Medical Expense Benefits

Exam: 412757RR Group Medical Expense Benefits

 

1. Because Dr. Roberts participates in an IPA, what type of patients can she see?

A. Only Medicare and Medicaid patients

B. Any patient she wants to see

C. Only patients who have been assigned to her

D. HMO patients

 

2. HMOs and PPOs are examples of

A. point-of-service plans.

B. Medicaid.

C. the Blues.

D. managed care plans.

 

3. Julio wants data to compare the performance of three different managed care plans before he chooses one for his company. Which one of the following tools will be most effective for his employees?

A. National Committee for Quality Insurance

B. Joint Commission on Accreditation of Healthcare Organizations

C. Health Plan Employer Data and Information Set

D. Utilization Review Accreditation Commission

 

4. An employer wishes to be assured that employees aren’t seeking unneeded medical care. One approach to that objective is

A. MSAs.

B. self-management.

C. HIPC.

D. indemnity.

 

5. Bob is insured under a managed care plan. Under this plan, which one of the following tasks is he allowed to do?

A. Go to any doctor he wants

B. Avoid review requirements

C. Participate in a weight-control program

D. Go to any hospital he wants

 

6. Continental Company employs 22 people—14 males and 8 females. According to the Model Act, this company would be considered a/an _______ employer.

A. discriminatory

B. growing

C. small

D. well-balanced

 

7. Mary goes to the doctor and pays for her visit. When she gets home, she must fill out a form and submit it to the insurance company so she can be reimbursed. What type of insurance does she most likely have?

A. Indemnity

B. HMO

C. POS

D. The Blues

 

8. Suppose an employee’s 18-year-old son has group medical insurance as a dependent of the employee. Which one of the following reasons would cause the son to lose his insurance benefits?

A. The son graduates from high school.

B. The employee becomes divorced.

C. The son joins the Navy.

D. The son moves away to attend college.

 

9. A new mom delivered her healthy baby by cesarean section on Monday at 8 p.m. Her insurance company has told her doctor that if she goes home on Wednesday by 8 p.m., it will provide a visit by a nurse to the new mom’s home. Is the insurance company able to offer this benefit? Why or why not?

A. Yes, the insurance company may offer this as an option to staying in the hospital because hospital expenses are remarkably high.

B. No, the Newborns’ and Mothers’ Health Protection Act makes it mandatory that a new mom with a cesarean section remain in the hospital for a full 96 hours after delivery.

C. No, the Newborns’ and Mothers’ Health Protection Act doesn’t allow extra benefits for patients who don’t use the 96 hours of allowed recovery time in the hospital unless those who stay receive it as well.

D. Yes, the new mom is allowed to have a nurse come to her home because she didn’t use the full allowed amount of recovery time in the hospital.

 

10. According to a Supreme Court ruling in 1949, _______ now have a role in employee benefits.

A. unions

B. insurance companies

C. senior management

D. all employees

 

11. In most states, if a woman who’s covered by a managed care plan believes that she’s pregnant, she should first call her

A. obstetrician, because she doesn’t need preauthorization for care.

B. primary care physician, to make an appointment so that she can be referred to an obstetrician.

C. managed care plan, to get preauthorization for care.

D. employer, to get preauthorization for care.

 

12. A contributing factor to the rise in health care costs from the 1970s to the 1990s is

A. Medicaid.

B. Medicare.

C. AIDS.

D. SIDS.

 

13. When Blue Cross first began, it was in the business of providing coverage for

A. prescriptions.

B. hospitalization.

C. physician care for the elderly.

D. physician care for the financially needy.

 

14. Which one of the following types of review is conducted after a patient has already been treated for the purpose of determining if the treatment was appropriate?

A. Reactive

B. Retrospective

C. Concurrent

D. Prospective

 

15. In the 1990s, the major shift in health care was that most employees were now

A. covered by an indemnity plan.

B. responsible for paying for their own insurance.

C. covered under a traditional insurance plan.

D. covered under a managed care plan.

 

16. In the 1960s, the main cause of the dramatic rise in health care coverage was the

A. introduction of Medicare and Medicaid.

B. introduction of HMOs.

C. increase in employee benefits.

D. Depression.

 

17. Which accrediting organization would be most likely to post HMO and POS reports on the Internet?

A. NCQA

B. URAC

C. JCAHO

D. HEDIS

 

18. Bev has an HMO that allows her to see a specialist without going through her primary care physician. What type of HMO does Bev have?

A. Group-model

B. Closed-panel

C. Direct-access

D. Staff-model

 

19. The medical expense insurance-like organizations that eventually came to be called Blue Cross plans were initially run by

A. physicians.

B. employers.

C. charity organizations.

D. hospitals.

 

20. Company X must make sure that it provides HMO coverage as an option in its benefit-selection process. What act would require Company X to do this?

