MINISTRY OF EDUCATION, SCIENCE, YOUTH, AND SPORTS OF UKRAINE
SUMY STATE UNIVERSITY MEDICAL INSTITUTE
Test Items for Licensing Examination Krok-1 General Medical Training
M E D I C A L B I O L O G Y
for medical students
Sumy
2012
Test Items for Licensing Examination: “Krok-1 General Medi- cal Training: Medical Biology” (For Medical Students) / Compiler O. Yu. Smirnov. – Sumy: Electronic Edition, 2012. – 82 pp.
Physiology and Pathophysiology Department Medical Institute of the Sumy State University
This book includes 319 test items in cytogenetics, classical genetics, mo- lecular genetics, medical genetics, population genetics, general biology, protozool- ogy, helminthology, and entomology.
All tests were received from the Testing center of Ministry of Public Health of Ukraine and selected from exam booklets (2003–2012).
Comments and notes are given to some test problems. Special attention is given to errors in tests.
© Compiling, revision, and comments. O. Yu. Smirnov, 2012 http://physiology.med.sumdu.edu.ua/
CONTENTS
Introduction 4
Cytology and Cytogenetics 4
Classical Genetics 16
Molecular Genetics 22
Medical Genetics 31
Population Genetics and Evolution 52
General Biology 53
Protozoans 59
Helminths 67
Arthropods 77
Mixed Questions on Parasitology 82
INTRODUCTION
All tests were received from the Testing center of Ministry of Public Health of Ukraine (http://testcentr.org.ua/), selected from the book named Collec- tion of tasks for preparing for test examination in natural science “Krok-1 General Medical Training” (V. F. Moskalenko, O. P. Volosovets, I. E. Bu- lakh, O. P. Yavorovskiy, O. V. Romanenko, and L. I. Ostapyuk, eds. – K.:
Medicine, 2006) and from exam booklets (2003–2012), and then were re- viewed and reorganized. Many mistakes in these tests were corrected (for example, we use the term “DNA repair” in this book instead of “repara- tion”, “Edwards' syndrome” instead of “Edward's syndrome” etc.). Comments and notes are given to some test problems. Special attention is given to er- rors in tests.
Five answers from which only one answer is correct are given to each test. Correct answer is marked by plus. Working with tests, close marks near answers by a piece of paper. Choose the answer (or finished state- ments) that fits best and then check your answer.
Year in which certain question was used during State examination
“Krok-1” is given in brackets. Exam test “Krok-1” contains 200 question;
approximately 16–18 of them are questions from medical biology.
Oleg Smirnov
CYTOLOGY AND CYTOGENETICS
1. The cell of the laboratory animal was overdosed with Roentgen rays. As a result albuminous fragments formed in the cytoplasm.
What cell organoid will take part in their utilization? (2003) + Lysosomes
– Endoplasmic reticulum – Ribosome
– Golgi complex – Cells centre
2. Moving of the daughter chromatids to the poles of the cell is observed in the mitotically dividing cell. At what stage of the mi- totic cycle is this cell? (2003, 2005, 2006)1
– Telophase + Anaphase – Prophase
1 In the book “Collection of tasks…” this question is written as follows: During the mitotic divi- sion in a cell we can observe the separation of chromatids towards the opposite poles. What stage of the cell cycle takes place in the cell?
– Metaphase – Interphase
3. During the postsynthetic period of mitotic cycle the synthesis of proteins – tubulins, which take part in the mitosis formation, was destroyed. It can cause the impairment of: (2004)
– duration of mitosis – chromosome spiralization – chromosome despiralization + chromosome separation – cytokinesis
4. Karyotyping of healthy man cells is carried out. A small acro- centric odd chromosome was found in the karyotype. What chro- mosome is it? (2004)
– Group A chromosome – Group B chromosome – X chromosome + Y chromosome – Group C chromosome
5. The study of mitotic cycle phases of an onion root revealed the cell, in which the chromosomes are situated in the equatorial plane, forming a star. What stage of the cell mitosis is it? (2004)1 – Interphase
+ Metaphase – Prophase – Anaphase – Telophase
6. Oval and round organelles with double wall are seen at the electron micrograph. The outer membrane is smooth, the inner membrane folded into cristae contains the enzyme ATP syn- thetase. These are: (2005)2
– ribosomes – centrioles – Golgi complex – lysosomes + mitochondria
1 In the book “Collection of tasks…” this question is written as follows: During the analysis of the mitotic stage in the onion root cells, a cell, in which spiralized chromosomes were placed in the equatorial zone, was revealed. What mitotic stage is the cell at?
2 In the book “Collection of tasks…” this question is written as follows: The electronograms of the rat's liver cells demonstrate some bimembraneous oval structures, the internal membrane of which forms cristae. What organelles are these?
7. A tissue sample of benign tumor was studied under the electron microscope. A lot of small (15–20 nm) spherical bodies, consist- ing of two unequal subunits were detected. These are: (2005, 2006) – microtubules
– Golgi complex – mitochondria + ribosomes
– smooth endoplasmic reticulum
8. Analysis of amniotic fluid that was obtained as a result of am- niocentesis (puncture of amniotic sac) revealed cells with the nu- clei that contain sex chromatin (Barr's body). What can it be evi- dence of? (2006)
– Genetic disorders of fetus development – Development of male fetus
– Polyploidy
+ Development of female fetus – Trisomy
9. In course of practical training students studied a stained blood smear of a mouse with bacteria phagocytosed by leukocytes.
What cell organella completes digestion of these bacteria? (2007) – Ribosomes
+ Lysosomes
– Granular endoplasmic reticulum – Golgi apparatus
– Mitochondrions
10. Golgi complex exports substances from a cell due to the fusion of the membrane saccule with the cell membrane. The saccule contents flows out. What process is it? (2008)
– Active transport – All answers are false – Facilitated diffusion + Exocytosis
– Endocytosis
11. Life cycle of a cell includes the process of DNA autoreduplica- tion. As a result of this process monochromatid chromosomes be- come bichromatid. This phenomenon is observed within the fol- lowing period of the cell cycle: (2008, 2009, 2012)
– G1
– G2 + S – G0
– M
12. While studying maximally spiralized chromosomes of human karyotype the process of cell division was stopped in the following phase: (2008, 2011)1
– prophase – anaphase – interphase + metaphase – telophase
13. Normal actively dividing cells of human red bone marrow are analyzed. What number of cell's chromosomes is typical for G1
period? (2010) + 46
– 48 – 23 – 45 – 47
14. A cell at the stage of mitotic anaphase was treated2 by colchi- cine that inhibits chromosome separation to the poles. What type of mutation will be caused? (2010, 2012)
– Duplication – Inversion – Translocation + Polyploidy – Deletion
15. On an electron micrograph a scientist has identified a struc- ture formed by eight histone proteins and a part of DNA molecule which makes about 1,75 revolutions around the molecules. Which structure has been identified? (2011)
– Chromosome – Elementary fibril + Nucleosome – Chromatid – Half-chromatid
16. The cell cycle is known to consist of several subsequent stages. At one of the stages the synthesis of DNA happens. What
1 In the book “Collection of tasks…” this question is written as follows: During the cell division we can see the maximum amount of condensed chromosomes. At what stage of the cell cycle is the process of the cell division stopped?
2 In the exam booklet the word "stimulated" was used. But colchicine does not stimulate cellu- lar processes!
do we call this period of the cell cycle?
– Presynthesis period (G1) of interphase + Synthesis period (S) of interphase – Mitosis
– Premitotic period of interphase
– Postsynthesis period (G2) of interphase
17. In a cell the chromosomes are in the condition of maximum spiralization and are placed along the equatorial zone. What pe- riod of mitosis is described?
– Prophase – Telophase + Metaphase – Anaphase – Prometaphase
18. An intensive aerobic process of energy formation and accumu- lation in the form of high energy ATP bonds takes place in the cells of muscular tissue. In which organelle does this process oc- cur?
– In the peroxisome
– In the endoplasmic reticulum – In the lysosome
+ In the mitochondrion – In the centriole
19. During mitotic cell division a scientist can see the phase when the nuclear envelope and nucleolus disappear, the centrioles are placed on the opposite poles of the cell and chromosomes are in the form of a thread ball freely placed in the cytoplasm. What stage of mitotic cycle is the cell at?
– Metaphase + Prophase – Anaphase – Interphase – Telophase
20. During the cell cycle regular changes in quantity of genetic material happen. What is the period, when the replication of DNA happens, called?
– Anaphase – Prophase – Metaphase + Interphase – Telophase
21. Microfilaments and microtubules are known to include tubulin proteins, which take part in the formation of the division spindle.
In what period of the mitotic cycle are tubulin proteins synthe- sized?
– Postmitotic period of interphase – Mitosis
– Synthesis period (S) of interphase + Postsynthesis period (G2) of interphase – Presynthesis period (G1) of interphase
22. In order to analyse the karyotype, a cell culture was influ- enced by colchicine, which destroys the spindle of division. At what stage was the mitosis stopped?
+ Metaphase – Prophase – Anaphase – Telophase – Prometaphase
23. During the whole life of a human in some adult cells mitosis is not observed, and the quantity of DNA stays permanent. What do we call these cells?
+ Neurons – Hepatocytes
– Eye cornea epitheliocytes – Red bone marrow cells – Germinal epithelium
24. During the examination of pancreatic gland cells under an electronic microscope there has been found an organelle which consists of cisterns, canals, closets and is connected with plas- malemma. What organelle is it?
– Centriole – Mitochondrion
+ Endoplasmic reticulum – Lysosome
– Peroxisome
25. During the examination of the cell structure, a globular monomembranous organelle, which contains hydrolytic enzymes, was found. This organelle is known to provide intracellular diges- tion and protective reactions of the cell. What organelle is it?1
1 In the book “Collection of tasks…” another similar question is also present: In a cell a ball- shaped monomembranous organelle that contains hydrolytic enzymes has been studied. What
– Endoplasmic reticulum – Centriole
+ Lysosome – Ribosome – Mitochondrion
26. There is an organelle near the nucleus which consists of two cylinders built of microtubules. The cylinders are situated perpen- dicularly to each other. The organelle is a component of the mi- totic spindle of division in animal cells. What organelle is this?
– Mitochondrion – Ribosome
– Endoplasmic reticulum + Centrosome1
– Lysosome
27. In the presynthesis period (G1) of the cell cycle the synthesis of DNA doesn't occur, that's why the number of DNA molecules is equal to the number of chromosomes. How many DNA molecules does any human somatic cell in the presynthesis period (G1) have?
– 23 – 92 + 46 – 69 – 48
28. During an experiment the culture of the cells divided by mito- sis was influenced by the substance which destroyed the spindle of division. Which substance was used in the experiment?
– Penicillin + Colchicine – Histamine – Methanol – Iodine
29. During anaphase chromosomes (each containing one chro- matid)2 are placed on the poles of the cell. How many chromo- somes does the cell have during the anaphase?
organelle is it?
1 There is a mistake in this question in the book “Collection of tasks…” – the term "centriole" is used in this book. Structure that contains a pair of cylinders (i. e. a pair of centrioles) is called the centrosome.
2 There is a mistake in this question in the book “Collection of tasks…” – incorrect word com- bination "monochromatic chromosomes" is used in this book.
– 96 – 46 – 23 – 69 + 92
30. According to the rule of the permanent chromosomes number, each animal species can be characterized by a specific and per- manent number of chromosomes. What mechanism provides this feature during sexual reproduction?
– Repair – Translation + Meiosis – Mitosis – Cytokinesis
31. To diagnose human chromosomal disorders in order to analyse the karyotype, a cell culture is influenced by colchicine – a sub- stance which destroys the spindle of division. At what mitotic stage is the karyotype studied?
– Telophase – Interphase – Prophase + Metaphase – Anaphase
32. A cell includes ball-shaped mono-membranous organelles that include proteolytic enzymes. Organelles size is 0.2–1 microme- ters. Their formation is connected with Golgi apparatus. What or- ganelles are these?
– Centrioles – Ribosomes – Plastids – Mitochondria + Lysosomes
33. In a nucleus there are non-constant structures that disappear at the beginning of cell division and appear again at the end of it.
They include protein and RNA. They take part in the formation of ribosome subunits. What are these structures called?
+ Nucleoli – Nucleosomes – Polysomes – Microfibrils – Microtubules
34. There is an organelle in human cells. The functions of this or- ganelle are the formation of lysosomes, the secretion of glycopro- teins, carbohydrates, lipids, and the formation of yolk granules during the oocytes maturation. What is this organelle called?
– Lysosome
– Endoplasmic reticulum + Golgi apparatus – Peroxisome – Ribosome
35. A cell was affected by a substance which broke the integrity of lysosome membranes. What can happen to the cell as a result?
– Specialization – Differentiation – Reproduction – Transformation + Autolysis
36. The nuclei of cells were affected by a substance which de- stroyed the histone structure. What components of the cells will change as a result of this intervention in the first place?
– Mitochondria – Nuclear membrane – Ribosomes
+ Chromosomes – Cell membranes
37. Under the influence of gamma-radiation a fragment of a chromosome has turned by 180°. What chromosomal mutation has taken place?
– Duplication – Deletion + Inversion
– Intrachromosomal translocation – Interchromosomal translocation
38. During the G2 phase (postsynthesis period) of the cell cycle the synthesis of tubulin proteins which take part in the production of the division spindle was impaired. What process can be dis- turbed?
– Chromosome despiralization – Chromosome spiralization
+ Disjunction1 of daughter chromosomes
1 In the book “Collection of tasks…” the term "divergence" is used (this is a mistake).
– Formation of ribosome subunits – Formation of nucleolus
39. A patient has an acute pancreatitis which can develop into pancreas autolysis. The dysfunction of what organelles can cause this pathology?
+ Lysosomes – Mitochondria – Ribosomes – Centrioles – Microtubules
40. Human karyotype is studied when a cell is at metaphase.
What do we call the substance that can stop the cell division at this stage?
– Methanol – Iodine + Colchicine
– Potassium chloride – Ethanol
41. During the inspection of a girl's karyotype a shortened arm of the 20th pair chromosome was found. What do we call this muta- tion?
– Duplication + Deletion – Inversion – Translocation
– Monosomy on the 20th chromosome
42. Mitosis is the basic mechanism of a cell that provides the de- velopment of organisms, their regeneration and reproduction. It is possible because this mechanism is responsible for:
– Formation of polyploid cells – Crossing-over
+ Equal distribution1 of chromosomes between daughter cells – Irregular distribution1 of chromosomes between daughter cells – Change of genetic information
43. Influenced by some chemical substances, the process of ribo- some subunits formation has been impaired in a cell. In conse- quence this will stop the synthesis of:
– carbohydrates + proteins
1 In the book “Collection of tasks…” the term "divergency" is used (this is a mistake).
– lipids – DNA – RNA
44. Under the influence of gamma-radiation a fragment of a chromosome was lost. What chromosomal mutation is it?
+ Deletion – Duplication – Inversion
– Intrachromosomal translocation – Interchromosomal translocation
45. In a cytogenetic laboratory the karyotype of a healthy man was studied. 46 chromosomes were seen in each somatic cell.
How many autosomes does each cell include?
– 23 – 22 + 44 – 46 – 92
46. The study of the female karyogram shows that the centromere in X chromosome is placed near the centre. What do we call such chromosome?
– Telocentric – Subacrocentric1 + Submetacentric – Acrocentric – Metacentric
47. Examination of a patient with hepatolenticular degeneration revealed that synthesis of ceruloplasmin protein has a defect.
What organelles is this defect connected with?
– Agranular endoplasmic reticulum – Mitochondrions
– Golgi complex
+ Granular endoplasmic reticulum – Lysosomes
48. As a result of expression of some genome components the embryo cells acquire typical morphological, biochemical and func- tional properties. Name this process:
– capacitation – reception
1 such term is not used in the world.
– determination + differentiation – induction
CLASSICAL GENETICS
49. Woman applied to the medico-genetic consulting centre for in- formation about the risk of haemophilia in her son. Her husband has been suffering from this disease since birth. Woman and her parents are healthy (don't have haemophilia). Is the boy likely to have the disease in this family? (2004)
– 25% of the boys will be ill – All boys will be ill
+ All boys will be healthy – 50% of the boys will be ill – 75% of the boys will be ill
50. A woman with O (I) blood group has born a child with AB blood group. Woman's husband has A blood group. What genetic interaction explains this phenomenon? (2006)
+ Recessive epistasis – Polymery
– Complementation – Codominance
– Incomplete dominance
51. A couple came for medical genetic counseling. The man has hemophilia, the woman is healthy and there were no cases of hemophilia in her family. What is the risk of having a sick child in this family? (2005)
– 25%
+ 0%
– 100%
– 75%
– 50%
52. It is known that the gene responsible for development of blood groups according to ABO system has three allelic variants.
If a man has IV blood group, it can be explained by the following variability form: (2010, 2012)
– phenocopy – phenotypic – genocopy – mutational + combinative
53. A woman with III (B) rh– blood group has born a child with II (A) blood group. The child is diagnosed with hemolytic disease of newborn as a result of rhesus incompatibility. What blood group
is the child's father likely to have? (2007) – I (O), Rh+
– I (O), rh– – II (A), rh– + II (A), Rh+ – III (B), Rh+
54. Hypertrichosis of auricles is caused by a gene that is localized in Y-chromosome1. Father has this feature. What is the probabil- ity that son will have this anomaly? (2006)2
– 25%
– 35%
– 0%
+ 100%
– 75%
55. Hartnup disease is caused by point mutation of only one gene which results in disturbance of tryptophane absorption in the bowels and its resorption in the renal tubules. It is the reason for disorder of both digestive and urination systems. What genetic phenomenon is observed in this case? (2008)
+ Pleiotropy – Semidominance3
– Complementary interaction – Codominance
– Polymery
56. A family of students who came from Africa got a child with anemia signs. The child died soon. Examination revealed that the child's erythrocytes have abnormal semilunar shape. Specify genotypes of the child's parents: (2010)4
– aa × aa
1 This information is out of date. According to more careful study, this trait is autosomal (some families hid their affected female members).
2 In the book “Collection of tasks…” another similar question is also present: An excessive ear pilosis (hypertrichosis) is determined by the gene, which is localized in Y chromosome. A man has got this feature. What is the probability of his having a son with such a feature? Answers:
75%; 0%; 25%; 35%; 100%. Authors propose the answer "100%" as correct but this is mis- take. When you ask about probability that parents will have a son with a feature, you should to calculate this probability among ALL children and correct answer must be 50%. Hence authors do not propose correct answer at all.
3 Incomplete dominance.
4 In the book “Collection of tasks…” this question is written as follows: In a family of students from Africa a child with signs of anemia was born. The child died within a short time. It was found that the child's erythrocytes were shaped like a sickle. What genotypes may the parents have if they have a light form of anemia?
– Aa × aa – Aa × AA – AA × AA + Aa × Aa
57. One of the parents is suspected of having phenylketonuria re- cessive gene. What is the risk of giving birth to a child with inborn phenylketonuria? (2011)
– 50%
+ 0%
– 100%
– 25%
– 75%
58. A boy has I (I0I0) blood group and his sister has IV (IAIB) blood group. What blood groups do their parents have? (2008)
+ II (IAI0) and III (IBI0) – II (IAIA) and III (IBI0) – I (I0I0) and III (IBI0) – I (I0I0) and IV (IAIB) – III (IBI0) and IV (IAIB)
59. Examination of newborns in one of the Ukrainian cities re- vealed a baby with phenylketonuria. The baby's parents don't suf- fer from this disease and have two other healthy children. Specify the most likely parents' genotype with phenylketonuria gene:
(2012) – AA x aa – Aa x AA – Aa x aa + Aa x Aa – aa x aa
60. Inclination to diabetes mellitus is provoked by the autosomal recessive gene. This gene becomes apparent only in 30% of ho- mozygous individuals. What genetic regularity is observed in this case?
– Discontinuity – Complementarity – Gene expressiveness + Incomplete penetrance – Pleiotropy
61. Children with normal hearing have been born by deaf and dumb parents with the genotype DDee and ddEE. What is the form of gene interaction between the genes D and E?
+ Complementarity – Complete dominance – Epistasis
– Polymery – Codominance
62. Some people with good clinical health can feel anemia symp- toms in the conditions of high mountains. During their blood test we can find sickle-shaped erythrocytes. What genotype can a person with occasional symptoms of sickle-cell anemia have?
– XcXc – aa – AA + Aa – XcY 1
63. The same genotype in a human can cause the development of a feature with different degrees of manifestation that depends on the interaction of this gene with the others and on the influence of environmental conditions. What do we call the degree of phe- notypic manifestation of the character controlled by a definite gene?
– Inheritance – Penetrance + Gene expression2 – Mutation
– Polymery
64. A husband is a homozygous by a dominant gene which causes polydactyly. His wife is a homozygote by recessive allele of this gene. Which of the below mentioned genetic regularities can be apparent in their children as for their having polydactyly?
– The law of segregation + The law of dominance3
– The law of independent assortment – Linkage of genes
– Sex-linked inheritance
65. Healthy parents have a son with phenylketonuria4, but owing
1 Other possible answers: XCXc, XCY.
2 This answer (in the book “Collection of tasks…”) is not good. The answer "expressivity" is better.
3 There is a mistake in this question in the book “Collection of tasks…” – incorrect word com- bination "The law of unit characters" is used in this book.
4 There is a mistake in this question in the book “Collection of tasks…” – incorrect phrase
to a special diet he has normal development. What type of vari- ability is his normal development connected with?
– Mutational variability – Combinative variability + Modificative variability – Genotype variability – Inherited variability
66. Parents with a normal phenotype gave birth to an ill with albi- nism child (the feature that is inherited by the autosomal reces- sive type). What genotype do the parents have?
– AA and aa – AA and AA – AA and Aa + Aa and Aa – aa and aa
67. Alcaptonuria1 is inherited as an autosomal recessive feature.
Parents with a normal phenotype have a baby with alcaptonuria.
What genotype do parents have?
– aa and aa – AA and AA – AA and Aa – Aa and aa + Aa and Aa
68. The intensity of human skin pigmentation is controlled by a few pairs of nonallelic dominant genes. It was found that if the quantity of the genes increases, the pigmentation become more intensive. What do we call this type of genes' interaction?
– Epistasis – Pleiotropy + Polymery – Codominance – Complementary
69. Endemic goiter is widespread among Transcarpathian popula- tion due to iodine deficiency in food. What form of variability is this case based on?
– Mutational + Modification
"Parents with normal health have an ill with phenylketonuria son" is used in this book.
1 In the book “Collection of tasks…” another similar question is also present: Galactosemia is an autosomal recessive character. What genotypes may healthy parents have if their baby has galactosemia?
– Combinative – Hereditary – Genotypical1
70. Very big teeth is an Y-linked sign. Mother's teeth are of nor- mal size, and her son's teeth are very big. Probability of father's having very large teeth is:
– 12,5%
– 25%
– 50%
– 75%
+ 100%
71. A wide cleft between incisors of both mother and father is the dominant feature. They are both homozygous. What genetic regularity will their children have?
+ Uniformity of first generation hybrids – Hybrid segregation by phenotype – Independent inheritance of traits – Non-linked inheritance
– Linked inheritance
72. Cystinuria in humans shows itself in form of cystine stones in kidneys (homozygotes) or else an increased rate of cystine in urine (heterozygotes). Cystinuria is a monogenic disease. Specify the type of interaction between cystinuria genes and normal rate of cystine in urine:
– Complete dominance + Semidominance – Codominance – Complementarity – Epistasis
1 Other possible incorrect answers: "Ontogenetic", "Correlative".
MOLECULAR GENETICS
73. RNA polymerase II is blocked due to amanitine poisoning (poi- son of death-cup). It disturbs: (2003, 2006)
+ synthesis of mRNA – primers synthesis – synthesis of tRNA – reverse transcription – maturation of mRNA
74. Genetic structure of eukaryote is "exon–intron–exon". This structure-functional organization of the gene caused transcription peculiarities. What will be pro-mRNA according to the scheme?
(2003, 2004) – Exon-exon-intron – Intron-exon + Exon-intron-exon – Exon-intron – Exon-exon
75. Part of the DNA chain turned 180 degrees as a result of gamma radiation. What type of mutation took place in the DNA chain? (2003, 2005, 2006)
+ Inversion – Deletion – Translocation – Doubling – Replication
76. An experiment proved that UV-radiated cells of patients with xeroderma pigmentosum restore the native DNA structure slower than cells of healthy individuals as a result of repair enzyme de- fect. What enzyme helps this process? (2006)
– Primase – RNA ligase
– DNA polymerase III + Endonuclease – DNA gyrase
77. Nowadays about 50 minor bases have been found in the tRNA structure besides the main four nitrogenous bases. Choose the minor nitrogenous base: (2006)
– cysteine + dihydrouracil – cytosine – uracil
– adenine
78. A patient's organism has decreased concentration of magne- sium ions that are necessary for attachment of ribosomes to the granular endoplasmic reticulum. It is known that it causes distur- bance of protein biosynthesis. What stage of protein biosynthesis will be disturbed? (2007, 2008, 2009, 2010)
– Amino acid activation – Termination
– Transcription – Replication + Translation
79. RNA of the AIDS virus penetrated into a leucocyte and forced a cell to synthetize a viral DNA by means of reverse transcriptase.
This process is based upon: (2007) – replication
+ reverse transcription – operon repression – reverse translation – operon depression
80. Labelled amino acids alanine and tryptophane were injected to a mouse in order to study localization of protein synthesis in its cells. The labelled amino acids will be accumulated near the fol- lowing organellas: (2007, 2008, 2010)
– agranular (smooth) endoplasmic reticulum – lysosomes
– Golgi apparatus + ribosomes – cell center
81. In some regions of South Africa there is a spread of sickle-cell anemia, in which erythrocytes have shape of a sickle as a result of substitution of glutamic acid by valine in the hemoglobin mole- cule. What is the cause of this disease? (2007)
– Genomic mutation – Crossing over – Transduction
– Disturbance of mechanisms of genetic information realization + Gene mutation
82. It was found that some compounds, for instance fungi toxins and some antibiotics can inhibit activity of RNA polymerase. What process will be disturbed in a cell in the case of inhibition of this enzyme? (2008, 2011)
+ Transcription – Replication – Translation – Processing – Repair
83. It was proved that a molecule of immature mRNA (precursor mRNA) contained more triplets than amino acids found in the synthesized protein. The reason for that is that translation is normally preceded by: (2008)
– initiation – replication + processing – repair – mutation
84. It was revealed that T lymphocytes were affected by HIV. Vi- rus enzyme – reverse transcriptase (RNA-dependent DNA poly- merase) – catalyzes the synthesis of: (2008, 2011)
– viral DNA on DNA matrix
+ DNA on the matrix of virus mRNA – mRNA on the matrix of virus protein – DNA on virus ribosomal RNA
– virus informational RNA on the matrix of DNA
85. You are studying functioning of the bacterial operon. The op- erator has been released from the repressor. Immediately after this the following process will start in the cell: (2009)
– processing + transcription – replication – translation – repression
86. According to the model of double DNA helix that was sug- gested by Watson and Crick, it was established that one of chains would not be lost during replication and the second chain would be synthesized complementary to the first one. What mechanism of replication is it? (2010)1
+ Semiconservative – Analogous – Dispersed
1 In the book “Collection of tasks…” this question is written as follows: DNA double spirals, which were formed as a result of replication, consist of one maternal chain and one daughter chain. What do we call this way of replication?
– Identical – Conservative
87. Protein synthesis includes several subsequent stages. It is preceded by the synthesis of immature mRNA. What do we call this process?
– Termination – Replication – Elongation – Translation + Transcription
88. It is known that the genetic code is degenerate and has triplet nature. What nucleotide can be changed in the coding triplet without loosing its sense?
– Second – First + Third
– Second or third – First or second
89. It was determined that the mRNA triplet sequence totally cor- responded to the amino acid sequence in the polypeptide chain.
What do we call this characteristic of the genetic code?
– Universality – Triplet nature – Specificity – Degeneracy + Collinearity1
90. Polypeptide which has been synthesized on the ribosome in- cludes 54 amino acids. How many codons did mRNA, used as a matrix during the synthesis, have?
– 44 – 27 – 108 – 162 + 54 2
91. Different physical and chemical factors can destroy the struc- ture of DNA. What do we call the ability of the cells to regenerate the DNA structure?
1 or colinearity.
2 There is a mistake in this question. Correct answer must be 55 because we must add the stop codon.
– Transduction – Transcription – Replication + Repair
– Transformation
92. An influenza virus penetrated into a cell. The mechanism of protein biosynthesis was reorganised for the virus protein synthe- sis to occur:
+ on the polyribosomes – in the nucleus
– in the lysosomes – in the peroxisomes – in the centriole
93. One of the main characteristics of a living being is an ability to reproduction. On what level of living organisms organization does this process happen on the basis of matrix biosynthesis?
–Organismic – Subcellular – Cellular – Tissue + Molecular
94. In the nucleus the molecule of immature mRNA transforms to the molecule of the mature mRNA, which is shorter than the im- mature mRNA. What do we call the combination of stages in this transformation?
– Replication + Processing – Recognition – Transmission – Termination
95. Some mRNA triplets (UAA, UAG, UGA) code no amino acids, but in the process of reading of information they serve as termi- nators, in other words, they are able to stop the translation. What are they?
+ Stop codons – Operators – Anticodons – Exons – Introns
96. According to the hypothesis of lactose operon (Jacob, Mono, 1961), in Escherichia coli the lactose, which gets into a cell from
the environment, acts as an inducer. In what way does the lac- tose induce the synthesis of enzymes that decompose it, that is turn on the operon?
+ It combines with the repressor protein – It combines with the operator1 gene – It combines with the regulator gene – It combines with the promoter – It combines with the structural gene
97. During the synthesis period (S) of the cell cycle, the redouble of DNA quantity takes place. This process occurs as a result of:
– denaturation of DNA – dissociation of DNA + replication of DNA – DNA repair
– coagulation of DNA
98. Under the influence of mutagen the composition of some trip- lets in a gene was changed but the cell continued the synthesis of the same protein. What characteristics of the genetic code can it be connected with?
– Specificity – Universality – Triplet nature + Degeneracy – Collinearity2
99. Protein-repressor has been found in a cell. What gene codifies the amino acid sequence of this protein?3
– Promoter – Terminator + Regulator – Modifier – Operator
100. The gene which codifies the polypeptide chain consists of 4 exons and 3 introns. When processing is over, the mature mRNA consists of nucleotides, which are complementary to:
– 3 introns
– 2 exons and 1 intron
1 In the book “Collection of tasks…” – the term "operator gene" is used. Correct term is "gene operator" or "operator".
2 or colinearity.
