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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available.
The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the
Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally.
What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
To determine: The chances of the couple having a child with two copies of the dominant mutant gene in the given case study.
Introduction: Autosomal genetic defects are inherited from one generation to next in two patterns, either in autosomal recessive form or in autosomal dominant form. For a disease that is autosomal recessive, presence of two copies of the defective alleles (homozygous condition) is required for the development of disease. In case of autosomal dominant disease, presence of even a single copy of the defective allele (heterozygous condition) can cause development of the disease.
Explanation of Solution
Information given in the case study is as follows:
- Achondroplasia is an autosomal dominant defect.
- Both the parents carry heterozygous allelic combination for the disease.
- Phenotypic effects of the disease are short stature, shortened arms and legs, abnormal facial features, and many more.
The parents wanted to know the chances of their child having homozygous dominant state for the disease. For this, genotype of the parents can be assumed as:
A represents gene for achondroplasia and a representing a healthy gene.
Both the parents are having heterozygous genotype for the disease, so their genotype will be Aa and the gametes produced by them will be having genotype as A and a.
Possible genotypes of the child produced from these parents can be found as:
| Gametes | A | a |
| A | AA | Aa |
| a | Aa | aa |
It is clear that only 25% (1 out of 4) chance is there that the child will have two copies of the dominant allele (AA) for the disease.
To determine: The chances of having a child with normal height in the given case study.
Explanation of Solution
Information given in the case study is as follows:
- Achondroplasia is an autosomal dominant defect.
- Both the parents carry heterozygous allelic combination for the disease.
- Phenotypic effects of the disease are short stature, shortened arms and legs, abnormal facial features and many more.
The parents in the given case study want to know the chances of their child having homozygous dominant state for the disease. For this, genotype of the parents can be assumed as:
A represents gene for achondroplasia and a representing a healthy gene.
Both the parents are having heterozygous genotype for the disease, so their genotype will be Aa and the gametes produced by them will be having genotype as A and a.
Possible genotypes of the child produced from these parents can be found as:
| A | a | |
| A | AA | Aa |
| a | Aa | aa |
Achondroplasia is a heterozygous dominant disease. So, from the possible genotypes of the child, AA and Aa will give dwarfism phenotypically, whereas only aa genotype will give a normal height phenotypically.
Thus, it can be seen that the child from the parents in the given case study will only have 25% chance of having normal height (possible only in aa genotype).
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Chapter 10 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
- A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?arrow_forwardClara has polydactyly, an autosomal dominant that is 80% penetrant. Clara inherited her polydactyly from her mother, her father had no polydactyly in his family. She has an extra toe on her left foot. Her husband Ralph has no polydactyly in his family. A) What is the chance that Clara will have a child that will inherit polydactyly? B) What is the chance that the child will express polydactyly. You can leave these as un-calculated equations.arrow_forwardCystic fibrosis (CF) is an autosomal recessive condition triggered by the overproduction of sticky mucus that clogs the lungs and pancreas. It is a life-threatening disease, but medical advances helped the afflicted to live through adulthood. The mother of Claudia died from cystic fibrosis, but her father was normal and never had any relative with CF. Her fiancé, Marcus, turned out to be a carrier of the CF allele. What are the genotypes of Claudia and Marcus? Claudia: ________________________ Marcus: _____________________ They planned to have four children. What is the probability that: a. all children will be normal b. at least two will be normal PLEASE SHOW COMPLETE SOLUTIONarrow_forward
- Cystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?arrow_forwardBelow is a pedigree chart for a family that has a history of Alkaptonuria. Individuals infected with this condition can have darkened skin, brown urine, and can suffer from joint damage and other complications. Given this pedigree answer the following questions. Given the data in the pedigree chart is this genetic condition autosomal dominant or autosomal recessive? What are the genotypes for #1, #2, and #3? If either of the 4th generation "aa" females were to mate with a homozygous dominant male would any of their offspring illustrate the phenotype? Why or why not?arrow_forwardSheree is referred to a genetic counselor because a cystic fibrosis (CF) test done as a routine part of her prenatal care indicated that she is a carrier of the most common mutant allele. Sheree is stunned, because no one in her family has the disease. She is 26 years old. The genetic counselor would most likely explain autosomal recessive inheritance and suggest that Sheree's husband be tested for the CF allele. explain autosomal dominant inheritance and suggest that Sheree's husband be tested for the CF allele. explain autosomal recessive inheritance and suggest that Sheree's parents be tested for the CF allele. advise Sheree to have amniocentesis to check the fetus for the CF genotype.arrow_forward
- Albinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?arrow_forwardA woman knows that her mother is a carrier of Kartagener’s syndrome (an autosomal recessive disorder). The woman does not know if either she or her husband are carriers. The couple wants to have a child, but is worried about whether or not they could have a child with Kartagener’s syndrome. Should the couple seek the advice of a genetic counselor? In other words, is there a chance they could have an affected child? If there is a chance, please make sure your answer includes the specific parental genotypes necessary to make this possible.arrow_forwardThe following pedigree illustrates the inheritance of Nance–Horan syndrome, a rare genetic condition in which affected people have cataracts and abnormally shaped teeth.arrow_forward
- A young couple went to see a genetic counselor because each had a sibling affected with cystic fibrosis. (Cystic fibrosis is a recessive disease, and neither member of the couple nor any of their four parents is affected.) a. What is the probability that the female of this couple is a carrier? b. What are the chances that their child will be affected with cystic fibrosis? c. What is the probability that their child will be a carrier of the cystic fibrosis mutation?arrow_forwardMrs. Dee (40 years old) and her husband have an amniocentesis for advanced maternal age. They already have four healthy children. They receive results indicating a 47,XXY karyotype. What is the phenotypic sex of the fetus? How many Barr bodies will be found in each somatic cell?arrow_forwardThe attached image is a pedigree of a family with a history of sickle cell anemia (the individuals with the filled-in symbols have the disease and no new mutations are occurring in any individual). Sickle cell anemia is inherited in an autosomal recessive manner. What is the probability that the individual with the question mark (?) will get the disease? a) 1/4 b) 1/2 c) 2/3 d) 1arrow_forward
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning