genetics Questions

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    1. True about Klinefelter’s syndrome are all, EXCEPT:(1) Increased frequency of occurrence of maternal age(2) Long limbs, long penis and gynaecomastia(3) Gonadotrophin levels are increased(4) Hypo or Azoospermia
    Ans. 2
    - Klinefelter’s syndrome occurs with karyotype 47 XXY. - Frequency increases with increasing age of mother. It is not identified in infancy and prepubertal boys as clinical features are absent. In post pubertal male, infertility occurs due to Azoospermia. Gonadotrophin levels are raised. Clinical features are long legs, short penis, small soft testis and gynaecomastia. (Ref. Nelson’s Essential Paediatrics, 4th Edition, Pg. 144)

    2. The pedigree described below is an example of what pattern of inheritance? Solid figure = Affected individualsOpen figure = Unaffected individuals(1) X-linked recessive inheritance(2) X-linked dominant inheritance(3) Autosomal recessive inheritance(4) Autosomal dominant inheritance
    Ans. 2
    The X-linked dominant inheritance pattern is characterized by having affected females in the heterozygote state. Affected females are twice as common as affected males, and the affected males are hemizygotes. In vitamin D-resistant rickets, both sexes are affected. However, the serum phosphate level is less depressed; hence, the rickets is less severe in the heterozygous female than in the hemizygous male. (Ref. Harrison’s, 15th Edition, Vol. 1, Pg. 376)

    3. The presence of which HLA type is thought to pose risk for type I diabetes mellitus:(1) B17 (2) DR3/DR4(3) A2/B7 (4) A3/B62
    Ans. 2
    HLA DR3/DR4 ® Diabetes mellitus type I HLA A2/B7 ® Haemochromatosis HLA A3/B62 ® Also, haemochromatosis HLA B17 ® Psoriatic arthritis (Ref. Harrison’s, 15th Edition, Vol. 2, Pg.240 )

    4. Autosomal recessive conditions are correctly characterized by which of the following statements?(1) They are often associated with deficient enzyme activity(2) Affected individuals are likely to have affected offspring(3) Both alleles contain the same mutation(4) They are more variable than autosomal dominant conditions
    Ans. 1
    Autosomal recessive conditions tend to have a horizontal pattern in the pedigree. Men and women are affected with equal frequency and severity. It is the pattern of inheritance, most often seen in cases of deficient enzyme activity (inborn errors of metabolism). Autosomal recessive conditions tend to be more severe than dominant conditions and are less variable than dominant phenotypes. Both alleles are defective but do not necessarily contain the exact same mutation. ‘ (Ref. Isselbacher, 13th Edition, Pg. 345-346)

    5. During DNA replication, the sequence 5’-TpApGpAp-3’ would produce which of the following complementary structures?(1) 5’-GpCpGpAp-3’(2) 5’-ApTpCpTp-3’(3) 5’-UpCpUpAp-3’(4) 5’-TpCpTpAp-3’
    Ans. 4
    DNA replication entails pairing of thymine with adenine and guanine with cytosine. The chains of the double helix are thus bonded in part by a hydrogen linkage between amino and keto groups. The strands themselves are synthesized in an anti-parallel direction, i.e. the 5’ ® 3’ sequence of phosphodiester bridges mentioned in the question specifies its complement in a 3’ ® 5’ direction. (Ref. Stryer, 4th Edition, Pg. 88-89)

    6. Baban arrives in your office for genetic counselling. Baban’s brother Tapan dies at a young age from Tay-Sach’s disease. Both he and his sister Suman are unaffected. Baban’s son Amar and Suman’s daughter Asha have recently married and are expecting their first child. What is the chance that the child would have Tay-Sach’s disease?(1) 1/4 (2) 1/9(3) 1/12 (4) 1/36
    Ans. 4
    In classic Mendelian genetics the unaffected siblings, Baban and Suman, each have a 2/3 chance of being a carrier for the Tay-Sach’s disease and hence a 50% chance of passing on the carrier state. Therefore, both Amar and Asha have a 1/3 chance of being a carrier. If both Amar and Asha are carriers, there is a ¼ chance that a child would be affected with the Tay-Sachs disease. Therefore, the cumulative probability is 1/3 ´ 1/3 ´ ¼, which equals 1/36. (Ref. Harrison’s, 15th Edition, Chapter 65, 68, Pg. 375, 407)

    7. A child with cleft palate, a heart defect, and extra fifth finger is found to have 46 chromosomes with extra material on one homologue of the chromosome 5. This chromosomal abnormality is best described by which of the following terms?(1) Balanced rearrangement(2) Ring formation(3) Mosaicism(4) Unbalanced rearrangement
    Ans. 4
    Extra material (i.e., extra chromatin) seen on chromosome 5 implies recombination of chromosome 5 DNA with that of another chromosome to produce a rearranged chromosome. Since this rearranged chromosome 5 takes the place of a normal chromosome 5, there is no change in number of the autosomes (non-sex chromosomes) or sex chromosomes (X and Y chromosomes). All cells karyotyped from the patient (usually 11-25 cells) had the same chromosomal constitution, ruling out mosaicism. The patient’s clinical findings are similar to those occurring in trisomy 13, suggesting that the extra material on chromosome 5 is derived from chromosome 13, producing an unbalanced karyotype called dup (13) or partial trisomy 13. (Ref. Isselbacher, 13th Edition, Pg. 369-371)

    8. Which of the following are due to micro deletion, EXCEPT:(1) Beckwith-Widman syndrome(2) Retinoblastoma(3) Prader-Willi syndrome(4) Angelman syndrome
    Ans. 1
    Beckwith-Widman syndrome is due to microduplication on ‘p’ arm of chromosome 11. Microdeletion is seen in: (a) WAGR complex (11p13) (b) Retinoblastoma (13q14) (c) Prader-Willi syndrome (15q11) (d) Angelman syndrome (15q11) (e) DiGeorge syndrome (22q11) (Ref. Harrison’s, 15th Edition, Vol. 1, Pg. 403)

    9. Which of the following would rule out hyperuricemia in a patient?(1) Lesch-Nyhan syndrome(2) Gout(3) Purine overproduction secondary to Von Gierke’s disease(4) Carbamoyl phosphate synthase deficiency
    Ans. 4
    Carbamoyl phosphate (CAP) synthase I is found in mitochondrial matrix and is the first step in urea synthesis, condensing CO2 and NH4+. Hyperammonemia occurs when it is deficient. CAP synthase II forms CAP as the first step in pyrimidine synthesis. Its complete deficiency would probably be a lethal mutation. When its activity is decreased, purine catabolism to uric acid is decreased, decreasing the possibility of hyperuricemia. In contrast, gout, Lesch-Nyhan syndrome, high xanthine oxidase activity and Von Gierke’s disease all lead to increased urate production and excretion. (Ref. Stryer, 4th Edition, Pg. 755-758)

    10. A 16-year-old adolescent is seen in your clinic. On physical examination, you note that he has small testis for his stated age and has poorly developed secondary sexual characteristics. In addition, on physical examination, there is notable gynaecomastia. He is rather tall, with abnormally long upper and lower limbs. A buccal smear is obtained and examined microscopically. How many chromatin positive inclusion bodies are seen?(1) 0(2) 1(3) 2(4) 3
    Ans. 2
    In somatic cells only one X chromosome is active; all other X chromosomes are inactivated. Chromosome inactivation typically begins by methylation of the DNA leading to downregulation of gene transcription. The inactivation of the X chromosome causes chromosome condensation and produces a Barr body. The clinical characteristics of the patient described are consistent with classic Klinefelter’s syndrome, and the karyotype is 47, XXY. Therefore one X chromosome will be inactivated and will be seen on a buccal smear as a Barr body. The normal male has a karyotype 46, XY and there is no Barr body seen; a normal female is 46, XX and again has one Barr body. (Ref. Harrison’s, 15th Edition, Chapter 66,Pg 396-404)

    11. A couple is referred to a physician because the first three pregnancies have ended in spontaneous abortion. Chromosomal analysis reveals that the wife has two cell lines in her blood, one with a missing X chromosome (45, X) and the other normal (46, XX). Her chromosomal constitution can be described as:(1) Chimeric(2) Monoploid(3) Trisomic(4) Mosaic
    Ans. 4
    The case described in the question represents one of the commoner chromosomal causes of reproductive failure, Turner’s mosaicism. Turner’s syndrome represents a pattern of anomalies, including short stature, heart defects, and infertility. Turner’s syndrome is often associated with a 45,X karyotype (monosomy X) in females, but mosaicism (i.e., two or more cell lines in the same individual with different karyotypes) is common. However, chimerism (i.e., two cell lines in an individual arising from different zygotes, such as fraternal twins who do not separate) is extremely rare. Trisomy refers to three copies of one chromosome; euploidy, to a normal chromosome number; and monoploidy, to one set of chromosomes (haploid in humans). (Ref. Isselbacher, 13th Edition, Pg. 369-371)

    12. Following is true about LHON (Leber’s Hereditary Optic Neuropathy):(1) Females > Males(2) Peripheral scotoma(3) Sensory neuropathy(4) Presents mostly during infancy
    Ans. 2
    LHON is mitochondrial disease with maternal inheritance. It is 4 times more common in males than female. The mean age of initiation of loss of vision is 20-22 years. It causes central scotoma, bilateral visual loss sensory neuropathy, ataxia and dystonia. Tobacco and alcohol are implicated in causation of genetically predisposed individual. (Ref. Harrison’s, 15th Edition, Vol. 1, Pg. 406)

    13. The standard karyotype is performed by photomicroscopy of cells at which mitotic stage?(1) Interphase(2) Prophase(3) Metaphase(4) Anaphase
    Ans. 3
    The standard karyotype is a arrangement of chromosomes from one cell that is undergoing division at metaphase. At other mitotic stages, the chromosomes are not sufficiently condensed or are too dispersed to allow counting and comparison of pairs under the microscope. After growth, metaphase arrest, separation, hypotonic treatment, and fixing of white blood cells, smearing on a slide yields only about 3 percent of cells that can be analyzed (metaphase spreads). In high resolution chromosome analysis, less condensed chromosomes in late prophase may be analyzed (prometaphase analysis); however, this process is extremely time consuming and usually requires focus on a particular chromosome region. (Ref. Gelehrter, 5th Edition, Pg. 201-229)

    14. A 3-month-old child arrives in your clinic who has profound hypotonia. On physical examination, in addition to hypotonia, he is noted to have a brachycephalic head with a flat occiput and a low nasal bridge. The hands are short and broad, and a single crease is noted on the fifth finger. The feet show a characteristic wide gap between the first and second toe and the furrow is extending along the proximal plantar surface. A full karyotype is performed and shows an abnormality. However, the total number of chromosomes is 46. What is the most likely explanation for this child’s clinical syndrome?(1) Fragile X syndrome(2) Down’s syndrome(3) Prader-Willi syndrome(4) Cri-du-chat syndrome
    Ans. 2
    The clinical characteristics of this child are consistent with Down’s syndrome; however, the child has a normal number of chromosomes. Infrequently a child with a Robertsonian translocation can develop a Down’s syndrome phenotype. For example, a Robertsonian translocation between the long arms of chromosomes 14 and 21 results in a t(14q;21q). In this example, the child has 46 chromosomes; however, the karyotype is effectively trisomic for chromosome 21. The phenotypic consequences of this are indistinguishable from those of the standard 47, XX or 47, XY, +21, which is the standard trisomic 21 genotype. A Robertsonian translocation involves two acrocentric chromosomes, which fuse at the centromere region and lose their short arms. A carrier of a balanced Robertsonian translocation has 45 chromosomes, including the translocation chromosome. (Ref. Harrison’s, 15th Edition, Chapter 65, 66, 68,Pg, 380-400,408

    15. In most patients with gout as well as those with Lesch-Nyhan syndrome, purines are overproduced and overexcreted. Yet the hypoxanthine analog, allopurinol, which effectively treats gout, has no effect on the severe neurological symptoms of Lesch-Nyhan patients because it does not:(1) Decrease do novo purine synthesis(2) Decrease de novo pyrimidine synthesis(3) Diminish urate synthesis(4) Increase phosphoribosyl pyrophosphate (PRPP) levels
    Ans. 1
    Most forms of gout are probably X-linked recessive with deficiencies in phosphoribosyl pyrophosphate (PRPP) synthetase, the first step of purine synthesis. Some patients may have a partial deficiency of hypoxanthine guanine phosphoribosyl transferase (HGPRTase) which salvages hypoxanthine and guanine by transferring the purine ribonucleotide of PRPP to the bases and forming iosinate and guanylate respectively. In all of these patients, the hypoxanthine analog allopurinol has two actions: (1) it inhibits xanthine oxidase, which catalyzes the oxidation of hypoxanthine to xanthine and then to uric acid (2) it forms an inactive allopurinol ribonucleotide from PRPP in a reaction catalyzed by HGPRTase, thereby decreasing the rate of purine synthesis. In contrast, because of the total loss of HGPRTase activity in Lesch-Nyhan patients, the allopurinol ribonucleotide cannot be formed. Thus, PRPP levels are not decreased and de novo purine synthesis continues unabated. The gouty arthritis caused by urate crystal formation is relieved in Lesch-Nyhan patients but their neurological symptoms are not. (Ref. Stryer, 4th Edition, Pg. 756-758)

    16. Which of the following is called an acrocentric chromosome?(1) Chromosome number 20(2) Chromosome number 17(3) Chromosome number 7(4) Chromosome number 13
    Ans. 4
    Groups A (#1-3) and F (#19-20) contain metacentric chromosomes (centromere approximately in middle), while groups B (#4,5) C (#6-12), and E (#16-18) contain submetacentric chromosomes (centromere toward one end). Groups (#13-15), G(#21-22) and the Y chromosome have centromeres very near their tips and are called acrocentric chromosomes. (Ref. Isselbacher, 13th Edition, Pg. 369-371)

    17. Chromosomal imbalance is most frequent during which of the following stages of human development?(1) Embryonic(2) Fetal(3) Childhood(4) Adult
    Ans. 1
    Chromosomal aberrations occur in approximately 1 in 200 live born infants. Although the exact frequency of chromosomal anomalies in human embryos (i.e., <8 weeks’ gestation) is unknown, the numbers above indicate a substantial frequency of at least 7.5 percent. (Ref. Isselbacher, 13th Edition, Pg. 343)

    18. The localization of a genetic locus to a particular chromosomal region using linkage or molecular analysis is known as:(1) Gene mapping(2) Gene cloning(3) Gene isolation(4) Gene splicing
    Ans. 1
    A major thrust of the human genome project is to identify every human gene and to localize it to a particular chromosome region. Gene mapping involves systematic isolation (cloning) of DNA segments by gene splicing. An order array of DNA segments are isolated from each chromosomal region, and their component genes are identified through DNA sequencing and reference to the genetic code. Application of these methods to all human chromosomes, with the goal of assembling a map for all human genes, is called the human genome project. Gene therapy is a set of techniques that employs gene isolation and gene splicing to replace or alter the expression of abnormal genes. (Ref. Isselbacher, 13th Edition, Pg. 356-360)

    19. Which of the following is an example of a balanced karyotype?(1) 46, XX, –14, + t(14;21) (p11;q11)(2) 46 XY, –5, +der5 (5;11) (q21l;q21) mat (3) 46, XY, +t(9;22) (q34;q11.2)(4) 46, X,–X, +t(X;21) (p21;p12)
    Ans. 4
    The karyotype listed in (4) is a result of a balanced translocation and has equal gene copies of all 46 chromosomes. Balanced translocations typically do not lead to an abnormal phenotype; however, translocations can lead to the formation of unbalanced gametes and therefore carry a high risk of abnormal offspring. (Ref. Harrison’s, 15th Edition, Chapter 66,Pg 396-404)

    20. The hydrolytic step leading to the release of a polypeptide chain from a ribosome is catalyzed by:(1) Stop codons(2) Peptidyl transferase(3) Release factors(4) Dissociation of ribosomes
    Ans. 2
    During the course of protein synthesis on a ribosome, peptidyl transferase catalyzes the formation of peptide bonds. However, when a stop codon is reached, such as UAA, UGA, or UAG, aminoacyl-tRNA does not bind to the A site of a ribosome. One of the proteins known as a release factor binds to the specific trinucleotide sequence present. This binding of the release factor activates peptidyl transferase to hydrolyze the bond between the polypeptide and the tRNA occupying the P site. Thus, instead of forming a peptide bond, peptidyl transferase catalyzes the hydrolytic step that leads to the release of newly synthesized proteins. (Ref. Stryer, 4th Edition, Pg. 901-902)

    21. What is the baseline risk for congenital malformation in the average pregnancy?(1) 2/10,000(2) 2/1000(3) 2/100(4) 2/10
    Ans. 3
    The incidence of major congenital anomalies or genetic diseases at birth is at least 2 to 3 percent of all pregnancies. If children are followed until age 7, the rate approaches 10 percent due to a variety of problems, such as hearing and heart defects, that are not appreciated early in life. Genetic disorders affect every organ system and cause a significant portion of infant mortality. (Ref. Thompson, 5th Edition, Pg. 2)

    22. All are seen in Marfan’s syndrome, EXCEPT:(1) Due to mutation of FBN I gene(2) Spontaneous pneumothorax(3) Autosomal recessive disorder(4) Flat cornea may be present
    Ans. 3
    Marfan’s syndrome is an autosomal dominant disease due to mutation of fibrillin protein on FBN I gene on chromosome 15. In addition to ectopia lentis, flat cornea may also be seen. Spontaneous penumothorax and apical blebs are pulmonary complications.

    23. True description of direct effects of AIDS drug azidothymidine (AZT) include all of the following, EXCEPT:(1) It prevents viral replication(2) It inhibits RNA synthesis(3) It inhibits DNA polymerase(4) It inhibits DNA provirus production
    Ans. 2
    The AIDS treatment drug azidothymidine (AZT) exerts its effect by inhibiting reverse transcriptase. Thus, it prevents replication of the human immunodeficiency virus. Reverse transcriptase is an RNA-directed DNA polymerase. The RNA of retroviruses utilize reverse transcriptase to synthesize DNA provirus, which, in turn, synthesizes new viral RNA. AZT does not directly inhibit synthesis of new viral RNA. (Ref. Stryer, 4th Edition, Pg. 58, 229, 384, 834

    24. Sickle cell disease is a result of a homozygous hemoglobin S (hemoglobin SS). States of decreased oxygen tension causes red blood cell sickling, which leads to occlusion of small vessels and results in “sickle crisis.” Patients typically present at birth but usually live into early adulthood. Despite the severe manifestations of this disease, the heterozygous frequency in West Africa is approximately 20%. Which of the following explanation best describes the reason for the maintenance of the hemoglobin S polymorphic phenotype in West Africa?(1) A selection advantage of both homozygous and heterozygous states(2) A selection advantage of the heterozygous state over both homozygous states(3) A selection advantage of bA/bS over bS/bS(4) A selection advantage of bA/bS over bA/bA
    Ans. 2
    Sickle cell disease arises from a single substitution in the second position of the triplet coding for glutamic acid and alters this to the code for valine. The carrier state in West Africa is approximately 20%, and it is generally agreed that the sickle cell mutation is a balanced polymorphism in areas with endemic malaria caused by Plasmodium falciparum. Deoxygenated erythrocytes with the sickle cell trait containing P. falciparum are more likely to undergo sickling than unparasitized cells. Hence this process probably accelerates removal of these cells from the circulation and therefore protects the individual from a malarial infections. Therefore, patients who carry the sickle cell trait (bA/bB) are at a distinct survival advantage over both homozygous states, including patients with the bA/bA phenotype. (Ref. Harrison’s, 15th Edition, Chapter 65)

    25. In 1991, it was discovered that the fragile X syndrome was caused by a mutation in the fragile X mental retardation-1 (FMR-1) gene. An area of CGG trinucleotide repeats just upstream of the coding area was found to be variable in size. All the following statements regarding the FMR1 gene are true, EXCEPT:(1) “Premutations” may expand to full mutations in future generations (2) Offsprings of male carriers inherit a premutation(3) Offsprings of female carriers may inherit a premutation or a full mutation(4) Individuals with premutation are likely to have mental retardation
    Ans. 4
    Several disorders have recently been found to be the result of expanding series of triplet repeats. These include the fragile X syndrome, myotonic dystrophy, and Huntington’s disease. Although the length of the region is variable in normal individuals, unaffected female carriers, and non-penetrant, transmitting males have “premutations” which are generally 50 to 230 repeats in length. Individuals with premutations are, therefore, phenotypically unaffected. Nonpenetrant males transmit only unstable premutations; female carriers may transmit either premutations or full mutations, which are associated with mental retardation and the other phenotypic features of the syndrome. (Ref. Isselbacher, 13th Edition, Pg. 347-349)

    26. All the following are true statements about what xanthine can be, EXCEPT:(1) A direct precursor of guanine(2) A substrate of the enzyme xanthine oxidase(3) A product of the enzyme xanthine oxidase(4) A result of oxidation of hypoxanthine
    Ans. 1
    Xanthine oxidase catalyzes the last two steps in the degradation of purines. Hypoxanthine is oxidized to xanthine, which is further oxidated to uric acid. Thus, xanthine is both product and substrate in this two-step reaction. In humans, uric acid is excreted via the urine. Allopurinol, an analogue of xanthine, is used in gout to block uric acid production and deposition of uric acid crystals in the kidney and joints. It acts as a suicide inhibitor of xanthine oxidase after it is converted to alloxanthine. Guanine can also be a precursor of xanthine. (Ref. Stryer, 4th Edition, Pg. 755-757)

    27. Match the following:1) Heterozygote a) Two locus for different allele
    2) Compound heterozygote b) One locus, one allele
    3) Double heterozygote c) One locus, one normal, one mutant allele
    d) One locus, two different, mutant allele
    (1) 1-b, 2-a 3-d(2) 1-c, 2-a 3-d(3) 1-c, 2-d 3-a(4) 1-b, 2-c 3-d
    Ans. 3
    A heterozygote, or in the case of an autosomal recessive disorder, a carrier, has one normal allele and one mutant allele at a given locus. A compound heterozygote has two different mutant alleles at same locus and a double heterozygote has one mutant allele at each of two different loci. (Ref. Isselbacher, 13th Edition, Pg. 345-346)

    28. On physical examination, the patient is noted to have some facial dysmorphism, including a long face, a prominent nose, and flattening in the malar region. In addition, the patient’s speech has an unusual quality. Which description best explains the patient’s condition?(1) Sequence(2) Syndrome(3) Disruption(4) Deformation
    Ans. 2
    The child described in the question has multiple independent anomalies that are characteristic of a syndrome. Although they are likely to be causally related, they do not appear to be sequential. These problems do not appear to be caused by the breakdown of an originally normal developmental process as in a disruption, nor do they appear to be related to a non-disruptive mechanical force as in a deformation. (Ref. Isselbacher, 13th Edition, Pg. 347-349

    29. Fluorescent insitu hybridization (FISH) analysis is useful in all the following situations, EXCEPT:(1) Determination of sex in cases of ambiguous genitalia(2) Determination of uniparental disomy(3) Rapid diagnosis of trisomies(4) Identification of submicroscopic deletions
    Ans. 2
    The availability of specific molecular probes allows the use of fluorescent in situ hybridization (FISH) analysis for the evaluation of specific chromosomal regions known to be associated with specific genetic syndromes. Probes specific for the X and Y chromosomes are used in determining sex in cases of ambiguous genitalia. The identification of three signals for specific chromosomes allows for the diagnosis of trisomies much more rapidly than standard karyotypic analysis. Submicroscopic deletions can be detected using FISH probes. Because the parental origin of chromosome cannot be determined with this technique, uniparental disomy cannot be detected. (Ref. Isselbacher, 13th Edition, Pg. 375-380)

    30. A patient has blood type AB. She has a sister with blood type O, the father has blood type A, and mother, blood type O. The maternal grandparents were first-degree cousins. What is the likely explanation for this patient’s blood type?(1) An error was made in typing of the patient’s blood type(2) The patient’s mother’s serum contains anti-A, anti-B, and anti-H antibodies(3) An error in typing occured in the mother’s blood type(4) A new mutation in the patient occurred, giving rise to the AB blood phenotype
    Ans. 2
    The H antigen is a substrate from which the A and B antigens are made. The A and B genes act on the H antigen, converting it to either the A or B antigen. The O gene is thought to be silent. Therefore, type O cells carry an unaltered H antigen. Anti-H antibodies are found in the serum of patients whose red blood cells lack the H antigen. These patients have an Oh phenotype. This was first identified in Indians living in Bombay and has been referred to as the Bombay phenotype. In this clinical situation there was consanguineous marriage between the patient’s grandparents. The mother has the Oh phenotype and therefore was unable to express the B phenotype despite the fact that she likely carries the B gene, which was then passed on to the patient. (Ref. Harrison’s, 15th Edition, Chapter 65, 114 Pg 377-396,733-36)

    31. All the following are true statements about translation, EXCEPT:(1) The genetic code can be overlapping (2) The last nucleotide in a codon has less specificity than the others(3) More than one group of nucleotides may code for a single amino acid(4) Specific nucleotide sequence signals termination of peptide chains
    Ans. 1
    Each group of three bases in a sequence codes for an amino acid. Thus, the genetic code is non-overlapping. The triplet genetic code is degenerate, which means, for most amino acids, there is more than one code word. The triplets of bases (codons) that specify the same amino acid usually differ only in the last base of the triplet Chain termination is determinated by three codons: UAA, UAG and UGA. (Ref. Stryer, 4th Edition, Pg. 104-112)

    32. All are true regarding Trisomy 21, EXCEPT:(1) Chromosomal non-dysjunction during maternal meiosis responsible for 80-90% of cases(2) Brushfield spots on iris(3) Epicanthal fold(4) Hypertonic at birth
    Ans. 4
    92% of Down’s syndrome have trisomy with an extra. 21 chromosome in all body cells. Chromosomal non- dysjunction during maternal meiosis is responsible for 90% of cases. Clinical features ® Mental retardation, Epicanthal fold, upturned nose, brushfield iris, hypotonia at birth. (Ref. Nelson Essential, Paediatrics, 4th Edition, Pg. 141)

    33. A 6-year-old girl is referred to a physician for evaluation. She is known to have mild mental retardation and a ventricular septal defect (VSD). The physician asks the parents for all the following information, EXCEPT:(1) Folate prior to and during pregnancy(2) Medications or drugs taken during pregnancy(3) Early developmental milestones (4) Parental consanguinity
    Ans. 1
    Taking a family history is a crucial ingredient in making the diagnosis of a genetic disease. It should be relevant to the clinical details of the case, and, if possible, should include information relating to multiple generations, ethnic and racial background, and a history of consanguinity. Evaluation of any child should always include a developmental history. Teratogen exposure is also important in assessing any problem that appears congential. Recently, folate deficiency in the periconceptional period has been associated with neural tube defects. (Ref. Isselbacher, 13th Edition, Pg. 342-350)

    34. A male child presents to your clinic with a history of multiple pulmonary infections. The child’s birth was complicated by meconium ileus. The child has had a recurrent cough with thick, difficult to mobilize, viscous sputum. There have been multiple episodes of recurrent pulmonary infections and abnormal chest X-rays. The child is also thin for his stated age and seems to be failing to thrive. Which of the following statements is correct concerning the mode of inheritance of this patient’s disease?(1) Most patients will have an affected parent(2) Males are more commonly affected than females(3) The recurrent risk is 1 in 4 for each subsequent sibling(4) The trait is never transmitted directly from father to son
    Ans. 3
    The patient’s clinical syndrome is consistent with cystic fibrosis inherited as an autosomal recessive disorder. Characteristically the trait appears only in siblings and not in their parents, offspring, or other relatives. On average, one-fourth of the siblings are affected. In other words, the recurrence rate for each subsequent child is 1 in 4. The parents of the affected child may be consanguineous. Males and females are equally affected. (Ref. Harrison’s, 15th Edition, Chapter 65, 68, 375-390,408)

    35. Feedback inhibition of pyrimidine nucleotide synthesis can occur by all the following means, EXCEPT:(1) Regulation of carbamoyl phosphate synthetase(2) Regulation of asparate transcarbamoylase(3) UMP allosteric effects(4) TTP allosteric effects
    Ans. 4
    Glutamine + CO2 + ATP Carbamoyl phosphate Asparate CTP UTP UMP Carbamoyl aspartate The first step in pyrimidine synthesis is the formation of carbamoyl phosphate. The enzyme of this step, carbamoyl phosphate synthetase (1), is feedback inhibited by UMP. The enzyme of the second step, aspartate transcarbamoylase (2), is feedback inhibited by CTP. ATP is an activator of this enzyme. Asparate transcarbamoylase is composed of catalytic and regulatory subunits. The regulatory subunit binds CTP or ATP. TTP has no role in the feedback regulation of pyrimidine synthesis. (Ref. Stryer, 4th Edition, Pg. 747-748)

    36. Following are true about Turner’s syndrome, EXCEPT:(1) Adult height < 150 cm(2) Coarctation of aorta(3) Cubitus varus of elbow(4) Horseshoe kidney
    Ans. 3
    Turner’s syndrome is due to functional monosomy of ‘p’arm of x-chromosome. Clinical features ® Short stature (<150cms) Sexual infantilism Bicuspid aortic valve CoA (Coarctation of Aorta) Low hairline, webbed neck, widely placed nipples. Horse shoe kidney, cubitus valgus of elbow (Ref. Nelson Essentials Paediatrics, 4th Edition, Pg. 143)

    37. Prenatal diagnosis can be done for:(1) Tay-Sach disease(2) Menke’s disease(3) Galactosemia(4) All of the above
    Ans. 4
    PRENATAL DIAGNOSIS DNA analysis a1-antitrypsin deficiency, thalassemia, sickle cell anemia, muscular dystrophy, hemophilia A, congential adrenal hyperplasia, phenylketonuria, many others
    Enzyme assay Tay-Sach’s disease, galactosemia, Hunter’s syndrome, Maple Syrup Urine disease, Wolman syndrome, Lesch-Nyhan syndrome, Gaucher disease, Menke’s disease
    Chromosome disorders Trisomy 13, 18, 21; chromosome deletions; Turner syndrome; Klinefelter’s syndrome; fragile X syndrome
    Elevated Twins, neural tube defects, intestinal obstruction, congenital hepatitis, congenital nephrosis, impending fetal demise, omphalocele
    Decreased Trisomy 21 and possibly other chromosomal disorders
    Chorionic gonadotropin
    Elevated Trisomy 21, triploidy
    Decreased Trisomy 13, trisomy 18
    Unconjugated Estriol
    Decreased Trisomy 21
    Ultrasonography Hydrops fetalis, hydronephrosis, neural tube defects, intestinal obstruction, congential heart disease, diaphragmatic hernia, gastroschisis, omphalocele, limb reduction anomalies, assessment of growth
    Cordocentesis Fetal anemia, fetal acid-base and oxygenation disorders, fetal hypoalbuminemia, thrombocytopenia, thalassemia
    Fetal skin biopsy Albinism, epidermolysis bullosa, xeroderma pigmentosum
    (Ref. Nelson Essential Paediatrics, 4th Edition, Pg. 150)

    38. A woman undergoes chorionic villus sampling (CVS), and a fetal karyotype of 46,XX/47,XX,+16 is found. The patient decides to continue the pregnancy, hoping that the abnormal cell line was an artefact. Subsequent fetal growth was delayed, but a karyotype on the small for gestational age infant was 46,XX. These findings are best explained by a:(1) True mosaicism in the embryo with later predominance of the normal cell line(2) True mosaicism in the early embryo with correction to uniparental disomy 16 in the fetus(3) Confined placental mosaicism with trisomy 16 in the fetus(4) Trisomy 16 in the newborn infant
    Ans. 2
    Trisomy 16 in the fetus or newborn is ruled out by the normal karyotype, although undetected trisomy 16 mosaicism is possible. Confined placental mosaicism does occur, and trisomy 16 cells in the placenta could cause growth retardation in the normal fetus through placental insufficiency. Culture mosaicism is common with chorionic villus sampling (CVS), because normal maternal tissue (i.e., uterine decidua) is included with the villi. Because trisomy 16 is a severe anomaly, found in spontaneous abortions but not in live births, mosaicism can easily arise through mitotic non-dysjunction to produce normal cells. Since two number 16 chromosomes in the trisomic cell derive from one parent, there is a 1/3 chance that the other parental chromosome will be lost, resulting in uniparental disomy 16. Growth retardation is a common effect of uniparental disomy and has been specifically reported with uniparental disomy 16. (Ref. Isselbacher, 13th Edition, Pg. 390)

    39. A 45-year-old man comes to your office with a history of stage I colon cancer. The patient’s father was diagnosed at age 49 with colon cancer, and his brother (the patient’s uncle) also had colon cancer diagnosed at age 47. The patient’s grandmother had endometrial cancer diagnosed at age 51. This patient most likely has a defect in which of the following genes?(1) p53(2) APC(3) MSH-2 gene affecting DNA mismatched repair(4) Retinoblastoma gene (Rb)
    Ans. 3
    The patient has Lynch syndrome, which is characterized by familial colon cancer typically affecting patients before the age of 50. In addition, a family history of endometrial or ovarian cancer has also been recognized. HNPCC mutations are caused by several different mismatched repair genes, including MSH-2 and MLH-1. These enzymes are involved in the detection of nucleotide mismatches and in the recognition of slipped strand trinucleotide repeats. Mutations of the p53 gene cause the Li-Fraumeni syndrome, which simply causes cancers involving the soft tissue or bone and, less frequently, leukemia, adrenal cortical carcinoma, or neuroblastoma. Patients are typically affected before age 30. The APC mutation causes familial polyposis syndrome and invariably leads to colon cancer before the age of 30 in affected individuals. Defects of the retinoblastoma gene cause retinoblastoma and, less frequently, osteosarcoma. The reciprocal translocation involving chromosomes 8 and 14 typically involves the c-myc that causes Burkitt’s lymphoma. (Ref. Harrison’s, 15th Edition, Chapter 65)

    40. Which of the following does not synthesize RNA?(1) RNA polymerase II(2) RNA polymerase III(3) Primase(4) Reverse transcriptase
    Ans. 4
    Reverse transcriptase is an RNA-dependent DNA polymerase that can synthesize first a single strand and then a double stranded DNA from a single strand RNA template. It was originally found in animal retroviruses. Primase' is a DNA-dependent RNA polymerase enzyme that synthesizes an RNA molecule, 10-200 nucleotides in length that initiates or “primes” DNA synthesis. RNA polymerase I synthesizes ribosomal RNA. RNA polymerase II synthesizes messenger RNA. Transfer RNA, 5SRNA, as well as other small RNAs of eukaryotes, are synthesized by RNA polymerase III. (Ref. Stryer, 4th Edition, Pg. 92, 136, 805, 835, 853-854)

    41. Neural tube defects, such as spina bifida or anencephaly, are best diagnosed by:(1) Maternal serum a-fetoprotein (MSAFP) levels and ultrasound at 16 weeks after conception(2) Amniotic fluid karyotype and ultrasound at 16 weeks after the LMP(3) Amniotic fluid a-fetoprotein (AFP) levels and ultrasound at 16 weeks after the LMP(4) Amniotic fluid acetylcholinesterase levels at sixteen weeks after conception
    Ans. 3
    Any defect of the fetal skin may elevate the amniotic a-fetoprotein (AFP) level, causing a parallel rise of this substance in the maternal blood. Neural tube defects such as anencephaly or spina bifida elevate the AFP including fetal kidney disease with leakage of fetal proteins into amniotic fluid. Mild forms of spina bifida or meningomyelocele may cover the skin so that the AFP is not elevated, and maternal serum AFP is less sensitive than amniotic fluid AFP for such cases. Ultrasound is required to detect covered neural tube defects that do not leak fetal AFP into the amniotic fluid and maternal blood. Acetylcholinesterase is an enzyme produced at high levels in neural tissue that is more specific than AFP for neural tube defects; it is used for confirmation rather than as a primary prenatal test. Chorionic villus biopsy is performed at about 10 weeks after the last menstrual period (LMP) and amniocentesis at 14-16 postmenstrual weeks. Since conception will often occur two weeks prior to the LMP, distinction between postconceptional and postmenstrual timing is important for early stages of pregnancy. Neural tube defects are usually localized, multifactorial anomalies rather than part of a malformation syndrome that can result from chromosomal aberrations. For this reason, documentation of the fetal karyotype by chorionic villus biopsy or amniocentesis will not influence the risk for neural tube defects. (Ref. Isselbacher, 13th Edition, Pg. 390)

    42. The following pedigree is an example of what pattern of inheritance? Solid figure = Affected individualOpen figure = Unaffected individuals(1) X-linked recessive inheritance(2) X-linked dominant inheritance(3) Autosomal recessive inheritance(4) Autosomal dominant inheritance
    Ans. 1
    X-linked recessive inheritance is marked by having the incidence of the trait much higher in males than in females. The genetic trait is passed from an affected male through all of his daughters to, on average, half of their sons. The trait is never transmitted directly from father to son. The trait may be transmitted through a series of carrier females; if so, the affected males are related to each other through the females, as is presented in this case. (Ref. Harrison’s, 15th Edition, Pg 376)

    43. Protein biosynthesis requires all the following, EXCEPT:(1) Ribosomal RNA(2) Messenger RNA(3) Primer protein(4) Peptidyl transferase
    Ans. 3
    Primer is not used in protein synthesis as it is in DNA synthesis. Prior to the start of protein synthesis, ATP is required for activation of amino acids. The activated aminoacyl-tRNAs then interact with ribosomes carrying mRNA. Peptidyl transferase catalyzes the formation of peptide bonds between the free amino group of activated aminoacyl-tRNA on the A site of the ribosome and the esterified carboxyl group of the peptidyl-rRNA on the P site; the liberated rRNA remains on the P site. (Ref. Stryer, 4th Edition, Pg. 893-902)

    44. Following about a child with phenylketonuria (PKU) is true, EXCEPT:(1) Mousy odour of urine(2) Normal intelligence is possible in most infants with PKU on diet restriction starting within first 10 days of life(3) Autosomal recessive(4) Adding 10% FeCl3 to urine of child with PKU gives deep blue colour
    Ans. 4
    PKU is autosomal recessive disorder resulting from defect of hydroxylation of phenylalanine to tyrosine due to absence of phenylalanine hydroxylase. Patient has severe mental retardation, IQ <30, mousy odour of urine, blue eyes, eczema. Diagnosis by ® normal or reduced tyrosine levels Increase level of phenylalanine Achievement of normal intelligence is possible by restriction of diet containing phenylalanine starting within 10 days of life. Phenylpyruvic acid is present is urine of the PKU after newborn period and is detected by adding few drops of 10% ferric chloride to fresh urine which gives deep green colour. (Ref. Nelson Essential Paediatrics, 4th Edition, Pg. 159-160)

    45. What is the frequency of pregnancy termination after prenatal diagnosis:(1) 50 percent(2) 25 percent(3) 10 percent(4) <5 percent
    Ans. 4
    With a risk ranging from about 1 to 2 percent at age 35 to a maximum of 5 percent at age 45, the overall frequency of elective pregnancy termination following prenatal diagnosis is only about 3 percent. Thus, 97 percent of couples are reassured by prenatal diagnostic procedures. (Ref. Isselbacher, 13th Edition, Pg. 390)

    46. A neonate is found to have an enzymatic deficiency in the conversion of pyruvate to pyruvate phosphate. The wild-type sequence includes the following:Lys -Arg -His -His -Tyr -LeuAAG -AAG -CAC -CAC -UAC -CUCThe sequence of the mutated enzyme isLys -Glu -Ala -Pro -Leu -ProAAG -GAA -GCA -CCA -CUA -CCUWhat kind of mutation is illustrated by the above amino acid sequence?(1) Point mutation(2) Frame shift mutation resulting in a nucleotide deletion(3) Chain termination mutation(4) Frame shift mutation resulting in the addition of a nucleotide
    Ans. 4
    Single base mutations change the DNA sequence, alter the code of the triplet base, and cause the replacement of one amino acid by another in the gene product. Since the code is degenerate, not all base substitutions will actually lead to an amino acid sequence alteration. Deletion or insertion of a base alters the whole reading frame. In the example listed, the addition of a single nucleotide alters the reading frame from that point onward. Chain termination mutations result in the presence of a stop codon, therefore causing a premature cessation of protein translation. Splice mutations are common and tend to affect the normal mechanism by which introns are excised and exons spliced together. Such changes typically lead to complete failure of synthesis of the gene product. (Ref. Harrison’s, 15th Edition, Pg. 376)

    47. All the following compose nucleosides, EXCEPT:(1) Phosphate groups(2) Deoxyribose sugar(3) Ribose sugar(4) Purine base
    Ans. 1
    A nucleoside consists of a purine or pyrimidine base linked to a pentose sugar. In DNA, deoxyribose sugars are used; in RNA, ribose sugars are used. On the other hand, nucleotides are phosphate esters of nucleosides with one to three phosphate groups, such as ATP, ADP or AMP. The nitrogenous bases are adenine, thymine, guanine, cytosine, or uridine. (Ref. Stryer, 4th Edition, Pg. 75-77, 740)

    48. In contrast to spermatogenesis, oogenesis is largely completed prior to birth. The ova develop from oogonia, which are cells derived from the ovarian cortical tissue. By the third month of gestation, the primary oocyte begins to develop. At which stage of cell division are the oocytes suspended prior to ovulation?(1) First metaphase(2) Second metaphase(3) First prophase(4) Second prophase
    Ans. 3
    By approximately the third month of prenatal development the oogonia have begun to develop into primary oocytes. The primary oocytes have already entered their first meiotic prophase. This process is not synchronized, however, and both early and late stages coexist within the fetal ovary. Primary oocytes remain in suspended prophase until sexual maturity is reached. As each individual follicle begins to mature, the meiotic division of the primary oocyte resumes. Meiosis I is completed about the time of ovulation. When the primary oocyte completes meiosis I, each daughter cell receives 23 chromosomes; one receives most of the cytoplasm and becomes the secondary oocyte, the other becomes a polar body. The second meiotic division commences almost immediately and proceeds as the ovum passes into and down the uterine tube. The second meiotic division usually takes place before the ovum reaches the uterus within the fallopian tube. The second meiotic division produces a mature ovum with virtually all the cytoplasm from the original primary oocyte, and a second polar body is formed. (Ref. Harrison’s, 15th Edition, Pg. 376, 408)

    49. Which of the following statements regarding a double-helical molecule of DNA is true?(1) All hydroxyl groups of pentoses are involved in linkage(2) Bases are perpendicular to the axis(3) Each strand is identical(4) Each strand is parallel
    Ans. 2
    In the classic double helical model of DNA proposed by Watson and Crick, the purine and pyrimidine bases attached to the sugar backbone are perpendicular to the axis and parallel to each other. They are paired and held together by hydrogen bonds. Each strand composing the double helix is different and antiparallel. The 3’ end of one strand is opposite the 5’ end of its complement and vice versa. It is this complementary nature of DNA that allows the strands to be templates for one another during DNA replication. (Ref. Stryer, 4th Edition, Pg. 788-791)

    50. Which of the following is most critical for the prenatal diagnosis of a Mendelian genetic disorder?(1) The disorder must be sufficiently common to warrant population screening and selective abortion(2) The laboratory must distinguish normal from abnormal cells(3) The laboratory must provide a result before 16 weeks of pregnancy(4) The couple must be willing to terminate the pregnancy if the fetus is affected
    Ans. 2
    In order to perform prenatal diagnosis of Mendelian disorder, the abnormal alleles must be detected through enzyme, protein, or DNA analysis of fetal tissues. Fetal karyotyping will not be helpful for demonstrating disease alleles but is usually performed as a routine screen for chromosomal disorders. Prenatal diagnosis has many uses besides a decision on abortion, allowing parents to prepare for a abnormal child or guiding the strategies for labor and delivery. Termination of fetuses afflicted with recessive disease will have little influence on the population incidence. (Ref. Isselbacher, 13th Edition, Pg. 390-392)

    51. In spermatogenesis, at which stage do the sex chromosomes segregate?(1) Meiosis II, primary spermatocyte(2) Meiosis I, primary spermatocyte(3) Meiosis II, secondary spermatocyte(4) Meiosis I, secondary spermatocyte
    Ans. 2
    Spermatogenesis occurs in the seminiferous tubules of the testes of the male from the time of sexual maturity onward throughout adulthood. The last stage of the developmental sequence is the primary spermatocyte. This cell undergoes meiosis I, the primary spermatocyte divides to form two secondary spermatocytes, each with 23 chromosomes. These cell rapidly undergo meiosis II, forming two spermatids. The spermatids mature without further division into spermatozoa and are released into the lumen of the tubule. The total time involved for all stages of spermatogenesis from the beginning of meiosis I to the formation of a mature sperm is approximately 64 days, and as many as 200 million sperm are produced per ejaculate. (Ref. Harrison’s, 15th Edition, Pg. 376, 396)

    52. The Ames test for potential chemical carcinogens(1) Utilizes nude mice(2) Determines mutagenic potential by measuring reversion of bacterial mutations(3) Requires back transplantation in genetically related rats(4) Measures mutagenicity by computer modeling and prediction
    Ans. 2
    The Ames test is a rapid and relatively inexpensive bacterial assay for determining mutagenicity of potential toxic chemicals. Dr. Bruce Ames developed a tester strain of Salmonella that has been modified not to grow in the absence of histidine because of a mutation in one of the genes for the biosynthesis of histidine. Toxic chemicals that are mutagens are placed in the center of the plate and result in reversion of the original mutations, so that histidine is synthesized and the mutated revertants multiply in histidine-free media. Essentially all chemicals known as carcinogens in humans cause mutagenesis in the Ames test. (Ref. Stryer, 4th Edition, Pg. 814)

    53. In a patient of homocystinuria, following is present, EXCEPT:(1) Neonatal screening test most commonly measures cystathione in whole blood(2) Dislocated lens, livedo reticularis(3) Increased risk of venous / arterious thrombosis(4) Due to deficiency of cystathione b-synthesis
    Ans. 1
    Homocystinuria occurs due to deficiency of above enzyme resulting in accumulation of homocystine in blood. Excess homocystiene is reconverted to methionine. Neonatal screening test measures levels of methionine in whole blood. Clinical features:- Marfanoid habitus, ectopia lentis, malar flush, livedo reticularis, arachnodactyly, genu valgum, increased risk of thrombosis.

    54. All of the following traits are X-linked, EXCEPT:(1) Sickle cell disease(2) Hemophilia A(3) Duchenne muscular dystrophy(4) Glucose-6-phosphate dehydrogenase deficiency
    Ans. 1
    X-linked recessive traits, such as hemophilia A, Duchenne muscular dystrophy, glucose-6-phosphate dehydrogenase deficiency, and color blindness, are inherited in an oblique fashion. They are much commoner in men than women. Sickle cell anemia is an autosomal recessive hemoglobinopathy, which is relatively common in African Americans. (Ref. Isselbacher, 13th Edition, Pg. 387-391)

    55. Some of the enzymes utilized in DNA replication are (1) DNA directed DNA polymerase, (2) unwinding proteins, (3) DNA polymerase I, (4) DNA-directed RNA polymerase, and (5) DNA ligase. What is the correct sequence of their use during DNA synthesis?(1) 2, 3, 4, 1, 5(2) 4, 2, 1, 5, 3(3) 4, 2, 1, 3, 5(4) 2, 4, 1, 3, 5
    Ans. 4
    Before DNA replication can actually begin, unwinding protein must open segments along the DNA double helix. DNA-directed RNA polymerase (primase) catalyzes the synthesis of a complementary RNA primer of approximately 50 to 100 bases on each DNA strand. Then DNA-directed DNA polymerase III adds deoxyribonucleotides to the 3’ end of the primer RNA, which replicates a segment of DNA, the Okazaki fragment. DNA polymerase I then removes the primer RNA and adds deoxyribonucleotides to fill the gaps between adjacent Okazaki fragments. The fragments are finally joined together by DNA ligase to create a continuous DNA chain. (Ref. Stryer, 4th Edition, Pg. 803-809)

    56. Galactosemic patient will have the following features, EXCEPT:(1) Normal anion gap acidosis(2) Galactosemic infants are at increased risk of severe staphylococcal sepsis(3) Diagnosis by extreme reduction in RBC galactose-1-phosphate uridyl transferase(4) Developmental cataract present
    Ans. 2
    Galactosemia is an autosomal recessive disease due to deficiency of galactose-1-phosphate uridyl transferase. Clinical features are most striking in neonate who when fed milk develops hyperbilirubinemia, hypoglycaemia, acidosis, glycosuria, amino aciduria and cataract. Ovarian failure is a late sequelae. Renal tubular dysfunction is evident by normal anion gap hyperchloremic acidosis. Treatment is dietary restriction. (Ref. Harper’s Biochemistry, 25th Edition)

    57. Indications for genetic counselling include all of the following, EXCEPT:(1) Consanguinity(2) Family history of cystic fibrosis(3) Family history of congenital infection(4) Advanced maternal age
    Ans. 3
    There are many indications for genetic counselling. These include advanced maternal age, family history of birth defects or other known or suspected genetic disease, unexplained mental retardation, and consanguinity. Although not technically a genetic problem, teratogen exposure is also generally accepted as an indication for genetic counseling. Although a history of congenital infection requires that medical information be given to the family, this is not a heritable disorder and, therefore, is not an indication for genetic counselling. However, should a pregnant woman herself contract an infection, such as rubella, which may be teratogenic, genetic counselling should be offered. (Ref. Isselbacher, 13th Edition, Pg. 343-347)

    58. An expression library of human DNA in E. coli can be screened by:(1) Western blotting(2) Southern blotting(3) Northern blotting(4) Nitrocellulose blotting
    Ans. 2
    Expression libraries, as opposed to genomic libraries, are often used to screen cDNA clones on the basis of their ability to direct the synthesis of a specific foreign protein of interest in E. coli. A 32P-Labelled DNA probe is used to note the presence of a sequence in spots on the replica when DNA is being screened. Southern blotting refers to the use of the 32P DNA probe of DNA, while northern blotting refers to the use of a 32P RNA. Nitrocellulose blotting simply refers to the transfer of bacteria from a master plate to a nitrocellulose filter for lysing prior to probing. (Ref. Stryer, 4th Edition, Pg. 60-63, 120-122, 173-174)

    59. In Maple Syrup Urine disease:(1) Aromatic amino aciduria exists(2) Mousy odour of urine is present(3) Hyperglycaemia and acidosis occur(4) Symptoms occur within 1st 4 weeks of life
    Ans. 4
    Maple syrup urine disease is autosomal recessive disorder due to deficiency of decarboxylase resulting in accumulation of branched chain amino acids. Leucine ® Isovaleryl CoA Isoleucine ® butryl CoA Valine ® Isobutryl CoA Lab investigations reveal hypoglycemia, ketonuria, increased level of branched chain amino acid in saving blood / plasma. Treatment is restricting intake of branched amino acid, haemodialysis. Hemofiltration can be life during crisis. (Ref. Nelson Essential Paediatrics, 4th Edition, Pg. 162-3)

    60. Which statement about the “genetic code” is the most accurate?(1) The code is degenerate (i.e., more than one codon may exist for a single amino acid)(2) Information is stored as set of trinucleotide repeats called codons(3) There are 64 codons, all of which code for amino acids(4) The sequence of codons that comprise a gene will exhibit an exact linear correspondence to the sequence of amino acids in the translated protein
    Ans. 1
    The “genetic code” uses 3-nucleotide “words” or codons, to specify the 20 different amino acids. The linear correspondence of codons in DNA and of amino acids in protein domains will be interrupted by the presence of introns in DNA. Codons comprise 64 different 3-base pair sequences (three positions with four possible nucleotides at each). It follows that the genetic code must be degenerate, i.e., different codons can specify the same amino acid. Three codons are reserved as “stop” signals that result in peptide chain termination. The genetic code is universal in the sense that codon/amino acid relationships are the same in all organisms. (Ref. Isselbacher, 13th Edition, Pg. 349-360)

    61. Which one of the following has catalytic activity?(1) Nucleotides(2) Phospholipids(3) Glycogen(4) RNA
    Ans. 4
    Until quite recently, it was thought that only proteins could act as biological catalyst. However, it has since been found that certain types of RNA are capable of undergoing self-splicing. The intron derived from precursor ribosomal RNA forms a 395-nucleotide RNA molecule that acts as a catalyst for the transformation of other RNA molecules. Such enzymatic RNA molecules are called ribozymes. Ribozymes (catalytic RNAs) like this catalyze the cleavage and joining of RNA chains at specific sites without being consumed. Thus, RNA as well as protein can act as a catalyst for enzymatic reactions. (Ref. Stryer, 4th Edition, Pg. 115-116, 864-870)

    62. Screening of PKU is done by Guthrie’s test. Reason for two false negative result is most commonly due to:(1) Sample of blood before adequate dietary intake due to early discharge(2) Bacteria on agar plate(3) Sample of blood after newborn period(4) Excess of infant blood on agar plate
    Ans. 1
    For Guthrie’s test, infant blood from heel or finger prick is placed on filter paper disk and mailed to central screening lab. Rapid screening of agar plates with filter disc is done and disc surrounded by bacterial colonies are counted as positive results. Most common cause of false negative result is lack of supplementation of diet containing phenylalanine before discharge due to early discharge requirement. (Ref. Isselbacher, 13th Edition, Pg. 390-392)

    63. The nature and utility of somatic cell hybridization are best described by which of the following?(1) Fusion of human muscle and embryonic cells to allow expression of embryonic muscle proteins(2) Fusion of human muscle and brain cells to assess complementation by tissue specific isozymes(3) Fusion of human and rodent cells to perform human gene mapping by selective chromosome loss(4) Fusion of human and rodent cells to assess similarities in the translational machinery
    Ans. 3
    Somatic cell hybridization involves the process of fusion of human and rodent somatic cells using viruses or chemicals that join their cell membranes. The hybrid cells retain their rodent chromosomes, but lose every human chromosome except one that is maintained by selective agents in the growth medium. A panel of somatic cell hybrids is constructed, each containing a different human chromosome. Assay of the panel for the presence of particular gene sequences or products will then identify the human chromosome that contains that genetic locus. (Ref. Isselbacher, 13th Edition, Pg. 361-363)

    64. Given that the chromosomes of mammalian cells may be 20 times as large as those of E. coli, how can replication of mammalian chromosomes be carried out in just a few minutes?(1) Eukaryotic DNA polymerase are extraordinarily fast compared with prokaryotic polymerases(2) The higher temperature of mammalian cells allows for an exponentially higher replication rate(3) Hundreds of replication forks work simultaneously on each piece of chromosomal DNA(4) The presence of histones speeds up the rate of chromosomal DNA replication
    Ans. 3
    Despite the great length of the chromosomes of eukaryotic DNA, the actual replication time is only minutes. This is because eukaryotic DNA is replicated bidirectionally from many points of origin. The hundreds of initiation sites of DNA replication on chromosomes share a consensus sequence called an autonomous replication sequence (ARS). Thus, while eukaryotic DNA, like E, coli DNA, is replicated semiconservatively, the use of hundreds of replication forks originating from the autonomous replication sequences allows for a rapid synthesis of chromosomal DNA. (Ref. Stryer, 4th Edition, Pg. 982-984)

    65. Following is true about MELAS, EXCEPT:(1) Maternal inheritance only(2) Stroke like episode(3) Deafness, diabetes, dystonia(4) None of the above
    Ans. 4
    Mitochondrial Encephalomyopathy, Lactic Acidosis, and Stroke-like episodes (MELAS); This syndrome is characterized by stroke-like events that cause subacute brain dysfunction, cerebral structural changes, seizures, and several other common clinical and laboratory features. Maternal inheritance of the MELAS syndrome may be obscured because of mild clinical features in relatives. A point mutation at nucleotide 3243 in the tRNA Leu(URR) gene accounts for 80% of MELAS cases. However, the clinical features of the 3243 mtDNA mutation are pleomorphic. It is also associated with nondeletion CPEO, myopathy, deafness, diabetes, and dystonia. (Ref. Harrison, 15th Edition, Vol. 1, Pg. 406)

    66. All are true about Zellweger’s syndrome, EXCEPT:(1) Peroxisomal defect(2) No treatment available(3) X-linked recessive(4) Hepatomegaly and hypotonia
    Ans. 3
    It is an autosomal recessive disorder also called cerebrohepatorenal syndrome. peroxisomes are virtually absent. Affected infants have high forehead, flat orbital ridge, hepatomegaly and hypotonia. No treatment is available and death occurs in 1st year of life. (Ref. Nelson Essential Paediatrics, 4th Edition, Pg. 172-3)

    67. Deletions of 11p13 may result in Wilm’s tumor, aniridia, genitourinary malformations, and mental retardation that is, the WAGR syndrome. In some patients, however, not all features are seen. Additionally, individual features of this syndrome may be inherited separately in a Mendelian fashion. Limited features may be seen in patients without visible chromosomal deletions. The most likely mechanism for this finding is:(1) Imprinting(2) Germ-line mosaicism(3) Uniparental disomy(4) Contiguous gene syndrome
    Ans. 4
    Contiguous gene syndromes, also known as microdeletion syndromes, occur when deletions result in the loss of several different closely linked loci. Depending on the size of the deletion, different phenotypes may result. Mutations in the individual genes may result in isolated features that may be inherited in a Mendelian fashion. (Ref. Isselbacher, 13th Edition, Pg. 375-380)

    68. New proteins destined for secretion are synthesized in the:(1) Golgi apparatus(2) Rough endoplasmic reticulum(3) Smooth endoplasmic reticulum(4) Free polysomes
    Ans. 2
    Protein synthesis occurs in the cytoplasm, on groups of free ribosomes called polysomes, and on ribosomes associated with membranes, termed the rough endoplasmic reticulum. However, proteins destined for secretion are only synthesized on ribosomes of the endoplasmic reticulum and are synthesized in such a manner that they end up inside the lumen of the endoplasmic reticulum. From there the secretory proteins are packaged in vesicles. The Golgi apparatus is involved in the glycosylation and packaging of macromoleculues into membranes for secretion. (Ref. Stryer, 4th Edition, Pg. 911-918)

    69. Which of the following Glycogen storage disorder shows cardiomegaly:(1) Forbe’s / Type III(2) Pompe’s / Type II(3) Her’s / Type VI(4) Tauri’s / Type VII
    Ans. 2
    GLYCOGEN STORAGE DISEASES Disease Affected enzyme Organs affected Neonatal manifestations
    Type I: Von Gierke Glucose-6-phosphatase Liver, kidney, GI tract, platelets Hypoglycemia, lactic academia, liver may not be enlarged
    Type II: Pompe Lysosomal alphaglucosidase All, notably striated muscle, nerve cells May have muscle weakness, cardiomegaly, or both
    Type III: Forbes Debranching enzyme Liver, muscles Usually none
    Type IV: Anderson Branching enzyme Liver, other tissues Usually none
    Type V: McArdle Muscle phosphorylase Muscle None
    Type VI: Her’s Liver phosphorylase Liver Usually none
    Type VII: Tauri’s Muscle phosphofructokinase Muscle None
    Type VIII Phosphorylase kinase Liver None
    (Ref. Nelson Essential Paediatrics, 4th Edition, Pg. 170)

    70. A patient suffers from Adenosine Deaminase (ADA) deficiency, an autosomal recessive immune deficiency in which bone marrow lymphoblasts cannot replicate to generate immunocompetent lymphocytes. The treatment option that would permanently cure the patient is:(1) Somatic cell gene therapy to replace one ADA gene copy in bone marrow lymphoblasts(2) Germ-line gene therapy to replace one ADA gene copy(3) Germ-line gene therapy to replace both ADA gene copies(4) Somatic cell gene therapy to replace both ADA gene copies in circulating lymphocytes
    Ans. 1
    Gene therapy refers to a group of techniques by which gene structure of expression is altered to ameliorate a disease. Because of ethical and practical difficulties, germ-line therapy involving alteration of genes in primordial germ cells is not being explored in humans. Although germ-line genetic engineering is being performed in animals with the goals of improved breeding or agricultural yield, it will alter the characteristics of offspring rather than the treated individual. Somatic cell gene therapy is targeted to an affected tissue or group of tissues in the individual, and is most effective if stem cells such as bone marrow can be treated. Somatic cell gene therapy offers the hope of replacing damaged tissue without the rejection problems of transplantation. For autosomal recessive disorders, only one of the two defective alleles must be replaced or supplemented. (Ref. Isselbacher, 13th Edition, Pg. 363-365)

    71. What is the correct order of the following steps in protein synthesis?(a) A peptide bond is formed(b) The small ribosomal subunit is loaded with initiation factors, messenger RNA, and initiation aminoacyl-transfer RNA(c) The intact ribosome slides forward three bases to read a new codon(d) The primed small ribosomal subunit binds with the large ribosomal subunit(e) Elongation Factors delivers aminoacyl-tRNA to bind to the “a” site.(1) b, d, e, a, c (2) a, b, c, d, e(3) b, c, d, e, a (4) c, b, d, e, a
    Ans. 1
    At the start of translation, initiation factors, m
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