(Test Bank) Textbook of Histology

Gartner, 4

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4th Edition

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Test Bank for Textbook of Histology, The downloadable file consists of 163 pages, and the total size of the file is 4.6 MB. Purchase now only for highly discounted price.

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Textbook of Histology 4th Edition Gartner Test Bank Gartner: Textbook of Histology, 4th Edition Chapter 2 – Cytoplasm Test Bank MULTIPLE CHOICE 1. The fluidity of the plasmalemma has well -established clinical significance. Which of the following molecules decreases the cell membrane fluidity? a. glycerol b. phosphate group c. cholesterol d. integral proteins e. peripheral proteins Explanation: The answer is c. Cholesterol decreases membrane fluidity. The polar heads of phospholipids molecules are composed of glycerol, to which a positively charged nitrogenous group is attached by a negatively charged phosphate group. Unsaturated fatty acyl molec ules increase memb rane fluidity. Integral aTndEpSeTriBphAeNraKl pSrEotLeiLnEs dRo.nCoOt hMave an effect on m embrane fluidity. 2. A patient suffering from asthma has difficulty breathing and is treated with albuterol, a drug that relaxes bronchiolar smooth muscles. Albuterol acts as a(n) a. cholinergic receptor antagonist b. cholinergic receptor antagonist c. adrenergic receptor antagonist d. adrenergic receptor agonist e. inhibitor of kinase phosphorylases Explanation: The answer is d. Albuter ol is an agonist that acts on β 2-adrenergic receptors that are located mainly on the cell membranes of bronchial tissues and, therefore, do not have many side effects. If it were an antagonist, then it would cause smooth muscle constriction. 3. A patient suf fering from asthma has difficulty breathing and is treated with albuterol and ipratropium, drugs that relax bronchiolar smooth muscles. Ipratropium acts as a(n) a. cholinergic receptor antagonist b. cholinergic receptor antagonist c. adrenergic receptor antagonist Test Bank 2 d. adrenergic receptor agonist e. inhibitor of kinase phosphorylases Explanation: The answer is a . Ipratropium is a cholinergic receptor antagonist and interferes with the binding of acetylcholine, a neurotransmitter, to its receptor sites on the cell membrane, thus preventing the contraction of bronchiolar smooth muscle. The two drugs are used in combination agonist -antagonist therapy to have the desired effect of maintaining an open airway. 4. A 9-month -old female Jewish baby, whose parents were born in Poland, is see n by her pediatrician because of neurologic problems, including generalized paralysis and blindness. The pediatrician should suspect that the child is suffering from a. spina bifida b. cranial nerve X (vagus nerve) disorder c. Tay-Sachs disease d. Fabry’s disease e. Refsum’s disease Explanation: The answer is c. Tay -Sachs disease affects mostly children of Eastern European Jewish ancestry as well as certain families in Lousiana of Cajun ancestry. Children suffering from this lysos omal -storage disease canTnEotScTatBaAboNliKzeSGEMLL2 EgaRn.glCioOsiMdes that accumulate in their lysosomes, and the enlarged lysosomes interfere with cell function. The interference is most problematic in the neurons of the central nervous system and causes death by the third year of life. Neither spina bifida nor vagus nerve disorders would cause gen eralized paralysis or blindness. Fabry’s disease, although a lysosomal -storage disorder, is due to an X chromosomal defect and affects only males. Refsum’s disease is due to the accumulation of phytanic acid, a by-product of chlorophyll metabolism, and aff ects older children who already eat fresh fruits and vegetables. 5. Acid hydrolases are synthesized on the rough endoplasmic reticulum and are delivered to the Golgi complex to be packaged and distributed to their final destination. In patients whose protooncogenes have mutated to form ras protein –related products known as ADP -ribosylation factor (ARF), this factor may interfere with a. coatomer II (COP II) synthes is b. coatomer II (COP II) assembly c. coatomer II (COP II) dissociation from vesicles d. clathrin synthesis e. clathrin assembly Explanation: The answer is b. Delivery of proteins manufactured on the rough endoplasmic reticulum to the Golgi complex is accomplished in COP -II coated vesicles. ARF catalyzes the assembly of Test Bank 3 COP -II molecules to form coatomer on the cytoplasmic aspect of these transfer vesicles. Synthesis of COP -II proteins as well as the disassembly of the coatom er are not influenced by ARF. Clathrin coats mediate the endocytotic pathway as well as the pathway that leads away from the trans Golgi network. 