3 It should be emphasised that promoter, terminator, and operator are NOT genes, but they are regulatory regions of genes!
– 1 exon and 1 intron + 4 exons
– 4 exons and 3 introns
101. The transcription is taking place in human cells. RNA poly- merase enzyme moving along the DNA molecule has reached a specific nucleotide sequence; after that the transcription ended.
What do we call this DNA site?
– Operator + Terminator – Promoter – Repressor – Regulator
102. The work of bacterium operon is being studied. A gene op- erator has been released from the protein-repressor. What do we call the process which begins right after that?
– Amino acids activation – Translation
– Replication – Processing + Transcription
103. It is known that the information about the amino acid se- quence in the protein molecule is written in the form of nucleotide sequence. There are 4 types of nucleotides in the DNA molecule.
Different amino acids are codified by a different number of triplets – from one to six. What do we call this property of the genetic code?
– Triplet nature – Universality – Collinearity1 + Degeneracy – Specificity
104. The operator is known to be responsible for joining the RNA polymerase enzyme and initiating the transcription. At that site deletion of two nucleotide pairs has taken place. What conse- quences could it have?
+ Lack of protein synthesis – Formation of abnormal proteins
– Synthesis of protein in unlimited quantities – Formation of normal protein
1 or colinearity.
– Short finish of protein synthesis
105. During translation a few ribosomes, placed along the mRNA molecule at a certain distance from one another, join each mRNA simultaneously. What do we call the translation complex that consists of one mRNA molecule and some ribosomes which are placed on it?
– Centrosome – Lysosome – Phagosome – Nucleosome + Polysome
106. Human cells were influenced by ultraviolet radiation, and as a consequence of this the DNA molecules had been changed1. Nev- ertheless, by means of specific enzymes the DNA structure was renewed. What do we call this phenomenon?
– Replication – Duplication + Repair – Initiation – Termination
107. Sickle-sell anemia, when erythrocytes are in the form of a sickle, is widespread among the population of some districts in tropic Africa. What biological phenomenon is this disease based on?
+ Gene mutation
– Chromosomal aberration – Modification
– Chromosomal mutation – Transduction
108. The students studied peculiarities of genetic code and found out that there are amino acids corresponded by 6 codons, 5 amino acids – 4 different codons. Other amino acids are codified by three or two codons and only two amino acids are codified by one codon. What peculiarity of genetic code did the students find out?
– Versatility – Collinearity + Redundancy2
1 In the book “Collection of tasks…” the word "destroyed" is used (this is a mistake).
2 i. e. degeneracy.
– Unidirectionality – Triplet code
109. In a genetical laboratory in course of work with DNA mole- cules of white rats of Wistar's line a nucleotide was substituted for another one. As a result only one amino acid was substituted in the peptide. This result is caused by the following mutation:
– Deletion – Duplication
– Displacement of reading frame + Transversion1
– Translocation
110. Blood of a child and putative father was referred to forensic medical examination for affinity2. What chemical components should be identified in the blood under study?
– Transfer RNA – Ribosomal RNA – Messenger RNA + DNA
– snRNA
111. In a cell the mutation of the first exon of structural gene took place. The number of nucleotide pairs has decreased – 250 pairs instead of 290. Determine the type of mutation:
– Inversion – Duplication + Deletion – Translocation – Nonsense mutation
112. DNA replication (selfreproduction) in human cells occurs ac- cording to semiconservative mechanism. Nucleotides of the new DNA chain are complementary to:
+ Maternal chain – Sense codons
– DNA polymerase enzyme – Introns
– RNA polymerase enzyme
1 In fact, this mutation is called "missense mutation" (variant of the point mutation) and can be both transversion and transition depending on the nature of base substitution.
2 In the collection of test question take from the Internet site of Testing center, incorrect term
"affiliation" is used.
MEDICAL GENETICS
113. Patient experienced increased susceptibility of the skin to the sunlight. His urine after some time became dark-red. What is the most likely cause of this? (2003)
– Pellagra – Albinism
– Hemolytic jaundice + Alkaptonuria – Porphyria
114. Examination of initial molecular structure of hemoglobin re- vealed substitution of the glutamic acid by valine. What inherited pathology is it typical for? (2003, 2004, 2005, 2006)
– Thalassemia + Sickle-cell anemia – Hemoglobinosis
– Minkowsky–Shauffard disease – Favism
115. 46 chromosomes were revealed on karyotype examination of the 5-year old girl. One of the 15th pair of chromosomes is longer than usual due to connected chromosome from the 21 pair1. What type of mutation does this girl have? (2003, 2012)
– Insufficiency2 – Deletion – Duplication – Inversion + Translocation
116. Healthy parents have got a fair-haired, blue-eyed girl. Irrita- bility, anxiety, sleep and feeding disturbance developed in the first months of the infant's life. Neurological examination revealed developmental lag. What method of genetic investigation should be used for the exact diagnosis? (2003, 2006)
– Population-statistical – Cytological
– Twin study (Gemellary) – Genealogical
+ Biochemical
117. The examination of a youth with mental retardation revealed
1 In the book “Collection of tasks…” (question No. 103) this question has the phrase "due to joining a part of chromosome of the 21 pair".
