A man with Huntington disease (he is heterozygous HD HD+) and a normal woman have two children.
a. | What is the probability that only the second child has the disease? |
b. | What is the probability that only one of the children has the disease? |
c. | What is the probability that none of the children has the disease? |
d. | Answer (a) through (c) assuming that the couple had 10 children. |
e. | What is the probability that 4 of the 10 children in the family in (d) have the disease? |
a.
To determine:
The probability of having the disease by the second child if a man is heterozygous for the Huntington disease mates with a normal woman.
Introduction:
Huntington’s disease is an inherited disorder which leads to the death of brain cells. This is an autosomal dominant disease mean a single copy of the defective gene may result in the disease.
Explanation of Solution
Only a copy of the defective gene can develop into Huntington’s disease because it is an autosomal dominant disease. The genotype of the male that is heterozygous for the disease is HD+HD. So, the possible gametes in males are HD+ and HD. The genotype of the normal female is HD+HD+ so; only one type of gamete will be formed in a female that is HD+.
The cross between a man who is heterozygous for the disorder and a normal female is as follows:
♂/ ♀ | HD+ | HD+ |
HD+ | HD+HD+ | HD+HD+ |
HD | HD+HD | HD+HD |
By observing the given cross, it can be concluded that 50% of the children will have the disease, and the remaining 50% will be normal. Thus, the probability of having the disease by a second child will be 25%.
b.
To determine:
The probability of having the disease by one of the children.
Introduction:
Symptoms of Huntington’s disease include jerking, muscle problems and impaired posture. The symptoms of this disease develop after the age of 30, and another name for this disease is Huntington’s chorea.
Explanation of Solution
The cross between a man who is heterozygous for the disorder and a normal female is as follows:
♂/ ♀ | HD+ | HD+ |
HD+ | HD+HD+ | HD+HD+ |
HD | HD+HD | HD+HD |
Two children are produced by the mating of heterozygous male and normal female. If the first child of the parents has a disease, then their second child will be normal and vice versa. Therefore, to calculate the probability of having a disease by one of the children multiplication and addition rule should be applied.
Thus, the formula for calculating the probability that only one child has the disease is as follows:
By putting values of probabilities in equation (I):
Thus, the probability that the only one has the disease is 50%.
c.
To describe:
The probability of having the disease by none of the children.
Introduction:
The cause of Huntington disease is the mutation in the HTT gene. This gene carries the information for producing a protein known as huntingtin. Huntingtin protein has an important role in nerve cells.
Explanation of Solution
The cross between a man who is heterozygous for the disorder and a normal female is as follows:
♂/ ♀ | HD+ | HD+ |
HD+ | HD+HD+ | HD+HD+ |
HD | HD+HD | HD+HD |
Thus, the formula for calculating the probability that none of the children has the disease is as follows:
By putting values of probabilities in equation (I):
Thus, the probability that none of the children has the disease is 25%.
d.
To describe:
The probability of having the disease by the second child and the probability of having the disease by none of the children if it is assumed that the couple had 10 children.
Introduction:
There is no cure for Huntington disease, but patients can use medication that will help them to manage the symptoms of the disease.
Explanation of Solution
If it is assumed that the couple has 10 children, then the formula for calculating the probability of 2nd child of the parents to have the disease is as follows:
By putting values of probabilities in equation (I):
Thus, the probability that the second child has the disease is 0.00097.
e.
To describe:
The probability of having the disease of 4 children out of 10 children in the family.
Introduction:
When the symptoms of the disease appear in a patient that after 15-20 years, the complication of this disease like pneumonia and infection increases, which causes the death of the patients.
Explanation of Solution
A binomial expression that can be used to calculate the probability of having the disease by 4 children out of 10 children is as follows:
Thus, the probability of having the disease by 4 children out of 10 is as follows:
Thus, the probability of having the disease by 4 children out of 10 is 0.20507.
Want to see more full solutions like this?
Chapter 2 Solutions
ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
- 18. Watch this short youtube video about SARS CoV-2 replication. SARS-CoV-2 Life Cycle (Summer 2020) - YouTube.19. What is the name of the receptor that SARS CoV-2 uses to enter cells? Which human cells express this receptor? 20. Name a few of the proteins that the SARS CoV-2 mRNA codes for. 21. What is the role of the golgi apparatus related to SARS CoV-2arrow_forwardState the five functions of Globular Proteins, and give an example of a protein for each function.arrow_forwardDiagram of check cell under low power and high powerarrow_forward
- a couple in which the father has the a blood type and the mother has the o blood type produce an offspring with the o blood type, how does this happen? how could two functionally O parents produce an offspring that has the a blood type?arrow_forwardWhat is the opening indicated by the pointer? (leaf x.s.) stomate guard cell lenticel intercellular space none of thesearrow_forwardIdentify the indicated tissue? (stem x.s.) parenchyma collenchyma sclerenchyma ○ xylem ○ phloem none of thesearrow_forward
- Where did this structure originate from? (Salix branch root) epidermis cortex endodermis pericycle vascular cylinderarrow_forwardIdentify the indicated tissue. (Tilia stem x.s.) parenchyma collenchyma sclerenchyma xylem phloem none of thesearrow_forwardIdentify the indicated structure. (Cucurbita stem l.s.) pit lenticel stomate tendril none of thesearrow_forward
- Identify the specific cell? (Zebrina leaf peel) vessel element sieve element companion cell tracheid guard cell subsidiary cell none of thesearrow_forwardWhat type of cells flank the opening on either side? (leaf x.s.) vessel elements sieve elements companion cells tracheids guard cells none of thesearrow_forwardWhat specific cell is indicated. (Cucurbita stem I.s.) vessel element sieve element O companion cell tracheid guard cell none of thesearrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage LearningHuman Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage LearningHuman Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
- Case Studies In Health Information ManagementBiologyISBN:9781337676908Author:SCHNERINGPublisher:Cengage