6. Patients with sickle cell anemia possess hemoglobin -s, a mutated form of hemoglobin. Hemoglobin -s is a. synthesized on free ribosomes b. synthesized on the rough endoplasmic reticulum c. modified in the Golgi apparatus d. packaged in the trans Golgi network e. stored in condensing vesicles Explanation: The answer is a. Hemoglobin, whether normal or mutated, is present freely in the cytosol; therefore, it is synthesized on free ribosomes. Only proteins that are to be packaged are synthesized on the rough endoplasmic reticulum. Because hemoglobin is not packaged, it does not enter the Golgi apparatus. 7. A neonatal baby boy is observed to have malformed hands, feet, and skull. Additional examination displays jaundice, hepatomegaly, and hypotonic muscles with psychomotor retardation. The pediatrician diagnos es the disorder as cerebrohepatorenal syndrome (Zellweger’s syndrom e). ThisTcEoSndTitBioAnNisKdSuEe LtoLpEroRb.leCmOsMwith which of the following organelles? a. mitochondria b. lysosomes c. early endosomes d. late endosomes e. peroxisomes Explanation: The answer is e. Children with cerebrohepatorenal syndrome die at an early age due to mutations in their peroxin genes that code for defective receptors that are unable to transfer enzymes destined for peroxisomes into these organelles. Mitochondria, lysos omes, and endosomes are not affected by this mutation. 8. A 20-year-old male seeking help from his ophthalmologist presents with deteriorating vision in both eyes, especially his central vision. The patient also evidences cardiac dysrythmia and an apparent demyelination of the optic nerve. The ophthalmologist suspects Leber’s hereditary optic neuropathy, a condition caused by a point mutation that is not X -linked but is transmitted only by females. This condition is due to problems with which of the following? a. lysosomes b. early endosomes Test Bank 4 c. late endosomes d. mitochondria e. peroxisomes Explanation: The answer is d. Leber’s hereditary optic neuropathy is due to a point mutation on the mitochondrial DNA. Because the spermatozoon does not contribute mitochondria to the zygote, only the female can transmit this condition to her offspring. Lysosomes, peroxisomes, and early and late endosomes do not possess DNA. 9. Which of the following refers to cadherins? a. they are peripheral proteins b. they are associated with cell-cell adhesion c. they are associated with adherence to the basal lamina d. they have binding sites for heparin sulfate e. they are associated with hemidesmosomes Explanation: The answer is b. Cadherins are integral proteins that are designed to assist cells to maintain contact with one another. Hemidesmosomes, heparin sulfate, and basal lamina are all associated with cell -to-extracellular matrix contact. 10. Dynein arms are always aTttEacSheTdBtAo NonKeSoEf LthLe EfoRll.owCiOnMg. Which one is that structure? a. central sheet b. singlets c. subunit A d. subunit B e. radial spoke Explanation: The answer is c. Dynein arms are microtubule -associated proteins that function in bending of the cilium. These molecular motors are attached to the subunit A of the axoneme. The central sheet and the radial spokes are both composed of an elastic protein that becomes stretched during the “cocking” of the axoneme and, as they return to their normal length, they effect ciliary motion. 11. The component of the erythrocyte cell membrane that is t he most essential in maintaining its biconcave disc shape is a. glycophorin b. spectrin c. ankyrin d. actin e. band 3 protein Test Bank 5 Explanation: The answer is b. Spectrin is a rod -shaped tetramer that, in association with actin and adducin, forms a hexagonal lattice that underlies the cytoplasmic surface of the red blood cell plasmalemma. Ankyrin and band 3 protein and band 1 protein anchor the hexagonal lattice to the cell membrane. Glycophorins and band 3 proteins are transmembrane proteins. 