2 Another possible answer – "aneuploidy".
eunuchoid body construction and genitals underdevelopment. The cells of the oral cavity contained chromatin. What method of ge- netic investigation should be performed to make more specified diagnosis? (2004)
– Population-statistic – Dermatoglyphics – Biochemical + Cytological
– Clinico-genealogical
118. A 18-year-old man with asthenic body constitution (tall, nar- row shoulders, broad pelvis) and with poor hair on his face came to the geneticist. There was marked mental retardation. The pre- liminary diagnosis was Klinefelter's syndrom. What method of medical genetics can confirm the diagnosis? (2004)1
– Dermatoglyphics – Population-statistic – Genealogic
+ Cytogenetic – Twin study
119. A 40-year-old pregnant woman underwent amniocentesis.
The examination of fetus karyotype revealed 47,XY+21. What pa- thology of the fetus was found out? (2004)
– Phenylketonuria – Patau's disease – Klinefelter's syndrome + Down's syndrome
– Schereschevsky–Turner's disease
120. A woman who was sick with rubella during the pregnancy gave birth to a deaf child with hare's lip and cleft palate. This congenital defect is an example of: (2005)
– genocopy – Down's syndrom – Edwards' syndrom
1 In the book “Collection of tasks…” this question is written as follows: An 18-year-old young man is tall and has narrow shoulders, a large pelvis, an adult woman pattern of hair distribu- tion, and oxyphonia. Mental retardation is also present. Based on these symptoms, the provi- sional diagnosis of Klinefelter's syndrome was made by a doctor. What genetic method can confirm the diagnosis? Answers: a) Cytogenetic; b) Pedigree analysis; c) Study of twins;
d) Biochemical; e) Population-statistical. Also in this book another similar question is present:
A teenager with the provisional diagnosis of Klinefelter's syndrome came for advice to a ge- netic consultation. What genetic method does the doctor have to apply to make a correct di- agnosis?
– Patau's syndrom + phenocopy
121. An individual is characterized by rounded face, broad fore- head, a mongolian type of eyelid fold, flattened nasal bridge, permanently open mouth, projecting lower lip, protruding tongue, short neck, flat hands, and stubby fingers. What diagnosis can be put to the patient? (2005, 2006)
– Alkaptonuria + Down's syndrome – Super male – Turner's syndrome – Klinefelter's syndrome
122. A 32 y.o. man is tall, he has gynecomastia, adult woman pat- tern of hair distribution, high voice, mental deficiency, sterility.
Provisional diagnosis is Klinefelter's syndrome. In order to specify diagnosis it is necessary to analyze: (2007)
– spermatogenesis – genealogy – blood group + karyotype – leukogram
123. Autopsy of a newborn boy revealed polydactylia, micro- cephaly, cheiloschisis and uranoschisis as well as hypertrophy of parenchimatous organs. These defects correspond with the de- scription of Patau's syndrome. What is the most probable cause of this pathology? (2007)1
– Partial monosomy
+ Trisomy of the 13th chromosome – Nondisjunction of sex chromosomes – Trisomy of the 21th chromosome – Trisomy of the 18th chromosome
124. Examination of cell culture got from a patient with lysosomal pathology revealed accumulation of great quantity of lipids in the lysosomes. What of the following diseases is this disturbance typical for? (2007)
1 In the book “Collection of tasks…” this question is written as follows: The pathoanatomic in- spection of a newborn boy's dead body showed the following abnormalities: polydactyly, mi- crocephaly, a cleft lip and cleft palate, hypertrophy of the parenchymal organs. These symp- toms are typical of Patau syndrome. What is the cause of this disease? Answers: a) Trisomy on the 21st chromosome; b) Trisomy on the 18th chromosome; c) Trisomy on the 13th chromo- some; d) Trisomy on X chromosome; e) Monosomy on X chromosome.
– Galactosemia – Phenylketonuria + Tay–Sachs disease – Gout
– Wilson disease
125. Examination of a 12-year-old boy with developmental lag re- vealed achondroplasia: disproportional constitution with evident shortening of upper and lower limbs as a result of growth disorder of epiphyseal cartilages of long tubular bones. This disease is:
(2008, 2011) – congenital – acquired
– inherited, sex-linked – inherited, recessive + inherited, dominant
126. Medical examination at the military registration and enlist- ment office revealed that a 15-year-old boy was high, with eunuchoid body proportions, gynecomastia, female pattern of pu- bic hair distribution. The boy had also fat deposits on the thighs, no facial hair, high voice, subnormal intelligence quotient. Which karyotype corresponds with this disease? (2009, 2011)
+ 47, XXY – 47, XXX – 46, XY – 46, XX – 45, XO
127. A 1.5-year-old child has mental and physical lag, decolorizing of skin and hair, decrease in catecholamine concentration in blood. When a few drops of 5% solution of trichloroacetic iron had been added to the child's urine it turned olive green. Such altera- tions are typical for the following pathology of the amino acid me- tabolism: (2009)1
– albinism – xanthinuria + phenylketonuria – alkaptonuria – tyrosinosis
1 In the book “Collection of tasks…” this question is written as follows: A few months after birth a child developed symptoms of the CNS disorder. The skin and hair became lighter. The solu- tion of 5% trichloroacetic ferric lactase, added to fresh urine, gives it olive-green coloring.
What kind of hereditary disorder is characterized by these manifestations? Answers: tyrosino- sis; alcaptonuria; fructosuria; phenylketonuria; albinism.
128. A 28-year-old female patient consulted a gynecologist about sterility. Examination revealed underdeveloped ovaries and uterus, irregular menstrual cycle. Analysis of the sex chromatin revealed two Barr's bodies in most somatic cells. What chromo- some disease is most likely? (2009, 2011)1
– Turner's syndrom + Triple X syndrom – Klinefelter's syndrom – Patau's syndrom – Edwards' syndrom
129. A couple had a child with Down's syndrom. Mother is 42 years old. This disease is most probably caused by the following impairment of prenatal development: (2010)
– blastopathy + gametopathy – embryopathy
– non-specific fetopathy – specific fetopathy
130. Cytogenetic examination of the patient with reproductive dysfunction revealed normal karyotype 46,XY in some cells, but most cells have karyotype of Klinefelter's syndrom – 47,XXY.
Such phenomenon of cell heterogeneity is called: (2010, 2011, 2012) + mosaicism
– duplication – inversion – heterogeneity2 – transposition
131. During a prophylactic medical examination a 7-year-old boy was diagnosed with daltonism. His parents are healthy and have normal color vision, but his grandfather on his mother's side has the same abnormality. What is the type of the abnormality inheri- tance? (2003, 2006, 2009, 2012)
– Autosomal dominant – Sex-linked dominant + Sex-linked recessive – Autosomal recessive
1 In the book “Collection of tasks…” this question is written as follows: A 28-year-old woman saw a physician because of infertility. Underdevelopment of the ovary and the womb, disorder of the menstrual cycle were diagnosed. During the test of buccal epithelium cells it appeared that most of their nuclei had two Barr bodies. The neutrophil nuclei had two "drumsticks" each.
