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The expression of antigen A or antigen B in red blood cells requires the help of an H antigen. A recessive mutation (h) that prevents the synthesis of the H antigen also prevents the expression of A and B antigens. This is called the Bombay effect. There is no ill effect in an individual with this mutation, but complications with blood transfusions or parental disputes may arise.
a. Individuals with the Bombay genotype (hh) produce anti-H antigen. How can this be a problem during blood transfusion?
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- . Mutations in an autosomal gene in humans cause aform of hemophilia called von Willebrand disease(vWD). This gene specifies a blood plasma proteincleverly called von Willebrand factor (vWF). vWFstabilizes factor VIII, a blood plasma protein specified by the wild-type hemophilia A gene. Factor VIIIis needed to form blood clots. Thus, factor VIII is rapidly destroyed in the absence of vWF.Which of the following might successfully be employed in the treatment of bleeding episodes in hemophiliac patients? Would the treatments workimmediately or only after some delay needed forprotein synthesis? Would the treatments have only ashort-term or a prolonged effect? Assume that allmutations are null (that is, the mutations result in thecomplete absence of the protein encoded by the gene)and that the plasma is cell-free.a. transfusion of plasma from normal blood into avWD patientb. transfusion of plasma from a vWD patient into adifferent vWD patientc. transfusion of plasma from a hemophilia A…Alpha-1 antitrypsin has codominant inheritance. M genes express normal levels. S and Z genes have low expression. Which of the following is most likely to develop emphyema? A person with: 1) two M genes who does smoke 2) two S genes who does smoke 3) one M and one S gene who does not smoke 4) one M gene and one S gene who does smoke 5) two M genes who does not smoke 6) two S genes who does not smokeThe terminal sugars of the antigenic determinants of A and B blood groups are: O a. N-acetyl-galactosamine and N-acetyl glucosamine O b. D-galactose and N-acetyl glucosamine O c. N-acetyl-D-glucosamine and L-fucose O d. N-acetyl-galactosamine and D-galactose
- Mutations in the genes for clotting factor VIII and IX cause hemophilia A and B, respectively. A woman may be heterozygous for mutations in both genes, with a mutated factor VIII allele on one X chromosome, and a mutated factor IX allele on the other. All of her sons should have either hemophilia A or B. However, on rare occasions, one of these women gives birth to a son who does not have hemophilia, and his one X chromosome does not have either mutated allele. Explain.Talk about the challenges involved in determining the genetic components of polygenic illnesses. Explain complementation groups and how the biochemical underpinnings of disease are determined using them. Hereditary illnesses of genomic instability include Werner syndrome, Bloom syndrome, XP, ataxia-telangiectasia, and Fanconi anemia. Which of these ailments has molecular mechanisms behind it? Which kind of genetic instability is connected to which disorder?.Chronic myelogenous leukemia (CML) is also particularly well-studied. Most people with CML have a particular genetic change. What is this change, and what two genes are affected by the change ?
- Define a Point mutation and give an example. What is sickle cell anemia and what causes it. What is nondisjunction? How does nondisjunction cause disorders? NUMER YOUR ANSWERSSeveral genes in humans in addition to the ABO gene () give rise to recognizable antigens on the surface of the red blood cells. The Rh marker is determined by positive (R) and negative alleles () of gene R, where R is completely dominant to r. The presence of M and N surface proteins are controlled by two codominant alleles of gene L (LM and LM. For each mother-child pair, choose the father of the child from among the males in the right column. (Assume that all mothers and fathers are HH; there is no involvement of the Bombay phenotype.) Paternal genotypes maybe used once, more than once, or not at all. Each mother-child pair matches with one or more than one paternal genotype. Maternal phenotype: Child phenotype: Paternal genotype: Reset A, M, Rh(neg) O, M, Rh(pos) B, N, Rh(neg) O, N, Rh(pos) O, M, Rh(neg) A, MN, Rh(pos) A, N, Rh(pos) AB, MN, Rh(pos) B, N, Rh(pos) A, MN, Rh(neg) Genotypes of possible fathers AiLMLN rr BiLMLN RR ii LNLN rr ii LMLM rr AALMLN RRName two ways in which loss of p53 function contributes to a malignant phenotype. Explain how benzo(a) pyrene can cause loss of p53 function. Hint: Loss of p53 function occurs in the majority of human tumors.
- Many antibodies exhibit dosage, discuss the antibodies that exhibit dosage and how it is demonstrated in heterozygous/homozygous cells. Additionally, in some cases, if an antibody is just starting to develop, it may not react with a heterozygous cell, discuss the reasons for this.The following chromosomal aberration is found in nearly 90-95% of all patients who have chronic myelogenous leukemia. This is because the change brings the BCR and ABL genes in close proximity. BCR is responsible for cell growth, and ABL is a proto-oncogene... this favors uncontrolled growth. Which is the most accurate description of the aberration... chio moso me 9 Philad elphia chromosome chromosome 22 BCR ABL 22q11.2 (BCR) 9934 1 (ABL) Deletion Translocation Inversion Duplication DELL O O O CGiven below is the electrophoretic profile of 2 proteins, a normal hemoglobin, HbA and the fetal hemoglobin, HbF. What information can be obtained from the profile shown? wwwwwww (+) HbF ww w (-) HbA ww ww A. HbF has a slightly different conformation compared with HbA B. HbF and HbA have different primary structures C. HbF has a higher affinity for oxygen than HbA D. HbF has a nonpolar amino acid residue in place of a basic amino acid. E. HbF has an acidic amino acid residue in place of a nonpolar amino acid