12. A six-month -old baby boy, when seen by his pediatrician,, exhibits an enlarged tongue and an enlarged liver. Additionally, the baby has problems with swallow ing. The pediatrician should suspect that the child is suffering from a. cranial nerve X (vagus nerve) disorder b. Pompe’s disease c. McArdle’s syndrome d. Tay-Sachs disease e. Tarui’s disease Explanation: The answer is b. Pompe’s disease is a lethal glycogen -storage disease that affects infants and usually causes death by the second or third year of life. The baby is missing lysosomal acid maltase, and as glycogen accumulates in the lysosomes of the liver, heart, and some s keletal muscles, these organs enlarge. Although cranial nerve X distributes through much of the body, its malfunction does not cause any of the symptoms present in this infant. McArdle’s syndrome has an adult ons et, and t he syTmEptSomTBs oAfNTKaSruEi’Ls dLiEseRas.eCuOsuMally are noted only a fter vigorous physical exercise. Tay-Sachs disease is not accompanied by swollen tongue or enlarged liver. 13. A patient with Kartagener’s syndrome has a genetic disorder that is known to a. affect microfilaments b. affect intermediate filaments c. have defective dynein d. have defective kinesin e. have defective α-tubulin Explanation: The answer is c. Kartagener’s syndrome is due to an autosomal recessive disorder that affects ciliary motion because the microt ubule -associated protein, dynein, is either defective or not present. The tubulins of microtubules are formed normally, and microtubule assembly, per se, is not affected. Microfilaments and intermediate filaments are not affected in Kartagener’s syndrome. 14. A patient who had surgery for ovarian cancer is placed on a combination of cisplatin and taxol therapy. Taxol is an antimitotic agent that a. affects micro filaments b. affects intermediate filaments Test Bank 6 c. has defective dynein d. prevents the polymerization of microtubules e. stabilizes microtubules Explanation: The answer is e. Unlike most antimitotic agents that interfere with polymerization or depolymerization of microtubules, taxol stabilizes microtubules, so they cannot shorten or lengthen. In this fashion the mitotic spindle apparatus cannot change its lengt h, thus preventing the movement of the chromosomes during the mitotic event. Dynein, microfilaments, and intermediate filaments are unaffected by taxol. 15. A patient with lymphoma that involves his small intestines is treated with a various chemotherapeutic agents, including vincristine. This antimitotic substance acts on a. α-actinin b. dynein c. tubulin d. myosin I e. actin Explanation: The answer is c. Vincristine interferes with polymerization of microtubules and halts mitosis, thus killing dividing cells. Vincristine has no effect on α-actinin, dynein, myosin I, or actin. 16. A two-week -old infant was seen by her pediatrician because of blistering of the fingers and the back of her hands. The blisters were quite large and the results of the biopsy indicated that blistering occurred in the basal layer of the epidermis. The diagnosis was epidermolysis bullosa simplex . This disease is caused by mutations in the a. tubulins b. neurofilaments c. G actins d. keratins e. chondroitins Explanation: The answer is d. Blistering in infants that occurs on the regions of the skin that are exposed to friction and pressure is a result of mutations in types 14 and 5 keratins. Tubulins, neurofilaments, G actins, and chondroitins are not affected. 17. Early in th e morning, a male patient sees his physician because he and his wife have been trying to start a family. According to the patient, his wife was told that she is able to become pregnant, and he wants to have the physician check him for sterility. The physician notices that the patient has a heavy cough. Upon questioning, the patient reveals that he also coughs in the evening, produces a lot of sputum, and has had numerous colds and even occasional Test Bank 7 pneumonia. The doctor asks for a sperm sample but already is thinking of a possible diagnosis of Kartagener’s syndrome, a genetic disorder that affects a. a number of keratins b. microtubule -associated proteins c. thin filaments d. intermediate filaments c. thick filaments Explanation: The answer is b. Kartagener’s syndrome is due to an autosomal recessive disorder that affects ciliary motion because a microtubule -associated protein, dynein, is either defective or not present. The tubulins of microtubules are formed normally and microtubule assembly, per se, is not affected. Thin filaments, thick filaments, and intermediate filam ents are not affected in Kartagener’s syndrome. 18. A 55 -year-old patient was placed on tetracycline therapy and a week later returns to the physician, indicating that the antibiotic did not alleviate his symptoms. During their conversation the physician determines that the patient took the tetracycline capsules with milk. Which of the following components of milk interfered with theaction of the drug? a. calcium b. lipid c. sodium d. rennin e. potassium Explanation: The answer is a. Calcium taken within an hour of oral tetracycline ingestion interferes with the absorption of the antibiotic through the membranes of the cells lining the stomach and small intestines. Lipid, sodium, and potassium do not inter fere with tetracycline absorption. Rennin, an enzyme that curdles milk protein, has no effect on the absorption of tetracycline. 19. A 77 -year-old female is having problems sleeping, and her physician prescribed flurazepam, a sleep aid. The patient reports that she feels “groggy” for about 2 days after having taken the drug. Flurazepam is a. destroyed by the acidic conditions in the stomach b. inactivated by pepsin in the stomach c. denatured by trypsin in the duodenum d. inactivated by the P-450 enzyme system of the liver e. destroyed by proteasomes in the jejunum Explanation: The answer is d. Many drugs are inactiv ated by the P -450 enzyme system of the liver. In this patient the level of this enzyme system is depressed; therefore, flurazepam is not inactivated in Test Bank 8 18 hours, as in most patients, but has a longer half-life, which accounts for the groggy feeling even 2 days later. Flurazepam is not affected by HCl, pepsin, or trypsin, nor does it enter the proteasome system of the cell. 20. In Fig. Img_001, the region at the arrow tip is the place where a. assembly of the small subunit of ribosomes occurs b. synthesis of proteins for the large ribosomal subunit occurs c. degradation of tRNA occurs d. amino acids are coupled to tRNA e. mRNA synthesis occurs Explanation: The answer is a. The arrow tip is in the nucleolus and it is here that the assembly of both small and large ribosomal subunits occur. rRNA, not mRNA, is transcribed in the nucleolus, and tRNA is not degraded here. Protein synthesis as well as the coupling of amino acids to tRNA occur in the cytoplasm, not in the nucleolus. 21. In Fig. Img_002, the collecting tubule of the kidney, water enters the lumen via Test Bank 9 a. pinocytosis b. endocytosis c. aquaporins d. ion channels e. carrier proteins Explanation: The answer is c. Water leaves the cells of the collecting tubule to enter its lumen by way of structures known as aquaporins. Pinocytosis and endocytosis both refer to substances entering the cell. Ion channels are designed to carry ions along a concentration gradient in or out of a cell, whereas carrier proteins can move ions as well as small molecules along and/or against a concentration gradient. 22. In Fig. Img_004, the structure at the tip of the arrow is responsible for which of the following? Test Bank 10 a. synthesis of ATP b. assembly of microtubules c. assembly of neurofilaments d. synthesis of lysosomal proteins e. assembly of neurotubules Explanation: The answer is d. The large cell in the center of this field is a multipolar neuron in the gray matter of the spinal cord. The arrow is pointing to a structure known as a Nissl body, which was noted by electron microscopy to be composed of rough ER, the region where proteins that are to be packaged, such as lysosomal proteins, are synthesized whereas microtubules (also known as neurotubules) and neurofilaments are assembled in the cytoplasm, but not involving the rough ER. 23. A patient who is unable to metabolize long-chain fatty acids has problems involving her a. early endosomes b. late endosomes c. lysosomes d. peroxisomes e. proteasomes Test Bank 11 Explanation: The answer is d. Peroxisomes function in the catabolism of long -chain fatty acids, in detoxifying noxious agents, and in killing bacteria. Early and late endosomes are organelles that participate in the endocytic and endolysosomal pathways, whereas proteasomes function in the catabolism of proteins. 24. Integrins are associated with which of the following structures? a. zonula occludens b. zonula adherens c. fascia occludens d. desmosomes e. hemidesmosomes Explanation: The answer is e. Integrins are transmembrane proteins that are restricted to interact with the extracellular matrix. In epithelial cells they are localized in the basal surfaces and assist hemidesmosomes in adhering to the basal lamina. They do not participate in cell -to-cell contact, and all the other choices involve intracellular contact. 