What provisional diagnosis can we make in this case?
2 During exams in 2010 and 2011 this answer was replaced by "monomorphism".
– Incomplete dominance1
132. A married couple consulted a specialist at the genetic consul- tation about probability of having children with haemophilia. Both spouses are healthy, but the wife's father has haemophilia. In this family hemophilia may be passed to: (2009)
– daughters only – all the children – half of daughters + half of sons
– both sons and daughters
133. The study of the genealogy of a family with hypertrichosis (hirsutism or pilosis) has demonstrated that this trait is mani- fested in all generations only in men and is inherited by son from his father. What is the type of hypertrichosis inheritance? (2003, 2004, 2005)2
– Autosomal-recessive – X-linked recessive + Y-linked3
– Autosomal-dominant – X-linked dominant
134. After the genealogy analysis a geneticist came to the conclu- sion: a feature is manifested in each generation, men and women inherit the feature with equal frequency, parents in the equal way give this feature to their offspring. What type of inheritance does the investigated feature have? (2004)4
+ Autosomal-dominant inheritance – X-linked dominant inheritance – X-linked recessive inheritance – Autosomal-recessive inheritance – Y-linked inheritance
135. A healthy woman has three sons affected by color blindness who were born after her two marriages. Children of her both hus- bands are healthy. What is the most possible pattern of inheri-
1 In the book “Collection of tasks…” and in some exam booklets – "semidominance".
2 In the book “Collection of tasks…” this question is written as follows: In a family pedigree hy- pertrichosis (excessive pilosis of the auricle) is observed. This feature appears in each gen- eration and is typical only of men. What type of inheritance does this feature have?
3 This information is out of date. According to more careful study, this trait is autosomal.
4 In the book “Collection of tasks…” this question is written as follows: Due to the results of the pedigree analysis a geneticist found out that a feature becomes apparent in each generation, a male and a female inherit this feature with the same frequency, both parents transmitting this feature to their children. What type of inheritance does this feature have?
tance of this disease? (2005) + X-linked recessive
– Autosomal-recessive – Y-linked
– Autosomal-dominant – X-linked dominant
136. A geneticist analyzed the genealogy of a family and found that both males and females may have the illness, not across all the generations, and that healthy parents may have ill children.
What is the type of illness inheritance? (2006) + Autosomal-recessive
– Y-linked
– X-linked recessive – Autosomal-dominant – X-linked dominant
137. As a result of prophylactic medical examination a 7 year old boy was diagnosed with Lesch–Nyhan syndrome (only boys fall ill with it). The boy's parents are healthy but his grandfather by his mother's side suffers from the same disease. What type of dis- ease inheritance is it? (2008)
– Autosomal recessive – Dominant, sex-linked – Autosomal dominant + Recessive, sex-linked – Y-linked
138. Sex chromosomes of a woman didn't separate and move to the opposite poles of a cell during gametogenesis (meiosis). The ovum was impregnated with a normal spermatozoon. Which chromosomal disease can be found in her child? (2011, 2012)
– Patau's syndrome + Turner's syndrome – Cat cry syndrome – Edwards' syndrome – Down's syndrome
139. A man suffering from a hereditary disease married a healthy woman. They got five children, three girls and two boys. All the girls inherited the father's disease. What is the type of the dis- ease inheritance? (2009, 2012)
– Autosomal recessive – Y-linked
– Recessive, X-linked
+ Dominant, X-linked – Autosomal dominant
140. According to the phenotypic diagnosis, a female patient has been provisionally diagnosed with X-chromosome polysomia. This diagnosis can be confirmed by a cytogenetic method. What karyotype will allow to confirm the diagnosis? (2012)
– 46,XX – 48,XXXY – 48, XXYY – 47,XXY + 47,XXX
141. During the examination of the woman's epithelium of the cheek mucosa it was established that in most cells the nuclei had two Barr bodies. What provisional diagnosis can we make in this case?
– Trisomy on the 13th chromosome – Trisomy on the 21st chromosome + Trisomy on X chromosome – Trisomy on the 18th chromosome – Monosomy on X chromosome
142. As a result of the abnormalities in the chromosomes disjunc- tion1 in meiosis, a secondary oocyte, which contains only 22 autosomes, has been formed. What disease can the baby have after the impregnation of this secondary oocyte by a normal spermatozoon?
– Klinefelter's syndrome + Turner's syndrome
– Trisomy on the X-chromosome – Down's syndrome
– Edwards' syndrome
143. During the observation of a baby boy a pediatrician noticed that the baby's crying was similar to a cat's cry. Besides, the baby had microcephaly and abnormality in heart development. By means of the cytogenetic method it was found that the baby's karyotype was 46, XY, 5p–. At what mitotic stage was the karyo- type of the baby examined?
+ Metaphase – Prometaphase – Prophase
1 In the book “Collection of tasks…” the term "divergence" is used (this is a mistake).
– Anaphase – Telophase
144. During the examination of the man's epithelium of the cheek mucosa it was established that in most cells the nuclei had Barr bodies. What syndrome is it typical of?
– Turner's syndrome + Klinefelter's syndrome – Trisomy on X chromosome – Down's syndrome
– Edwards' syndrome
145. A human has galactosemia – a disease of accumulation.
Which genetic method can we use to diagnose the case?
– Cytogenetic + Biochemical
– Population-statistical – Study of twins – Pedigree analysis
146. One of the forms of rickets is inherited in the autosomal dominant way. This disease is a result of:1
– Aneuploidy
– Changes in the number of chromosomes – Chromosomal mutations
– Polyploidy + Gene mutations
147. A baby was born with abnormalities of the external and in- ternal organs development. During the check up the following ab- normalities were found: epicanthus, shortened extremities, a small skull, impaired development of the cardiovascular system.
On these grounds the provisional diagnosis of Down's syndrome was made. What genetic method can confirm this pathology?
– Pedigree analysis – Population-statistical – Study of twins + Cytogenetic – Biochemical
148. Both a mother and a father are phenotypically healthy. They have a sick baby in whose blood and urine phenylpyruvic acid has been found, which indicates phenylketonuria. What is the type of
1 This question has bad answers because aneuploidy and polyploidy are changes in the num- ber of chromosomes, and these three types of mutations are chromosomal mutations.
