Concept explainers
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).
Want to see more full solutions like this?
Chapter 10 Solutions
Human Heredity: Principles and Issues
- Artificial Selection: Explain how artificial selection is like natural selection and whether the experimental procedure shown in the video could be used to alter other traits. Why are quail eggs useful for this experiment on selection?arrow_forwardDon't give AI generated solution otherwise I will give you downwardarrow_forwardHello, Can tou please help me to develope the next topic (in a esquematic format) please?: Function and Benefits of Compound Microscopes Thank you in advance!arrow_forward
- Identify the AMA CPT assistant that you have chosen. Explain your interpretation of the AMA CPT assistant. Explain how this AMA CPT assistant will help you in the future.arrow_forwardwhat is the difference between drug education programs and drug prevention programsarrow_forwardWhat is the formula of Evolution? Define each item.arrow_forward
- Define the following concepts from Genetic Algorithms: Mutation of an organism and mutation probabilityarrow_forwardFitness 6. The primary theory to explain the evolution of cooperation among relatives is Kin Selection. The graph below shows how Kin Selection theory can be used to explain cooperative displays in male wild turkeys. B When paired, subordinant males increase the reproductive success of their solo, dominant brothers. 0.9 C 0 Dominant Solo EVOLUTION Se, Box 13.2 © 2023 Oxford University Press rB rB-C Direct Indirect Fitness fitness fitness gain Subordinate 19 Fitness After A. H. Krakauer. 2005. Nature 434: 69-72 r = 0.42 Subordinant Dominant a) Use Hamilton's Rule to show how Kin Selection can support the evolution of cooperation in this system. Show the math. (4 b) Assume that the average relatedness among male turkeys in displaying pairs was instead r = 0.10. Could kin selection still explain the cooperative display behavior (show math)? In this case, what alternative explanation could you give for the behavior? (4 pts) 7. In vampire bats (pictured below), group members that have fed…arrow_forwardExamine the following mechanism and classify the role of each labeled species in the table below. Check all the boxes that applyarrow_forward
- 1. Define and explain the two primary evolutionary consequences of interspecific competitionarrow_forward2 A linear fragment of DNA containing the Insulin receptor gene is shown below, where boxes represent exons and lines represent introns. Assume transcription initiates at the leftmost EcoRI site. Sizes in kb are indicated below each segment. Vertical arrows indicate restriction enzyme recognition sites for Xbal and EcoRI in the Insulin receptor gene. Horizontal arrows indicate positions of forward and reverse PCR primers. The Horizontal line indicates sequences in probe A. Probe A EcoRI Xbal t + XbaI + 0.5kb | 0.5 kb | 0.5 kb | 0.5kb | 0.5 kb | 0.5 kb | 1.0 kb EcoRI On the gel below, indicate the patterns of bands expected for each DNA sample Lane 1: EcoRI digest of the insulin receptor gene Lane 2: EcoRI + Xbal digest of the insulin receptor gene Lane 3: Southern blot of the EcoRI + Xbal digest insulin receptor gene probed with probe A Lane 4: PCR of the insulin receptor cDNA using the primers indicated Markers 6 5 4 1 0.5 1 2 3 4arrow_forward4. (10 points) woman. If both disease traits are X-linked recessive what is the probability A man hemizygous for both hemophilia A and color blindness mates with a normal hemophilia A nor colorblindness if the two disease genes show complete that a mating between their children will produce a grandson with neither a. linkage? (5 points) that a mating between their children will produce a grandson with both hemophilia A and colorblindness if the two disease genes map 40 cM apart? (5 points)arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningBiology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning