
Concept explainers
To determine: The treatment options for adenosine deaminase (ADA)-deficient severe combined immunodeficiency disease (SCID).
Introduction: Mr. and Mrs. J were referred to a clinical geneticist because their 6-month-old daughter was not growing appropriately. After the examination, the doctor observed that their daughter had a constant cough, wheeze, and extensive yeast infection. The doctor also took a blood test and determined that their daughter has suffered from SCID. It was also found that their daughter had inherited mutant allele for SCID from the parents.

Explanation of Solution
According to the case study, both the parents have the heterozygous mutant allele for ADA deficiency. The daughter has received both the mutant alleles from both father and mother and has become homozygous for the condition. ADA-deficient SCID is an immunodeficiency disorder. This condition causes a reduction in the synthesis of immune cells. In the absence of immune cells, the person would become immunocompromised. A person with ADA-SCID would become susceptible to infections. The possible treatment of this disease can be done by medicating the person with immune-modulating drugs.
Gene therapy is also a good option for the treatment of ADA-SCID. Using gene therapy, the gene of adenosine deaminase deficiency is replaced with normal genes. This would stimulate the immune system and improve the condition of the patient. However, the success rate of gene therapy is very low. In order to prevent the patient from infections, the patient must be kept in a sterile environment. However, there is no appropriate treatment for ADA-SCID that can cure this disease.
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Chapter 17 Solutions
HUMAN HEREDITY (LL)-W/MINDTAP ACCESS
- You’re curious about the effect that gap genes have on the pair-rule gene, evenskipped (eve), so you isolate and sequence each of the eve enhancers. You’re particularly interested in one of the enhancers, which is just upstream of the eve gene. Describe an experimental technique you would use to find out where this particular eve enhancer is active.arrow_forwardFor short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…arrow_forward3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Propertiesarrow_forward
- 4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2arrow_forwardplease answer (A,B,C,D,E) questions with the asnwer choice provided below. thank you!arrow_forwardplease fill out empty spots. thank youarrow_forward
- There is a species of eagle, which lives in a tropical forest in Brazil. The alula pattern of its wings is determined by a single autosomal gene with four alleles that exhibit an unknown hierarchy of dominance. Genetic testing shows that individuals 1-1, 11-4, 11-7, III-1, and III-4 are each homozygous. How many possible genotypes among checkered eagles in the population?arrow_forwardwhat is this called?arrow_forwardcan you help me identify this it's based on onion rootarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