25. In Fig. Img_019, the core of the structures at the arrow tip is composed of Test Bank 12 a. thin filaments b. intermediate filaments c. thick filaments d. microtubules e. actin filaments Explanation: The answer is d. The photomicrograph depicts a pseudostratified, ciliated columnar epithelium and the structures at the arrow tip are cilia. The core of each cilium is consists of an axoneme. Axonemes are composed of nine doublet microtubules surrounding t wo central singlet microtubules. Thin filaments (actin filaments) constitute the core of microvilli. Intermediate filaments and thick filaments are not associated with cilia. Gartner: Textbook of Histology, 4th Edition Chapter 3 – Nucleus Test Bank MULTIPLE CHOICE 1. A 2-month old infant is brought to the physician’s office to be examined, and the physician notices that he has flaccid muscles, a smaller head than usual, a large tongue, and a short nose and broad face. The parents say that the baby is very quiet and hardly ever cries. The pediatrician suspects aneuploidy. The baby probably has which of the following chromosomal configurations? a. monosomy b. trisomy c. normal diploid complement d. haploid complement e. tetraploid complement Explanation: The answer is b. The baby most probably has trisomy 21 (Down syndrome), in which chromosome 21 is present insTteEadSTofBtwANo KchSroEmLoLsoEmRe.sC, oOnMe from the mother and the other from the father. During meiosis, nondisjunction occurs, so that one cell has the extra chromosome 21 and the other cell is monosomic —that is, it is missing chromosome 21. 2. A patient with a chromosomal complement of XXY has a. trisomy of the autosomes b. Turner’ s syndrome c. monosomy of the sex chromosomes d. Klinefelter syndrome e. a female phenotype Explanation: The answer is d. The patient has Klinefelter syndrome and exhibits a male phenotype. Since there are three sex chromosomes, this is not a monosomy nor is it trisomy of the autosomes. Turner’s syndrome is exhibited by females who lack the second X chromosom e. 3. A patient with a chromosomal complement of XO has a. trisomy of the sex chromosomes b. Turner’s syndrome Test Bank 2 c. monosomy of the autosomes d. Klinefelter syndrome e. a male phenotype Explanation: The answer is b. Patients with Turner’s syndrome are females who lack the second X chromosomes; therefore, they are monosomal for the sex chromosomes but not for the autosomes. Turner’s syndrome is a male phenotype with an extra X chromosome. 4. A patient with a chromosomal complement of XXX has a. a male phenotype b. a female phenotype c. a very short life span d. a higher IQ than his/her siblings e. trisomy of autosomes Explanation: The answer is b. The patient is a female who has the triple X syndrome; thus she has an extra sex chromosome rather than trisomy of the autosomes. These patients usually have a lower IQ than their siblings but have a normal life span and can give birth to children with a normal chromosomal complement. 5. During a routine examinatiTonEoSfTaB3A-yNeKarS-oEldLcLhEildR,.thCeOpMediatrician notes that the color of the pupil is white and that the child is cross -eyed. She asks the child’s father if there are any cases of retinoblastoma in the family, and when the answer is in the affirmative she calls in a pediatric ophthalmologist. The ophthalmologist knows that r etinoblastoma is due to the mutation of the Rb gene (retinoblastoma gene) and that it is a recessive trait. Retinoblastoma can metastasize to the brain via the a. optic nerve b. superior division of the oculomotor nerve c. trochlear nerve d. abducent nerve e. inferior division of the oculomotor nerve Explanation: The answer is a. The retina is attached to the brain via the optic nerve, and malignant cells can use the optic nerve as a passageway to the brain. The superior and inferior divisions of the oculomotor nerve, the trochlear nerve, and the abducent nerve all innervate the external muscles of the eye and do not contact the retina. 6. During a routine examination of a 3-year-old child, the pediatrician notes that the color of the pupil is white and that the child is cross -eyed. She asks the child’s father if there are any cases of retinoblastoma in the family, and when the answer is in the affir mative she calls in a pediatric ophthalmologist. The ophthalmologist knows that retinoblastoma Test Bank 3 is due to the mutation of the Rb gene (retinoblastoma gene) and that it is a recessive trait. Not all cases of retinoblastoma have a familial component. In the sporadic form of retinoblastoma, the child’s normal genetic complement does not predispose the child to have retinoblastoma. If this child had the sporadic form of retinoblastoma, then a. one copy of the Rb gene mutated b. both copies of the Rb gene mutated c. the patient’s future children will most probably will have one mutated Rb gene d. the patient’s future children will most probably have two mutated Rb genes e. the mutated Rb gene is known as a protooncogene Explanation: The answer is b. Because retinoblastoma is a recessive trait, both copies of the Rb gene would have mutated. The normal gene, which codes for the Rb protein, is known as a protooncogene, whereas its mutated form is known as an oncogene. Because the mutations did not occur in the gametes, the patient’s future children will not have any mutated Rb genes. 7. Which RNA synthesis is catalyzed by RNA polymerase II? a. tRNA b. rRNA that codes for the large ribosomal subunit c. rRNA that codes for the small ribosomal subunit d. mRNA e. tRNA that carries the star t codon Explanation: The answer is d. RNA polymerase II catalyzes the synthesis of messenger RNA (mRNA). The synthesis of all tRNA is catalyzed by RNA polymerase III, whereas all rRNA synthesis is catalyzed by RNA polymerase I. 8. Chiasmata formation a. occurs during metaphase II b. results in the exchange of genetic material between nonhomologous chromosomes c. occurs during the pachytene phase d. results in nondisjunction e. results in the formation of the synaptonemal complex Explanation: The answer is c. Chiasmata formation occurs during the pachytene phase of prophase I of the first meoitic event. Homologous pairs of chromosomes lined up during the zygotene phase of prophase I, forming the synaptonemal complex, thus permitting crossing over to occur during the pachytene phase so that there could be a random exchange of genet ic material that contributes to a more diverse gene pool. Nondisjunction occurs when the homologous pairs of chromosomes separate during anaphase I. Test Bank 4 9. During meiosis some chromosomes do not separate from each other. This is called a. diplotene b. nondisjunction c. diakinesis d. zygotene e. leptotene Explanation: The answer is b. During anaphase I of meiosis, the homologous chromosomes may remain attached to each other and the two newly formed cells will have unequal chromosomal numbers. This process is known as nondisjunction. Diplotene, zygotene, diakinesis, and leptotene are phases of prophase I of the first meiotic division. 10. The binding of cyclin B to CDK1 (cyclin -dependent kinase 1) permits the cell to progress from a. S phase into G2 phase b. G2 phase into M phase c. G1 phase into G2 phase d. G1 phase into S phase e. prometaphase into metaphase Explanation: The answer is b. The formation of the cyclin B -CDK1 complex allows the cell to enter the mitotic cycle —that is, progress from G2 phase into the M phase. The cell cycle cannot gofrom G1 to G2 directly; it has to go through the S phase first. Cyclins D and E are required to enter S from G1. Cyclins A and B are both required for the entry from S phase into G2 phase. 11. In Fig. Img_009, the cell at the tip of the pointer is in which of the following phases of mitosis? Test Bank 5 a. prophase b. prometaphase c. metaphase d. anaphase e. telophase Explanation: The answer is d. The sister chromatids of this cell are being pulled apart to proceed toward the opposite poles of the cell; therefore, this cell is in anaphase. 12. In Fig. Img_010, the cell at the tip of the pointer is in which of the following phases of mitosis? Test Bank 6 a. prophase b. prometaphase c. metaphase d. anaphase e. telophase Explanation: The answer is c. The sister chromatids of this cell are located at the equator of the mitotic spindle (metaphase plate configuration); therefore, this cell is in metaphase. Note that the mitotic spindle microtubules are attached to the kinetochores of the sister chromatids and will pull the sister chromatids apart during anaphase of the mitotic event. 13. In the Fig. Nucleus, the cell depicted in this electron micrograph synthesize Test Bank 7 Test Bank 8 a. short -chain lipids b. proteins destined for the cytosol c. cholesterol d. proteins destined to be packaged e. long-chain lipids Explanation: The answer is d. The nucleus of this cell has a high level of euchromatin and a rich supply of rough ER. Both of these characteristics indicate that this cell functions in protein synthesis. The presence of rough ER is indicative that the proteins being synthesized have to be packaged by the cell, so the newly -formed protein is not destined for the cytosol. Cholesterol and lipid synthesis requires the presence of smooth ER as well as a rich supply of mitochondria. 14. A patient with cancer possesses cells that are undergoing mitosis at a very rapid rate. The contractile rings that are formed during cytokinesis are composed of a. thin filaments b. intermediate filaments c. thick filaments d. microtubules e. neurofilaments Explanation: The answer is a. Contractile rings, composed of thin (actin) filaments form around the site where the cleavage furrow forms during telophase. Just prior to complete separation of the two daughter cells, the midbody forms, which is then incorporated into on e of the two cells. Intermediate filaments, thick filaments, and neurofilaments (intermediate filaments of neurons) do not participate in the formation of contractile rings. The midbody is composed of polar microtubules and a bridge of cytoplasm. Gartner: Textbook of Histology, 4th Edition Chapter 4 – Extracellular Matrix Test Bank MULTIPLE CHOICE 1. A patient whose symptoms include discolored, violet -colored skin with large blisters, fever, malaise, rapid heart rate, and a great deal of pain is diagnosed with necrotizing fasciitis. The rapid spread of Staphylococus aureus and gr oup A streptococci that ravage the connective tissue of this patient is due to the ability of these organisms to destroy one of the components of the extracellular matrix of the skin. This component is a. hyaluronic acid b. keratan sulfate c. dermatan sulfate d. heparan sulfate e. chondroitin -4-sulfate Explanation: The answer is a. Hyaluronic aTcEidSisTaBvAeNryKlSarEgeLgLlyEcRo.saCmOinMoglyc an that is not sulfated and is not bound to a protein core. It is able to bind a large number of proteoglycans, thus forming a very large molecule with an enormous domain. This macromolecule complex is responsible for the “gel state” of the extracellular matrix. Staphylococus aureus and group A streptococci possess hyalurinidase that cleaves hyaluronic acid and converts the gel to a “sol state,” making it much easier for the microorganisms to spread rapidly through the connective tissue element, destroying the tissues in its path. Some ca ll these microorganisms “flesh -eating bacteria”. The other four glycosaminoglycans are much smaller and, by binding to a protein core, form proteoglycans. They are not large enough by themselves to form a gel. 2. An African -American patient had surgery, and a very thick, elevated growth has formed at the site of the skin incision. This elevated growth is known as a keloid and is due to the excessive accumulation of a. hyaluronic acid b. proteoglycans c. type I collagen d. elastic fibers e. oxytalan fibers Explanation: The answer is c. African -Americans are especially predisposed to the excessive accumulation of type I collagen at the site of wound healing. Test Bank 2 3. The tensile strength of the healing wound about a week after the cut surfaces of the skin are sutured to each other is about 10% of the normal tensile strength of skin. The initial weakness of the dermis during wound healing is due to the material elaborated by fibroblasts to close the wound in a very rapid fashion. This material is a. type I collagen b. type II collagen c. type III collagen d. type V collagen e. type VII collagen Explanation: The answer is c. The first collagen to be formed by fibroblasts to ef fect initial closure of a wound is type III collagen (reticular fibers). This slender collagen type has a very weak tensile strength and is replaced by type I collagen within 4 weeks of the injury, and at that point, the tensile strength of the healed tissue is about 80% of the tensile strength of intact skin. 4. A male patient is seen for the first time by a physician; the patient has very thin, slender legs and unusually long, slender fingers. The patient’s history indicates that he has been diagnosed with a connective tissue disorder known as Marfan syndrome. The mutation that causes this syndrome affects a.tropomyosin b.tropo nin c.microfibrils d. elastin e. fibrillin Explanation: The answer is e. Marfan syndrome is due to a mutation that involves the gene that codes for the component of elastic fibers known as fibrillin. Individuals with this condition frequently have cardiovascular problems, including aortic aneurysm and high blood pressure, and dislocation of the lens of the orb. Their collagen fibers, muscle fibers , and actin filaments are normal. 5. A patient who is suffering from leukocyte adhesion deficiency will have a/an a. autosomal dominant disorder b. sex-linked chromosomal disorder c. dysfunctional adhesins d. dysfunctional integrins e. dysfunctional cadherins Explanation: Test Bank 3 The answer is d. Leukocyte adhesion deficiency is an autosomal recessive disorder that results in the inability of white blood cells to synthesise the beta chains of their integrin molecules. Therefore, the leukocytes are incapable of adhering to the endothelial cells and are unable to migrate to the site of inflammation. Patients with this disease have difficulty resisting bacterial infections. Adhesins are surface antigens of many microorganisms that permit them to adhere to mammalian cell surfaces, whereas cadherins are responsible for cell–cell adhesions. 6. A 9-month -old infant is seen by her pediatrician, who notices a lack of weight gain as well as thin, knobby joints. The parents tell the physician that they are making their own baby food and that their daughter has been very irritable and cries a lot when she is crawling on the floor. The baby’s gums appear to be overly red. The pediatrician suspects lack of vitamin C in the baby’s diet and recommends that the parents switch to commercially prepared baby foods and also urges them to offer the child citrus fruits. Additionally, the child is to receive high doses of vitamin C for 7 days and a lower dose for an additional 30 days. Vitamin C is necessary for the synthesis of normal a. elastin b. fibrillin c. tropocollagen d. troponin e. hyaluronic acid Explanation: The answer is c. Vitamin C is necessary for the hydroxylation of proline residues. Hydroxyproline is required to hold the alpha chains of tropocollagen molecules to each other. The synthesis of elastin, fibrillin, and hyaluronic acid is not affected by vit amin C. Troponin is a protein present in skeletal and cardiac muscle and is not affected by the lack of vitamin C. 7. A 12-year-old child complains to his physician that he frequently dislocates his joints. Upon examination the physician notes that the patient has hyperextensive joints, highly stretchable skin, and kyphoscoliosis (hump back). The physician should suspect a. scurvy b. Ehler -Danlos syndrome c. Marfan syndrome d. cutis laxa e. mucopolysaccharidosis Explanati on: The answer is b. The physician should suspect Ehler -Danlos syndrome because individuals afflicted with this genetic disorder have very flexible joints, occasional kyphoscoliosis, and skin that can be stretched a couple of inches and that when released returns to its previous position. A patient with scurvy would be recognized by bleeding gums and possible disfiguration of the long bones of the leg. Patients with Marfan Test Bank 4 syndrome usually are tall and thin, with especially long fingers. Cutis laxa is characterized by stretchy skin that hangs loosely in folds. Patients with mucopolysaccharidosis die at a very young age. 8. In Fig. Basallamina2, the arrow points to an area that is rich in a. type I collagen b. type II collagen c. type III collagen d. type IV collagen e. type VII collagen Explanation: The answer is d. The lamina densa of the basal lamina is formed mostly of a network of type IV collagen. Type I collagen is the most common collagen present in the body and it forms thick bundles of collagen fibers. Type II collagen is present in the matrix of cartilage. Type III collagen, also known as reticular fibers, is a thin fibrous collagen with Test Bank 5 many sugar groups that allow it to bind silver stain. Type VII collagen is located in the lamina reticularis of the basement membrane. 9. In Fig. Basallamina3, the clear area indicated by the tip of the arrow is rich in a. hyaluronic acid b. fibronectin c. laminin d. perlacans e. cadherins Explanation: The answer is c. The clear area in this electron micrograph represents the lamina lucida component of the basal lamina. It is rich in laminin, a glycoprotein that facilitates adhesion between the basal lamina and epithelial cells. F ibronectin is located on the connective tissue side of basal lamina. Cadherins are transmembrane proteins that permit