A. Financial Services Modernization Act

B. Health Insurance Portability and Accountability Act

C. Health Maintenance Organization Act

D. Americans with Disabilities Act

 

Describe WHY carbohydrates are the desired source of energy and sparing protein is important for life and homeostasis.

1. Describe WHY carbohydrates are the desired source of energy and sparing protein is important for life and homeostasis.  How would you convey this to someone who wants to be healthy and go on a high protein, low carbohydrate diet?

2.  Describe what happens when a friend goes on a crash diet to metabolism the first couple of days.  Weight drops on the scale but what is that weight?  If the diet continues after 3-4 days what happens to metabolism (what energy source is used) and how does the body respond with amount ant type ofweight loss?

3.  Why would someone argue, using the information about metabolism and anabolism of fat, that fat is the most fattening for the body and that a low fat diet would be the best approach to lose weight?  HINT: this is about metabolism and not just calories per gram.

Which of the following is a scientific conclusion based on knowing that humans and bacteria share a common genetic language?

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

1) What is a gene? A) a type of eukaryotic cell B) an organelle that houses DNA C) a type of prokaryotic cell D) a type of animal cell E) a unit of heredity

2) Which of the following is a scientific conclusion based on knowing that humans and bacteria share a common genetic language?

A) Humans and bacteria have the same number of genes. B) Humans and bacteria share a common ancestor. C) The cells of both humans and bacteria store their DNA in a

nucleus. D) The same genetic code was created for humans as for bacteria. E) Bacteria will eventually develop into humans.

3) What is the difference between discovery science and hypothesis-driven science?

A) There is no difference between them. B) Discovery science involves predictions about outcomes, whereas

hypothesis-driven science involves tentative answers to specific questions.

C) Discovery science is based on deductive reasoning, whereas hypothesis-driven science is based on inductive reasoning.

D) Discovery science leads from the specific to the general, whereas hypothesis-driven science leads from the general to the specific.

E) Discovery science “discovers” new knowledge, whereas hypothesis-driven science does not.

4) What are eukaryotic genes composed of? A) A B) RNA C) C D) G E) DNA

5) Which of the following is a producer? A) dog B) sun C) cat D) earthworm E) house plant

6) Which of these is a hypothesis? A) My car is too old to function properly. B) If my car does not start and I recharge the battery, then my car will

start. C) What is wrong with my car? D) My car’s battery is dead. E) My car will not start.

7) What is the difference between a tissue and an organ system? A) A tissue cannot exist unless it is a component of an organ system,

whereas an organ system can exist independently of tissues. B) Tissues are not considered to be living, whereas organ systems are

considered to be living. C) The tissue level of organization is more inclusive than the organ

system level. D) Tissues are not composed of cells; organ systems are composed of

cells. E) An organ system includes tissues.

8) Adjacent water molecules are connected by the ______. A) sharing of electrons between the hydrogen of one water molecule

and the oxygen of another water molecule B) sharing of electrons between hydrogens of adjacent water

molecules C) electrical attraction between the hydrogens of adjacent water

molecules D) sharing of electrons between adjacent oxygen molecules E) electrical attraction between the hydrogen of one water molecule

and the oxygen of another water molecule

9) An atom with an electrical charge is a(n) ______. A) ion B) compound C) molecule D) radioisotope E) isotope

10) Sugar dissolves when stirred into water. The sugar is the ______, the water is the ______, and the sweetened water is the ______.

A) solution . . . solute . . . solvent B) solvent . . . solute . . . solution C) solution . . . solvent . . . solute D) solute . . . solvent . . . solution E) solvent . . . solution . . . solute

Please read the following scenario to answer the following question(s).

The last few miles of the marathon are the most difficult for Heather, her hair plastered to her head, sweat clinging to her arms, and her legs already feeling as if they had nothing left, just dead weight. After grabbing a cup of ice water, she feels the ice cubes smash against her nose as she gulps some cool refreshment and keeps on running. In these last few miles, the breeze kicks up and she finally feels some coolness against her skin. Drips of sweat, once clinging to her forehead, now spill down, and Heather feels more pain as the sweat flows into her eyes.

11) Sweat remained on Heather’s forehead and arms because of the ______. A) high evaporative cooling effect of water B) cohesive nature of water C) ability of water to act as a solvent D) high salt content of sweat E) ability of water to moderate heat

12) The hydrogens and oxygen of a water molecule are held together by _____ bonds.

A) osmotic B) ionic C) hydrogen D) covalent E) hydrolytic

13) In the following reaction, what type of bond is holding the two atoms together? K + Cl → K+ + Cl— → KCl

A) hydrophilic B) ionic C) hypertonic D) covalent E) hydrophobic

14) Which of the following elements, essential to life, is a trace element? A) sulfur B) iodine C) calcium D) hydrogen E) phosphorus

15) The consumption of sugar is a major cause of ______. A) gout B) cancer C) rheumatoid arthritis D) acne E) tooth decay

16) A friend of yours appears to have put on a lot of muscle very quickly, and at the same time you notice that your friend has become very irritable and depressed. It is reasonable for you to suspect that your friend has begun to take ______.

A) ephedra B) amino acid supplements C) protein powder D) creatine E) an anabolic steroid

17) A protein’s function is dependent on its ______. A) size B) temperature C) shape D) pH E) weight

18) The linear sequence of monomers in a polypeptide chain is referred to as its ______ structure.

A) tertiary B) pentamerous C) secondary D) primary E) quaternary

19) Amino acids consist of ______. A) a central hydrogen, a nitrogen atom, an amino group, and a

carboxyl group B) a central nitrogen, a carbon atom, an amino group, and a carbonyl

group C) a central hydrogen, a nitrogen atom, a hydroxyl group, and a

carbonyl group D) a central carbon, a hydrogen atom, a hydroxyl group, and a

carbonyl group E) a central carbon, a hydrogen atom, an amino group, and a carboxyl

group

20) Saturated fats are saturated with ______. A) hydrogen B) oxygen C) carbon D) phosphorus E) nitrogen

21) Proteins are polymers constructed from ______ monomers. A) 5-carbon ring B) nucleotide C) hydrocarbon D) amino acid E) peptide

22) Tay-Sachs disease results from ______ lacking a specific type of lipid- digesting enzyme.

A) the Golgi apparatus B) mitochondria C) lysosomes D) the endoplasmic reticulum E) the plasma membrane

23) When mixed with water, phospholipids spontaneously form membranes because they ______.

A) have hydrophilic phosphate groups that are attracted to water and hydrophobic fatty acid tails that avoid water

B) are capable of violating the second law of thermodynamics C) do not spontaneously form membranes when mixed with water D) have hydrophilic phosphate groups that are attracted to their

hydrophobic fatty acid tails E) have hydrophilic fatty acid tails that are attracted to water and

hydrophobic phosphate groups that avoid water

24) When using a light microscope to view a cell you obtained from scraping under your fingernails, you notice that the cell lacks a nucleus; therefore, you conclude that the cell must be a type of ______ cell.

A) fungal B) plant C) prokaryotic D) animal E) eukaryotic

25) Based on its function in detoxifying drugs, you would expect to find a large amount of smooth ER in ______ cells.

A) lung B) brain C) muscle D) liver E) intestinal

26) In plant cells, ______ contain organic nutrients, pigments, and poisons. A) ribosomes B) mitochondria C) chloroplasts D) lysosomes E) central vacuoles

27) In eukaryotic cells, what name is given to the region between the nucleus and the plasma membrane?

A) nucleoplasm B) cytosol C) gene D) phospholipid bilayer E) cytoplasm

Read the following scenario to answer the following question(s).

The earliest cells detectable in fossils were different from the cells in animals, plants, fungi, and protists living today. These first prokaryotic cells gave rise to eukaryotic cells approximately 1.7 billion years ago. The structure of eukaryotic cells today suggests how they might have evolved from their prokaryotic ancestors. Scientists examining mitochondria and chloroplasts now think that these organelles were probably free-living prokaryotes before becoming a part of eukaryotic cells long ago.

28) What evidence suggests that mitochondria might have evolved before chloroplasts?

A) Some mitochondria have chloroplasts inside of them. B) Almost all eukaryotes have mitochondria but only some cells have

chloroplasts. C) A double membrane surrounds mitochondria and a single

membrane surrounds chloroplasts. D) Only mitochondria have their own DNA. E) Mitochondria can sometimes divide to produce chloroplasts.

29) A cell that neither gains nor loses water when it is immersed in a solution is ______.

A) isotonic to its environment B) metabolically inactive C) hypertonic to its environment D) hypotonic to its environment E) dead

30) Ozygen crosses a plasma membrane by ______. A) phagocytosis B) pinocytosis C) active transport D) osmosis E) passive transport

31) Facilitated diffusion across a biological membrane requires ______ and moves a substance ______ its concentration gradient.

A) energy and transport proteins . . . down B) transport proteins . . . against C) transport proteins . . . down D) energy and transport proteins . . . against E) energy . . . down

32) Which component of the following reaction is the substrate? lactose + lactase + water → lactase + glucose + fructose

A) lactose B) fructose C) glucose D) lactase E) There is no substrate in this reaction.

33) The energy of motion is ______ energy. A) potential B) created C) stored D) conserved E) kinetic

34) In a hypotonic solution, an animal cell will _____. A) experience turgor B) shrivel C) neither gain nor lose water D) lyse E) lose water

35) An enzyme’s function is dependent on its ______. A) size B) shape C) weight D) pH E) temperature

36) The final electron acceptor of aerobic respiration is ______. A) ATP B) lactic acid C) oxygen D) carbon dioxide E) NAD+

37) Aerobic means with ______. A) oxygen B) carbon dioxide C) ATP D) carbohydrate E) light

38) Where in the cell does glycolysis occur? A) cytosol B) within the fluid just inside the inner mitochondrial membrane C) along the outside of the outer mitochondrial membrane D) ER E) between the inner and outer mitochondrial membrane

39) In metabolic terms, dogs are best described as ______. A) obligate aerobes B) facultative anaerobes C) obligate anaerobes D) facultative aerobes E) producers

40) The functioning of an electron transport chain is analogous to ______. A) a canoe going over a waterfall B) playing Ping-Pong C) a person leaping from the top to the bottom of a flight of stairs in

one jump D) a Slinky toy going down a flight of stairs E) a person climbing a flight of stairs one step at a time

41) Some friends are trying to make wine in their basement. They’ve added yeast to a sweet grape juice mixture and have allowed the yeast to grow. After several days they find that sugar levels in the grape juice have dropped, but there’s no alcohol in the mixture. The most likely explanation is that ______.

A) the mixture needs less sugar; high sugar concentrations stimulate cellular respiration, and alcohol is not a by-product of cellular respiration

B) the mixture needs more sugar; yeast need a lot of energy before they can begin to produce alcohol

C) the mixture needs less oxygen; yeast only produce alcohol in the absence of oxygen

D) the mixture needs more oxygen; yeast need oxygen to break down sugar and get enough energy to produce alcohol

E) none of the above

42) Large amounts of oxygen gas first appeared in Earth’s atmosphere about ______ years ago.

A) 4.5 billion B) 500 million C) 1.5 billion D) 2.5 billion E) 3.5 billion

Please refer to the accompanying figure to answer the following question(s).

43) One of the compounds that is a direct output of the Calvin cycle is ______.

A) C6H12O6 B) G3P C) O2 D) ATP E) NADPH

44) C4 plants conserve water by ______. A) keeping their stomata closed most of the time B) shuttling CO2 from the Calvin cycle to the water-splitting

photosystem C) growing very deep roots D) incorporating CO2 into RuBP E) running the Calvin cycle at night

45) Which of these colors contributes the least energy to photosynthesis? A) blue B) orange C) violet D) red E) green

46) What is responsible for the yellow-orange coloration of leaves in the fall? A) RuBP B) chlorophyll b C) carotenoids D) chlorophyll a E) phycoerythrin

47) How many molecules of glucose are produced by each cycle of the light reaction?

A) four B) three C) two D) zero E) one

48) Oxygen is highly corrosive; therefore, which of the following would provide definitive evidence of large amounts of O2 in the atmosphere?

A) the presence, in the fossil record, of cells with chloroplasts B) the presence, in the fossil record, of cells with mitochondria C) mass extinction D) rust E) the presence of cyanobacteria in the fossil record

49) In photosynthesis, an H+ ion gradient forms across the ______. A) thylakoid membrane B) inner chloroplast membrane C) endomembrane D) stromal membrane E) outer chloroplast membrane

50) Which of the following is an example of a polysaccharide? A) sucrose B) starch C) maltose D) fructose E) glucose

X-linked Agammaglobulinemia

CASE 1 X-linked Agammaglobulinemia

An absence of B lymphocytes.

One of the most important functions of the adaptive immune system is the production of antibodies. it is estimated that a human being can make more than one million different specific antibodies. This remarkable feat is accom­ plished through a complex genetic program carried out by B lymphocytes and their precursors in the bone marrow (Fig. 1.1). Every day about 2.5 billion (2.5 x 109) early B-ceU precursors (pro-B cells) take the first step in this genetic program and enter the body’s pool of pre- B cells. From this pool of rapidly dividing pre-B cells 30 billion daily mature into B cells, which leave the bone marrow as circulating B lymphocytes, while 55 billion fail to mature success­ fully and undergo programmed cell deqth. This process continues throughout life, although the numbers gradually decline with age.

Mature Circulating B cells proliferate on encounter with antigen and differen­ tiate into plasma cells, which secrete antibody. Antibodies, which are made by the plasma cell progeny of B cells, protect by binding to and neutralizing toxins and viruses, by preventing the adhesion of microbes to cell surfaces, and, after binding to microbial surfaces, by fixing complement and thereby enhancing phagocytosis and lysis of pathogens (Fig. 1.2).

‘Ihis case concerns a young man who has an inherited inability to make anti­ bodies. His family history reveals that he has inherited this defect in antibody synthesis as an X-linked recessive abnormality. This poses an interesting puzzle because the genes encoding the structure of the immunoglobulin polypeptide chains are encoded on autosomal chromosomes and not on the X chromosome. Further inquiry reveals that he has no B cells, so that some gene on the X chromosome is critical for the normal maturation of B lymphocytes.

This case was prepared by Raif Geha, MD, in collaboration with Ari Fried, MD.

Topics bearing on this case:

Humoral versus cell· mediated immunity

Effector functions of antibodies

Effector mechanisms of humoral immunity

Actions of complement and complement receptors

.–­ B-cell maturation

Methods for measuring T-cell function

D Case 1: X-linked Agammaglobul inemia Early pro-B cell 1lI1e pro-B cell Large pre-B cellStem cell Small pre-B cell

pre-B receptor

© l1′ It­ (5) (cW r ®a ~~ H-chain genes

L-chain genes

Su rface Ig

V-OJO-J Germline rearrangingrearranging

Germline Germline Germline

Absent Absent Absent

Fig. 1_1 The development of B c ells proceeds through several stages marked by the rearrangement of the immunoglobulin genes. The bone marrow stem cell that gives rise to the B-Iymphocyte lineage has not yet begun to rearrange its immunoglobulin genes; they are in germline configuration_ The first rearrangements of 0 gene segments to JH gene segments occur in the early pro-B cells, generating late pro-B cells. In the late pro-B cells, a VH gene segment becomes joined to the rearranged DJ H, producing a pre-B cell that is expressing both low levels of

;3 s-~-b~-~-·\

The case of Bill Grignard: a medical student with scarcely any antibodies.

Bill Grignard was well for the first 10 months of his life. In the next year he had pneu­ monia once, several episodes of otitis media (inflammation of the middle ear), and on one occasion developed erysipelas (streptococcal infection of the skin) on his right cheek. These Infections wereall treated successfully with antibiotics but it seemed to his mother, a nurse, that he was constantly on antibiotics.

His mother had two brothers who had died, 30 years prior to Bill’s birth, from pneu­ monia in their second year of life, before antibiotics were available. She also had two sisters who were well ; one had a healthy son and daughter and the other a healthy daughter.

Bill was a bright and active child who gained weight, grew, and developed normally but he continued to have repeated infections of the ears and sinuses and twice again had pneumonia. At 2 years 3 months his local pediatrician tested his serum immunoglobulins. He fou nd 80 mg dl-‘ IgG (normaI60Q-1500 mg dl-‘), no IgA (normal 50- 125 mg dl-‘), and only 10 mg dl-‘ IgM (normal 75-150 mg dl-‘ ).

Bill was started on monthly intramuscular injections of gamma globulin; his serum IgG level was maintained at 200 mg dl-1. He started school at age 5 years and per­ formed very well (he was reading at second grade level at age 5 years) despite pro­ longed absences because of recurrent pneumonia and other infections.

At 9years of age he was referred to the Children’s Hospital because of atelectasis (col­ lapse of part of a lung) and a chron ic cough. On physical ~xamination he was found to be a well-developed, alert boy. He weighed 33.5 kg and was 146 cm tall (height and

VOJ rearranged

Germline

VO,I rearranged

V-J rearranging

f.l chain transiently at surface as part of pre-B-cell receptor. Mainly intracellular

‘—–

Intracellular f.l chain

surface and high levels of cytoplasmic ~ heavy chain. Finally, the light-chain genes are rearranged and the cell, now an immature B cell, expresses both light chains (L chains) and /l heavy chains (H chains) as surface IgM molecules. Cells that fail to generate a functional surface immunoglobulin, or those with a rearranged receptor that binds a self antigen, die by programmed cell death. The rest leave the bone marrow and enter the bloodstream.

Immature S cell Mature B cell

19M

6 l1’ ~ VOJ

rearranged

VJ rearranged

19M expressed on cell surface

~

VJ rearranged

IgO and 19M made from alternatively spliced H-chain

transcripts

IgD 19M

co VO,I

rearranged

• • • • • •

Bacterial toxins I I Bacteria in extracellular space I I Bacteria in plasma ,.. , , , –, ,

\’,.

• cell with receptors

JY for toxin I

Neutraliza~ Opsonlzatioll Complement activation

tp. .~ h I I ~J!./

()

I I

~1 ~tf

Y +

I~p~mem I

~ ~ ~ e ‘.· 0/, v ,e

.,,’, !} •.~t;::; •• 0 . ‘ ~ e D’

weight normal for his age), The doctor noted that he had no visible tonsils (he had never had a tonsillectomy). With a stethoscope the doctor also heard rales (moist crackles) at both lung bases.

Further family history revealed that Bill had one younger sibling, John, a 7-year-old brother, who also had contracted pneumonia on three occasions. John had a serum IgG level of 150 mg dl-l,

Laboratory studies at the time of Bill’s visit to the Children’s Hospital gave a white blood cell count of 5100111-‘ (normal), of which 45% were neutrophils (normal), 43% were lymphocytes (normal), 10%were monocytes (elevated), and 2% were eosinophils (normal),

Flow cytometry (Fig. 1.3) showed that 85% of the lymphocytes bound an antibody to C03, aT-cel! marker (normal); 55% were helper T cells reacting with an anti-C04 antibody;and 29%were cytotoxic T cells reacting withan anti-C08 antibody (normal). However, none of Bill’s peripheral blood lymphocytes bound an antibody against the B-cell marker e019 (normal 12%) (Rg. 1.4),

T-cell proliferation indices in response to phytohemagglutinin, concanavalin A, teta­ nus toxoid, and diphtheria toxoid were 162, 104, 10, and 8, respectively (all normal), Serum IgG remained low at 155 mg dl-\ and serumIgA and IgMwere undetectable.

Case 1: X-linked Agammaglobulinemia D Fig. 1.2 Antibodies can participate in host defense in three main ways. The left-hand column shows antibodies binding to and neutralizing a bacterial toxin, preventing it from interacting with host cel ls and from causing pathology. Unbound toxin can react with receptors on the host cell, whereas the toxin:antibody complex cannot. Antibodies also neutralize complete virus particles and bacteri al cells by binding to them and inactivating them. The antigen:antibody complex is eventually scavenged and degraded by macrophages. Antibodies coating an antigen render it recognizable as foreign by phagocytes (macrophages and polymorphonuclear leukocytes), which then ingest and destroy it; this is called opsonization. The central column shows the opsonization and phagocytosis of a bacterial cell. The right-hand column shows the activation of the complement system by antibodies coating a bacterial cell. Bound antibodies form a receptor for the first protein of the complement system, which eventually forms a protein complex on the surface of the bacterium that favors its uptake and destruction by phagocytes and can, in some cases, directly kill the bacterium. Thus, antibodies target pathogens and their products for disposal by phagocytes.

G Case 1: X-linked Agammaglobulinemia Fig. 1.3 The FACSTM allows individual cells to be identified by their cell-surface antigens and to be sorted. Cells to be analyzed by flow cytometry are first labeled with fluorescent dyes (top panel). Direct labeling uses dye-coupled antibodies specific for cell-surface antigens (as shown here), whereas indirect labeling uses a dye-coupled immunoglobulin to detect unlabeled cell-bound antibody. The cells are forced through a nozzle in a single-cell stream that passes through a laser beam (second panel). Photo-multiplier tubes (PMTs) detect the scattering of light, which is a sign of cell size and granularity, and emissions from the different fluorescent dyes. This information is analyzed by computer (CPU). By examining many cells in this way, the number of cells with a specific set of characteristics can be counted and levels of expression of various molecules on these cells can be measured. The bottom part of the figure shows how these data can be represented, using the expression of two surface immunoglobulins, IgM and IgO, on a sample of B cells from a mouse spleen. The two immunoglobulins have been labeled with different-colored dyes. When the expression of just one type of molecule is to be analyzed (lgM or IgO), the data are usually displayed as a histogram, as in the left-hand panels. Histograms display the distribution of cells expressing a single measured parameter (such as size, granularity, fluorescence color). When two or more parameters are measured for each cell (lgM and IgO), various types of two-color plot can be used to display the data, as shown in the right-hand panel. All four plots represent the same data. The horizontal axis represents the intensity of IgM fluorescence, and the vertical axis the intensity of IgO fluorescence. Two­ color plots provide more information than histograms; they allow recognition, for example, of cells that are ‘bright’ for both colors, ‘dull’ for one and bright for the other, dull for both, negative for both, and so on. For example, the cluster of dots in the extreme lower left portions of the plots represents cells that do not express either immunoglobulin; these are mostly T cells. The standard dot plot (upper left) places a single dot for each cell whose fluorescence is measured. It is good for picking up cells that lie outside the main groups but tends to saturate in areas containing a large number of cells of the same type. A second method of presenting these data is the color dot plot (lower left), which uses color density to indicate high-density areas. A contour plot (upper right) draws 5% ‘probability’ contours, with 5% of the cells lying between each contour providing the best monochrome visualization of regions of high and low density. The lower right plot is a 5% probability contour map that also shows outlying cells as dots.

Bill was started on a preparation of gamma globulin rendered suitable for Intrave­ nous administration. He was given a dose of gamma globulin Intravenously to main­ tain his IgG level at 600 mg dl-l, He Improved remarkably. The rales at his lung bases disappeared. He continued to perform well in school and eventually entered medical school. Except for occasional bouts of conjunctivitis or sinusitis, which respond well

Mixture of cells Is labeled with fluorescent antibody

t ~ ~

~~;! ~ ~

~ o CPU

u B o

Laser Q Forward scatler

Analysis of cells atalned with labeled antibodies

IgM Dot plots Contour maps

}.~ I I I

1000 1 Standard I 5% probability

/ J \\ 100

I I 10

\ 0.1″-J 0.1 10 100 1000 0. 1 10 100 1000 0.1 10 100 1000

IgO

l W0

1 ;~~l~:l~rSity I With outliers -,–i~~_~ ~k~· ;. 100

10 I

l~ I. “,,-/ ,’ ) IlgO

0 1~

0. 1 10 1W 1000 IL ~~M 1 10 100 1000 0.1 10 100 1000

Case 1: X-linked Agammaglobulinemia ~

Normal

…;.” ,’. I’ .

0> I~.~~:Ei … ,~I ‘ U I).:.., …. . .. ‘

.. ;~&;~..; f~ .:, ,,.~ . . f ;iW!!!/j•.. ~ ‘.. :~~;. I. ‘~'” .”.’, … .

CD3

XLA

0>

Ei (.)

. ‘:’J ····I?~;.f:.,)…. “” !:’i •:.• I~~:. .

,’J. ~•,c …. ,’~’. ~:: o t ‘ ~ • • I t I, . .. ,. ; ‘.

CD3

Fig . 1.4 Clinical FACS™ analysis of a normal individual (top panel) and a pat ient with X-linked agammaglobulinemia (XLA) (bottom panel). Blood lymphocytes from a normal individual bind labeled antibody to both the B-cell marker CD19 and the T-cell marker CD3 (see top panel). However, blood lymphocytes from an individual such as Bill with X-linked agammaglobulinemia show only binding to antibodies against the T-cell marker CD3. This indicates an absence of B cells in these patients.

to oral antibiotic treatment, he remains In good health and leads an active life. He became skilled at inserting aneedle into a vein on the back of his hand and he infuses himself with 10 g of gamma globulin every weekend.

X-linked agammaglobulinemia.

Males such as Bill with a hereditary inability to make antibodies are subject to recurrent infections. However, the infections are due almost exclusively to common extracellular bacterial pathogens-Haemophilus inJIuenzae, Streptococcus pneumoniae, Streptococcus pyogenes, and Staphylococcus aureus. An examination of scores of histo ries of boys with this defect has established that they have no problems with in tracellular infections, such as thqse caused by the common viral diseases of childhood. T-cell number and function in males witb X-linked agammaglobulinemia are normal, and these individuals therefore have normal cell-mediated responses, which are able to terminate viral infections and infections with intracellular bacteria such as those caus­ ing tuberculosis.

The bacteria that are the major cause of infection in X-linked agammaglob­ ulinemia are all so-called pyogenic bacteria. Pyogenic means pus-forming, and p us consists largely ofneutrophils. The normal host response to pyogenic infections is the production of antibodies that coat the bacteria and fix com­ plement, thereby enhancing rapid uptake of the bacteria into phagocytic cells such as neutrophils and macrophages, which destroy them. Since antibiot­ ics came into use, it haS’ been possible to treat pyogenic infections success­ fully. However, when they recur frequently, the excessive release of proteolytic enzymes (for example elastase) from the bacteria and from the host phago­ cytes causes anatomical damage, particularly to the airways of the lung. The bronchi lose their elasticity and become the site of chronic inflammation (this is called bronchiectasis) . If affected males do not receive replacement ther­ apy-gamma globulin-to prevent pyogenic infections, they eventually die of chronic lung disease.

Gamma globulin is prepared from human plasma. Plasma is pooled from approximately 1000 or more blood donors and is fractionated at very cold tem­ peratures (-5°C) by adding progressively increasing amounts of ethanol. This method was developed by Professor Edwin J. Cohn at the Harvard Medical School during the Second World War. The five plasma fractions obtained are still called Cohn Fractions 1, II, III, Tv, and V. Cohn Fraction I is mainly com­ posed of fibrinogen. Cohn Fraction II is almost pure IgG and is called gamma globulin. Cohn Fraction III contains the beta globulins, including IgA and IgM; Fraction Tv, the alpha globulins; and Fraction V, albumin. Cohn Fraction II, or gamma globulin, is commercially available as a 16% solution of IgG. During the processing of the plasma some of the gamma globulin aggregates, and for this reason the 16% solution cannot be given intravenously. Aggregated gamma globulin acts like tmmune complexes and causes a reaction of shak­ ing chills, fever, and low blood pressure when given intravenously. Gamma globulin can be dis aggregated with low pH or insoluble proteolytic enzymes. It can then be safely administered intravenously as a 5% solution. In newer preparations, fractionation is followed by a further purification step using anion-exchange (DEAE) chromatography to get rid of trace contaminants. To decrease the risk of transmitting infection, the current commercially available products have several virus removal and inactivation steps incorporated into the manufacturing process.

The gene defect in X-linked agammaglobulinemia was identified when the gene was mapped to the long arm of the X chromosome at Xq22 and

G Case 1: X-linked Agammaglobulinemia

1°10 I I I ~ ~ OlD OlD 0,0

)~fZl OD ° ~ affected male 0 normal female o normal male II propositus

subsequently cloned. The gene, BTK, encodes a cytoplasmic protein tyrosine kinase called Bruton’s tyrosine kinase (Btk), which is found in pre-B cells, B celis, and neutrophils. Btk is activated at different stages of B-cell develop­ ment by the engagement of both the pre-B-cell receptor and the B-cell recep­ tor. Btk is required to mediate the survival and further differentiation of the progenitor B celis in which successful rearrangement of their heavy-chain genes has occurred. It is also required for the survival of mature B cells.

Fig. 1.5 Bill’s family tree.

Questions.

0: Fig. 1.5 shows Bill’s family tree. It can be seen that only males are affected and that the females who carry the defect (Bil l’s mother and maternal grandmother) are normal. This inheritance pattern is characteristic of an X-linked recessive trait. We do not know whether Bill’s aunts are carriers of the defect because neither of them has had an affected male child. Now that the BTK gene has been mapped, it is possible in principle to detect carriers by testing for the presence of a mutant BTK gene. But there is a much simpler test that was already available at the time of Bill’s diagnosis, which is still used routinely. Can you suggest how we could have determined whether Bill’s aunts were carriers?

~ Bill was well for the first 10 months of his life. How do you explain this?

@] Patients with immunodeficiency diseases should never be given live viral vaccines! Several male infants with X-linked agammaglobulinemia have

been given live oral polio vaccine and have developed paralytic poliomyelitis. What sequence of events led to the development of polio in these boys?

1)& Bill has a normal number of lymphocytes in his blood (43% of a normal concentration of 5100 white blood cells per 111). Only by phenotyping these lymphocytes do we realize that they are all T cells (CD3′) and that he has no B cells (CD19+). What tests were performed to establish that his T cells function normally?

@] Bill’s recurrent infections were due almost exclusively to Streptococcus and Haerr1 0philus species. These bacteria have a slimy capsule composed primarily of polysaccharide polymers, which protect them from direct attack by phagocytes. Humans make IgG2 antibodies against these polysaccharide polymers. The IgG2 antibodies ‘opsonize’ the bacteria by fixing complement on their surface, thereby facilitating the rapid uptake of these bacteria by phagocytic cells (Fig. 1.6). What other genetic defect in the immune system might cfinically mimic X-linked agammaglobulinemia?

Encapsulated bacteria resist uptake by neutrophil! and avoid

engulfment

o

Binding of IgG2 antibodies to the bacterlel 8uriace leads to activation of complement and binding of C3b

Case 1: X-linked Agammaglobulinemia ~

Factor Hand Factor I together cleave C3b to form IC3b

Uptake of bacteria into neutrophil phagosomes mediated by lC3b

receptors

b b b4

[6l fhe doctor noted that Bill had no tonsils even though he had never had his tonsils removed surgically. How do you explain this absence of tonsils, an important diagnostic clue in suspecting X-linked agammaglobulinemia?

[7J It was found by trial and error that Bill would stay healthy and have no signincant infections if his IgG level were maintained at 600 mg drJ

of plasma. He was told to take 10 g of gamma globulin every week to maintain that level. How was the dose calculated?

@] Females with a disease exactly mimicking X-linked agammaglobulinemia have been found. Explain how this might happen.

Fig. 1.6 Encapsulated bacteria are efficiently engulfed by phagocytes only when they are coated with complement. Encapsulated bacteria resist ingestion by phagocytes unless they are recognized by antibodies that fix complement. IgG2 antibodies are produced against these bacteria in humans, and lead to the deposition of complement component C3b on the bacterial surface, where it is cleaved by Factor H and Factor I to produce iC3b, still bound to the bacterial surface. iC3b binds a specific receptor on phagocytes and induces the engulfment and destruction of the iC3b-coated bacterium. Phagocytes also have receptors for C3b, but these are most effective when acting in concert with Fc receptors for IgG1 antibodies, whereas the iC3b receptor is potent enough to act alone, and is the most important receptor for the phagocytosis of pyogenic bacteria.

 

Case Study: “X-linked Agammaglobulinemia”

All responses must be in your own words. Answers that have been copied and pasted will not receive credit.

1. Bill was well for the first 10 months of his life. State a plausible explanation for this. Be specific.

2. Based on information in the case study, state one reason why specific immunity diminishes with advancing age.

3. What laboratory test results indicate that Bill has no B-cells?

4. Based on information in the case study, what is Bill’s brother John’s likely diagnosis? Why?

5. Does Bill’s innate immune system appear to be normal? What lab results support your answer?

6. This is the second case study we’ve reviewed that mentions bronchiectasis secondary to repeated bacterial respiratory infections.According to this case study, what is the etiology of the bronchiectasis?

7. What are gamma globulins (as found in plasma)? Where does the name ‘gamma globulin’ originate? [Internet]

8. What is Immune Globulin (as used in intravenous therapy)? [Internet]

9. Name 3 other situations where Immune Globulin is administered. [Internet]