) and purple (red + blue). Red, yellow, and blue feathers result from production of one colored pigment, and white results from absence of pigment production. What is the genotype of a pure-breeding purple parakeet strain? Express your answer as combination of allelic symbols. Example: AaBBccDd.arrow_forward
- Ch. 13-7 Mutations in the TYR gene may render its enzyme product—tyrosinase—nonfunctional.Individuals homozygous for such mutations cannot make the pigment melanin. Albinism, the absence of melanin, results. Humans and many other organisms can have this phenotype (right). Mutated tyrosinase alleles are recessive when paired with the normal allele in heterozygous individuals.In the following situations, what are the probable genotypes of the father, the mother, and their children?a. Both parents have normal phenotypes; some of their children have the albino phenotype and others are unaffected.b. Both parents and children have the albino phenotype.c. The mother and three children are unaffected; the father and one child have the albino phenotype.arrow_forwardMaple syrup urine disease (MSUD) is an autosomal recessive metabolic disorder that presents in newborns, typically within the first 48 hours after birth. As the name suggests, a key indicator of the disease is the presence of a sweet odor in the urine that smells like maple syrup. Left untreated, MSUD can result in failure of central neurological function and the respiratory system and can be fatal. MSUD is caused by mutations in components of the branched-chain alpha-keto acid dehydrogenase complex (BCKDC). These mutations result in the inability for cells to break down branched-chain amino acids (BCAAs). BCAAs and their byproducts accumulate and are excreted in the urine, giving rise to the maple syrup scent. QUESTIONS: Draw the structures of the three branched-chain amino acids (BCAA). Draw the structures of the three BCAA products released by functional BCKDC. Indicate the original amino acid.arrow_forwardFor the following genotype, indicate if beta-galactosidase will be made in the presence of lactose and absence of lactose. P₁+1+PL+OcZ/ P₁1 P₁O+Z+ O O O beta-galactosidase will be made in both the presence and absence of lactose (+; +) beta-galactosidase will be made in the presence but not in the absence of lactose (+;-) beta-galactosidase will not be made in the presence but will be made in the absence of lactose (-; +) beta-galactosidase will not be made in either the presence or absence of lactose (-;-)arrow_forward
- Phenylketonuria (PKU) is a genetic disorder that causes the abnormal metabolism of the amino acid called phenylalanine. PKU is an autosomal recessive disease due to a mutation in the gene encoding the enzyme phenylalanine hydroxylase. Phenylalanine hydroxylase (PAH) usually converts excess phenylalanine into tyrosine. In persons with PKU, the remaining phenylalanine is instead converted into phenylpyruvate (also known as phenylketone). This results in a poisonous build-up of phenylketone in the blood and urine that is why it is called phenylketonuria. When PKU is untreated, it can lead to brain damage, mental retardation and other serious medical problems. Babies with PKU are normal at birth due to the mother’s ability to break down phenylalanine during pregnancy. PKU can be diagnosed through a simple blood test for elevated phenylalanine levels shortly after birth.Answer the following questions:1. What is phenylketonuria (PKU)? How important is the synthesis of correct proteins?2. How…arrow_forwardTay Sachs is a rare autosomal recessive disorder that causes mental and physical disabilities leading to death in infants. Affected individuals are lacking the enzyme hexosaminidase, causing lipids to build up in the brain.The HEXA gene on chromosome 15 codes for hexosaminidase, and a four base pair insertion in the gene results in an altered reading frame and non-functional enzyme being produced. Individuals who are carriers (heterozygotes) of the Tay-Sachs allele are not affected by the disease but appear to have increased protection against tuberculosis.The incidence of Tay-Sachs disease is much higher among Ashkenazi Jews originating from Eastern Europe than the general population of the United States. About 1 in 3 500 babies of Ashkenazi Jewish heritage are born with Tay-Sachs disease and about 1 in 30 Ashkenazi Jews are carriers compared to about 1 in 320 000 babies born with the disease and about 1 in 300 carriers in the general United States population. Ashkenazi Jews living in…arrow_forwardPlease help me with this questionarrow_forward
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning