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1. A. B. C. D. E. * 2. A. B. C. * D. E. 3. A. B. C. D. E. * 4. A. B. C. D. E. * 5. A. B. C. D. E. * 6. A. B. C. D. E. * 7. A. B. C. D. E. * Two percentage of recombination equals a distance between the loci of ... 200 map units 200 centimorgans 20 map units 20% of crossovers 2 centimorgans Which of the following is not true about crossing over? it occasionally regenerates the extreme phenotype it involves a physical exchange of genetic material between homologous chromosomes it occurs before DNA synthesis it involves a physical exchange of genetic material between non-sister chromatids it leads to a recombination of genetic traits In hemoglobin Tocucci there was a replacement of the amino acid histidine to tyrosine. What kind of mutation is this? Genomic mutation. Aneuploidy. Polyploidy. Inversion. Gene (point) mutation. In hemoglobin Tocucci there was a replacement of the amino acid histidine to tyrosine. What kind of mutation is this? Genomic mutation. Nonsense mutation. Silent mutation. Chromosomal mutation. Missense mutation. Sometimes things may go wrong in protein synthesis. A cell may receive too much ultraviolet light causing an extra base to be inserted in a protein. This type of occurrence is called a: Mistake. Replication. Chromosome aberration. Mitosis. Gene mutation. In protein there was a replacement of the amino acid histidine to glutamine. What kind of mutation is this? Genomic mutation. Nonsense mutation. Silent mutation. Chromosomal mutation. Missense mutation. How many Barr bodies are present in the somatic cell of a female with karyotype 45, XO: 1 2 3 4 0 8. A. B. C. D. E. * 9. A. B. C. D. E. * 10. A. B. C. D. E. * 11. A. B. C. D. E. * 12. A. B. C. D. E. * 13. A. B. C. D. E. * 14. A. B. How many Barr bodies are present in the somatic cell of a female with karyotype 47, XXX: 1. 3 0. 4. 2. How many Barr bodies are present in the somatic cell of a male with karyotype 47, XXY: 2 3 0 4 1 Which methods of human genetics you have to use for diagnosis of Edwards syndrome? Genealogy method. Twin method Dermatoglyphics Pedigree Analysis. Cytogenetical method. Which methods of human genetics you have to use for diagnosis of Turner syndrome? Genealogy method. Twin method Dermatoglyphics Pedigree Analysis. Chromosomes Analysis. A baby was born with absence of the eyes, cleft palate and cleft lip, polydactyly, and micrognathia (micromandibula), low-set, malformed ears and abnormalities of the heart. Chromosome analysis has shown that the karyotype 47, XX, 13+. What disorder does this child have? Down syndrome. Turner syndrome. Edwards syndrome. Klinefelter syndrome Patau syndrome. A baby was born with scaphocephaly (a condition in which the skull is abnormally long and narrow), low-set, malformed ears, micrognathia, short digits and rocker-bottom feet, heart defects. Chromosome analysis has shown that the karyotype 47, XY, 18+. What disorder does this child have? Down syndrome. Patau syndrome. Turner syndrome. Klinefelter syndrome. Edwards syndrome. A baby was born with small skull, round face and a long protruding tongue, short, flat-bridge nose, a mongolian type of eyelid fold (epicanthal fold), short neck, short phalanges (fingers), flat hands, transverse palm crease. Chromosome analysis has shown that the karyotype 47, XY, 21+. What disorder does this child have? Patau syndrome. Turner syndrome. C. D. E. * 15. A. B. C. D. E. * 16. A. B. C. D. E. * 17. A. B. C. D. E. * 18. A. B. C. D. E. * 19. A. B. C. D. E. * 20. A. B. C. D. E. * 21. A. B. C. D. E. * 22. Edwards syndrome. Klinefelter syndrome. Down syndrome. Give the karyotype formula of a female with cri-du-chat syndrome. 46, XX. 47, XXY. 45, X0. 47, XXX. 46, XX, 5р-. Give the karyotype formula of a female with Turner syndrome. 46, XX. 47, XXY. 47, XXX. 46, XX, 5р-. 45, X0. Give the karyotype formula of a male with Patau syndrome. 46, XX, 9р+; 46, XX, 5р-; 47, XY, 18+; 47, XY, 21+. 47, XY, 13+. Give the karyotype formula of a male with Edwards syndrome. 46, XX, 9р+; 46, XX, 5р-; 47, XY, 21+. 47, XY, 13+. 47, XY, 18+; Give the karyotype formula of a male with Down syndrome. 46, XX, 9р+; 46, XX, 5р-; 47, XY, 18+; 47, XY, 13+. 47, XY, 21+. Define 2pq from the Hardy-Weinberg principle: Dominant allele frequency. Recessive allele frequency. Homozygous dominant genotype frequency. Homozygous recessive genotype frequency. Heterozygous genotype frequency. Define p from the Hardy-Weinberg principle: Recessive allele frequency. Homozygous dominant genotype frequency. Homozygous recessive genotype frequency. Heterozygous genotype frequency. Dominant allele frequency. Define p2 from the Hardy-Weinberg principle: A. B. C. D. E. * 23. A. B. C. D. E. * 24. A. B. C. D. E. * 25. A. B. C. D. E. * 26. A. B. C. D. E. * 27. A. B. C. D. E. * 28. A. B. C. D. E. * 29. A. B. Dominant allele frequency. Recessive allele frequency. Homozygous recessive genotype frequency. Heterozygous genotype frequency. Homozygous dominant genotype frequency. Define q from the Hardy-Weinberg principle: Dominant allele frequency. Homozygous dominant genotype frequency. Homozygous recessive genotype frequency. Heterozygous genotype frequency. Recessive allele frequency. Define q2 from the Hardy-Weinberg principle: Dominant allele frequency. Recessive allele frequency. Homozygous dominant genotype frequency. Heterozygous genotype frequency. Homozygous recessive genotype frequency. Define homozygous dominant genotype frequency in mathematical term from the Hardy-Weinberg principle: 2pq q2 p q p2 Define homozygous recessive genotype frequency in mathematical term from the Hardy-Weinberg principle: 2pq p2 p q q2 Define heterozygous genotype frequency in mathematical term from the Hardy-Weinberg principle: p2 q2 p q 2pq Define dominant allele frequency in mathematical term from the Hardy-Weinberg principle: 2pq p2 q2 q p Define recessive allele frequency in mathematical term from the Hardy-Weinberg principle: 2pq p2 C. D. E. * 30. A. B. C. D. E. * 31. A. B. C. D. E. * 32. A. B. C. D. E. * 33. A. B. C. D. E. * 34. A. B. C. D. E. * 35. A. B. C. D. E. * 36. A. B. C. D. E. * q2 p q Allelic genes ... Have the different loci in the homologous chromosomes. Are situated in the non-homologous chromosomes Are situated in the homologous chromosomes. Have the different loci in the heterochromosomes. Have the same loci in the homologous chromosomes. Which form of gene interactions is characterized that one gene masks the phenotypic effect of the gene from different pair? Complete dominance. Incomplete dominance. Superdominance Codominance Epistasis Which type of the human ABO blood group is example of codominance? I (O) II (A) III (B) All of the above. IV (AB) How many types of gametes does the person with NnffPp genotype produce? 1 2 6 8 4 How many types of gametes does the person with AaBbCC genotype produce? 1 2 6 8 4 How many types of gametes does the person with NnFfPp genotype produce? 1 2 4 6 8 How many types of gametes does the person with AaBbCc genotype produce? 1 2 4 6 8 37. A. B. C. D. E. * 38. A. B. C. D. E. * 39. A. B. C. D. E. * 40. A. B. C. D. E. * 41. A. B. C. D. E. * 42. A. B. C. D. E. * 43. A. B. C. D. E. * 44. What will be the phenotypic ratio in the first filial generation during test cross, if an analyzing person is heterozygous by one pair of genes? 3:1. 9:3:3:1. 4:1. Segregation is absent. 1:1. What will be the phenotypic ratio in the first filial generation during test cross, if an analyzing person is heterozygous by two pairs of genes? 3:1. 9:3:3:1. 4:1. Segregation is absent. 1:1:1:1. In a cross, an AABBCC individual is paired with an aabbcc individual. What will be the expected frequency of aabbcc individuals in the F2 generation? 16/64 8/64 4/64 2/64 1/64. What is the genotype of man with the rhesus-negative III group of blood? IAIA RhRh. IAiRhrh. IAIB Rhrh. IAIB rhrh. IBirhrh. What is the genotype of man with the rhesus-negative IV group of blood? IAIA RhRh. IAiRhrh. IBirhrh. IAIB Rhrh. IAIB rhrh. What is the genotype of man with the rhesus-positive IV group of blood? IAIA RhRh. IAiRhrh. IBirhrh. IAIB rhrh. IAIB Rhrh. Linkage between two genes was interrupted by crossing-over when: Genes stay together at a very short distance on the same chromosome. Genes stay on the two non-homologous chromosomes. Genes stay together at distance 60 centimorgans. Genes stay together at distance 70 centimorgans. Genes stay together at distance 20 centimorgans. Linkage between two genes was interrupted by crossing-over when: A. B. C. D. E. * 45. A. B. C. D. E. * 46. A. B. C. D. E. * 47. A. B. C. D. E. * 48. A. B. C. D. E. * 49. A. B. C. D. E. * 50. A. B. C. D. E. * 51. A. B. C. D. Genes stay together at a very short distance on the same chromosome. Genes stay on the two non-homologous chromosomes. Genes stay together at distance 60 centimorgans. Genes stay together at distance 70 centimorgans. Genes stay together at distance 30 centimorgans. How many types of gametes does the person with AABBCC genotype produce? 2 4 6 8 1 How many types of gametes does the person with AaBBCC genotype produce? 1 4 6 8 2 Which sex-linked trait do you know? Baldness Beard growth. Polydactyly Albinism Hemophilia Which sex-linked trait do you know? Baldness Beard growth. Polydactyly Albinism Colour-blindness. Which sex-limited trait do you know? Baldness Hemophilia Polydactyly Albinism Beard growth. Which sex-limited trait do you know? Baldness Hemophilia Polydactyly Albinism Breast size. Which sex-influenced trait do you know? Breast size. Hemophilia Polydactyly Albinism E. * 52. A. B. C. D. E. * 53. A. B. C. D. E. * 54. A. B. C. D. E. * 55. A. B. C. D. E. * 56. A. B. C. D. E. * 57. A. B. C. D. E. * 58. A. B. C. Baldness Ichthiosis is inherited as Y-linked trait (d- allele of the ichthiosis). Which types of gametes does a man with ichtiosis produce? D, d. D. d. Xd, Y. X, Yd In human Hemophilia is a sex-linked recessive disorder. A healthy couple have a single child with hemophilia. Who passed the gene of hemophilia to the child? Father Aunt Uncle Grandfather Mother Molecular disorders can be caused by: Changes in chromosome structure. Changes in chromosome number. Genomic mutations. All of the above. Gene mutations. A segment of DNA of normal diploid cell has the sequence of the nucleotides: AACGTTA. After ionizing radiation mutant DNA have the sequence of the nucleotides: AATGCTA. What kind of mutation does take place? Deletion Duplication Insertion Translocation Inversion A segment of DNA of normal diploid cell has the sequence of the nucleotides: AACGTTA. After ionizing radiation mutant DNA have the sequence of the nucleotides: AACGT. What kind of mutation does take place? Duplication Inversion Insertion All of the above. Deletion What is the cause of phenylketonuria? The lack of an enzyme tyrosinase. The lack of an enzyme hexosaminidase. The lack of an enzyme halactose. All of the above. The lack of an enzyme phenylalaninehydroxilase. How many Barr bodies are present in the somatic cell of a female with triplo-X syndrome? 1 3 0 D. E. * 59. A. B. C. D. E. * 60. A. B. C. D. E. * 61. A. B. C. D. E. * 62. A. B. C. D. E. * 63. A. B. C. D. E. * 64. A. B. C. D. E. * 65. A. B. C. D. 4 2 How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome (karyotype is 47, XXY)? 2 3 0 4 1 How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome (karyotype is 48, XXXY)? 1 3 0 4 2 How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome (karyotype is 49, XXXXY)? 1 2 0 4 3 How many Barr bodies are present in the somatic cell of a female with karyotype 48, XXXX? 1 2 0 4 3 How many Barr bodies are present in the somatic cell of a female with karyotype 49, XXXXX? 1 2 0 3 4 How many Barr bodies are present in the somatic cell of a female with Down syndrome? 2 3 0 4 1 How many Barr bodies are present in the somatic cell of a female with Patau syndrome? 2 3 0 4 E. * 66. A. B. C. D. E. * 67. A. B. C. D. E. * 68. A. B. C. D. E. * 69. A. B. C. D. E. * 70. A. B. C. D. E. * 71. A. B. C. D. E. * 72. A. B. C. D. 1 Height is quantitative (polygenic) trait in humans. Height is controlled by three pairs of genes: A1a1, A2a2, A3a3, localed in different chromosomes. A very tall person has 6 recessive genes (a1a1a2a2a3a3) and very shot person has 6 dominant genes (A1A1A2A2A3A3). What is the genotype of a medium-height person? A1A1A2a2A3a3. A1a1a2a2A3a3. A1a1a2a2a3a3. a1a1a2a2a3a3 A1a1A2a2A3a3. What is the genotype of man with the rhesus-positive II group of blood? IAirhrh. IBirhrh. IAIB Rhrh. IAIB rhrh. IAIARhRh. Complete linkage occurs when ... Genes stay on the two non-homologous chromosomes. Genes stay together at distance 10 centimorgans. Genes stay together at distance 20 centimorgans. Genes stay together at distance 60 centimorgans. Genes stay together at a very short distance on the same chromosome. Non-allelic genes: Are situated in the sex chromosomes. Are situated in the homologous chromosomes. Have the same loci in the homologous chromosomes. Have the same loci in the heterochromosomes. Are situated in the non-homologous chromosomes. How many types of gametes does the person with mmBBCc genotype produce? 4 8 3 7 2 Which holandric trait do you know? Baldness Beard growth. Polydactyly Hemophilia Hairy pinnae. In human colour-blindness is a sex-linked recessive disorder (d- allele of the colour-blindness, D allele of the normal sightedness). Which types of gametes does normal-sighted man produce? D, d. D. d. Xd, Y. E. * 73. A. B. C. D. E. * 74. A. B. C. D. E. * 75. A. B. C. D. E. * 76. A. B. C. D. E. * 77. A. B. C. D. E. * 78. A. B. C. D. E. * 79. A. B. * C. D. E. XD, Y. What is the cause of albinism? The lack of an enzyme phenylalaninehydroxilase. The lack of an enzyme hexosaminidase. The lack of an enzyme halactose. The lack of an enzyme helicase. The lack of an enzyme tyrosinase. Hairy ears is inherited as Y-linked trait (d- allele of the hairy ears). Which types of gametes does a man with hairy ears produce? D, d. D. d. Xd, Y. X, Yd In human Hemophilia is a sex-linked recessive disorder. How many types of gametes does a man with hemophilia produce? 1 3 4 5 2 How many linkage groups does human have? 46 44 2 1 23 In human genes A and B located in the same autosome. What types of spermatozoons does a man with genotype AB//ab produce in the case of complete linkage? AB, ab, Ab, aB. Ab, aB. AB. ab. AB, ab How many linkage groups does human female have? 46 24 2 1 23 How many linkage groups does human have? 46 23 2 1 24 80. A. B. C. D. E. * 81. A. B. C. D. E. * 82. A. B. C. D. E. * 83. A. B. C. D. E. * 84. A. B. C. D. E. * 85. A. B. C. D. E. * 86. A. B. Which type of chromosomal aberration is characterized by separation and loss of a chromosome segment? Duplication Translocation Inversion Polyploidy Deletion A permanent transmissible change in the genetic material (modification in chromosomes and genes) is: Modification Gametogenesis Fertilization Phenotypic variation. Genetic variation State the way of combinatorial variation: Homologous chromosomes on the equatorial platter of the meiotic spindle in a random arrangement, thus facilitating genetic mixing; Homologous chromosomes pair making physical contact (synapsis) and forming a tetrad in the prophase of meiosis. Separation of sister chromatids in the two daughter cells formed in meiosis I and their distribution as chromosomes into two daughter cells. All of these. Random chromosomes combination during fertilization. What type of mutations, which occur in humans, have the greatest probability will be shown in the following generation? Recessive, linked with a sex, which arises in X-chromosome. Recessive, which arises in autosomes. Which arises in Y-linked. Dominant, linked with a sex, which arises in X-chromosome. Dominant autosoma mutation. Genomic mutations are: Changes in structure of chromosomes. Changes in structure of the nucleotides. None of these. All of these. Changes in number of chromosomes in karyotype. Otosclerosis (a pathological condition of the bony labyrinth of the ear, resulting in hearing loss) determines by autosomal dominant gene. Gene’s penetrance is 50%. What’s the probability of these disease in the family with both heterozygous parents? 0%. 25%. 75%. 100%. 37,5%. Which type of chromosomal aberration is characterized by multiplication of a chromosome segment? Translocation Deletion C. D. E. * 87. A. B. C. D. E. * 88. A. B. C. D. E. * 89. A. B. C. D. E. * 90. A. B. C. D. E. * 91. A. B. C. D. E. * 92. A. B. C. D. E. * 93. A. B. C. D. Inversion Polyploidy Duplication For which women would an amniocentesis be recommended? For all pregnant women. For pregnant women who are over 30 years of age. For pregnant women who had a late menarche. For non-pregnant women who had a late menarche. For pregnant women who have a family history of certain genetic disorders. In human genes A and B located in the same autosome. What types of spermatozoons does a man with genotype AB//ab produce in the case of incomplete linkage? AB, ab Ab, aB. AB. ab. AB, ab, Ab, aB. Name 3 sex chromosome trisomies. Down syndrome, Patau syndrome, Edwards syndrome Patau syndrome, Turner syndrome, Cri-du-chat syndrome Turner syndrome, Edwards syndrome, triplo-X- syndrome Cri-du-chat syndrome, Klinefelter syndrome, triplo-X- syndrome Klinefelter syndrome, triplo-X- syndrome, Supermale syndrome Incheritance of a dominant autosomal disorder differs from incheritance of an autosomal recessive disorder in that: A Dominant disorder may be passed on only if both parents are affected A Dominant disorder is evident only if the offspring is homozygous for the allele A Dominant disorder is more seen in females A Dominant disorder is more seen in males A Dominant disorder may be passed on even only one parent is affected Which one of these is inconsistent with X-linked recessive incheritance? Both parents do not have the disorder Only females in the pedigree chart have the disorder Both males and females in the pedigree chart have the disorder Rare females in the pedigree chart have the disorder Mostly males in the pedigree chart have the disorder If an affected male has all affected daughters but no affected sons, the trait is likely to be an autosomal recessive trait autosomal dominant trait X-linked recessive trait Y-linked X-linked dominant trait Klinefelter syndrome: may have one chromosome as an isochromosome. is usually associated with severe mental retardation is observed only in females cannot be detected at birth E. * 94. A. B. C. D. E. * 95. A. B. C. D. E. * 96. A. B. C. D. E. * 97. A. B. C. D. E. * 98. A. B. C. D. E. * 99. A. B. C. D. E. * 100. A. B. C. D. is observed only in males Turner syndrome: is observed only in males may have one chromosome as an isochromosome is usually associated with severe mental retardation cannot be detected at birth is observed only in females Which of the following conditions are chromosome aberrations? Turner syndrome. Penylketonuria. Albinism. Patau syndrome. Cri-du-chat syndrome. A nineteen year old female with short stature, wide spaced nipples, and primary amenorrhea most likely has the karyotype of: 47,XX,+18 46,XY 47,XXY 46,XX 45,X0 A boy, 9 years of age, is admitted to the pediatric ward with hemophilia A. He inherited this condition through a ___________ trait. X-linked dominant Y-linked Autosomal dominant Autosomal recessive X-linked recessive Consider the cross Aa x Aa. If the alleles for both genes exhibit incomplete dominance, what phenotypic ratio is expected in the resulting offspring? 1:1:1:1. 9:3:3:1. 3:6:3:1:2:1. 3:1. 1:2:1 Consider the cross Aa x Aa. If the alleles for both genes exhibit complete dominance, what phenotypic ratio is expected in the resulting offspring? 1:1:1:1. 9:3:3:1. 3:6:3:1:2:1. 1:2:1 3:1. How many chromosomes does a triploid human karyotype have? 23 46 92 44 E. * 101. A. B. C. D. E. * 102. A. B. C. D. E. * 103. A. B. C. D. E. * 104. A. B. C. D. E. * 105. A. B. C. D. E. * 106. A. B. C. D. E. * 107. A. B. C. D. E. * 69 How many chromosomes does a person with Edward syndrome have? 45 46 48 44 47 How many chromosomes does a person with Down syndrome have? 45 46 48 44 47 How many chromosomes does a person with Turner syndrome have? 46 47 48 44 45 How many chromosomes does a person with Klanfelter syndrome have? 45 46 48 44 47 In human ichthyosis is holandric disorder. A couple have a single child with ichthyosis. Who passed the gene of ichthyosis to the child? Mother Aunt Uncle Grandfather Father Genes A, B and C belong to one group of linkage. It is known, that the distance between the genes A and B is equal to 5 centimorgans and betweeen the genes B and C - 3 centimorgans. What is distance between the genes A and C? 5 centimorgans. 3 centimorgans. 10 centimorgans. 13 centimorgans. 8 centimorgans. Sicle-cell anaemia is caused by an _____ gene. X-linked dominant X-linked recessive Y-linked Autosomal recessive Autosomal dominant 108. A. B. C. D. E. * 109. A. B. C. D. E. * 110. A. B. C. D. E. * 111. A. B. C. D. E. * 112. A. B. C. D. E. * 113. A. B. C. D. E. * 114. A. B. C. D. E. * Newborn screening for _______is example of a screening test Down syndrome. Patau syndrome. Albinism. Sicle-cell anaemia. Phenylketonuria. Color blindness is an X-linked trait in humans. If a color-blind woman marries a man with normal vision, the children will be all color-blind daughters, but normal sons. all normal sons, but carrier daughters. all color-blind children all normal children. all color-blind sons, but carrier daughters What is the cause of phenylketonuria? The lack of an enzyme tyrosinase. The lack of an enzyme hexosaminidase The lack of an enzyme halactose. The lack of an enzyme DNA-polymerase. The lack of an enzyme phenylalaninehydroxilase. What is the term for an error in which homologous chromosomes fail to separate during meiosis or mitosis? Aneuploidy Polyploidy Translocation Deletion Nondisjunction Phenylketonuria is caused by an _____ gene X-linked dominant X-linked recessive Y-linked Autosomal dominant Autosomal recessive What is the optimal time for ultrasonography of a fetus for been required defect abnormal growth of head or limbs? 2-3 weeks gestation 4-6 weeks gestation 7-8 weeks gestation 9-10 weeks gestation 16-18 weeks gestation Which of the following disorders has a mode of inheritance similar to hemophilia A? Cri du chat syndrome Polycystic kidney Down syndrome Sickle-cell anaemia Duchenne muscular dystrophy (DMD) 115. A. B. C. D. E. * 116. A. B. C. D. E. * 117. A. B. C. D. E. * 118. A. B. C. D. E. * 119. A. B. C. D. E. * 120. A. B. C. D. E. * 121. A. B. C. D. E. * Match the disease with the chromosomal abnormality: single X chromosome with no homologous X or Y chromosome Down syndrome. Patau syndrome. Edwards syndrome. Klinefelter syndrome. Turner syndrome. What is the most clinically useful technique for prenatal diagnosis of chromosomal abnormalities at 2 months’ gestation? Gene mapping Linkage analysis Amniocentesis Ultrasonography Chorionic villi biopsy People who have neurofibromatosis have a varying degree of the condition because of the genetic principle of: Penetrance Dominance Recessiveness Pleiotropy Expressivity Cystic fibrosis is caused by an _____ gene X-linked dominant X-linked recessive Y-linked Autosomal dominant Autosomal recessive Match the disease with the chromosomal abnormality: two X chromosomes and one Y chromosome Down syndrome. Patau syndrome. Turner syndrome. Edwards syndrome. Klinefelter syndrome. Match the disease with the chromosomal abnormality: three X chromosomes and one Y chromosome Down syndrome. Patau syndrome. Turner syndrome. Edwards syndrome. Klinefelter syndrome. Match the disease with the chromosomal abnormality: four X chromosomes and one Y chromosome Down syndrome. Patau syndrome. Turner syndrome. Edwards syndrome. Klinefelter syndrome. 122. A. B. C. D. E. * 123. A. B. C. D. E. * 124. A. B. C. D. E. * 125. A. B. C. D. E. * 126. A. B. C. D. E. * 127. A. B. C. D. E. * 128. A. B. C. D. E. * Breast development in mammals is typically only seen in females and the trait is inherited on the autosomes. This is an example of a Y-linked trait X-linked trait sex-influenced trait autosomal trait sex-limited trait The theory that the Barr Body is an inactivated X-chromosome is the cell theory the chromosome theory of inheritance the genetic balance theory none of these the Lyon hypothesis Human males are neither heterozygous or homozygous for alleles on the X chromosome, they are heterozygous homogametic dominant recessive hemizygous The crosses involving the white-eyed and red-eyed alleles on the X chromosome in fruit flies proved to be a test of the Cell Theory Lyon Hypothesis Rule of Segregation Mendel’s laws of inheritance Chromosome Theory of Inheritance If 83 individuals are homozygous for the recessive trait, eyeless, and 67 of those individuals are eyeless, then this is an example of expressivity pleiotropy incomplete dominance comlementaryty incomplete penetrance In humans, many more males are affected by Duchene muscular dystrophy than females and these affected males typically come from unaffected parents. The likely inheritance pattern is autosomal dominant autosomal recessive X-linked dominant Y-linked X-linked recessive Which of the following syndromes are associated with the chromosome complement of 47,XXY? Down Syndrome Prader Willi Syndrome Angelman Syndrome Turner Syndrome Klinefelter Syndrome 129. A. B. C. D. E. * 130. A. B. C. D. E. * 131. A. B. C. D. E. * 132. A. B. C. D. E. * 133. A. B. C. D. E. * 134. A. B. C. D. E. * 135. A. B. C. D. E. * 136. A. B. Which of the following syndromes are associated with the chromosome complement of 45,X? Down Syndrome Prader Willi Syndrome Angelman Syndrome Klinefelter Syndrome Turner Syndrome Which of the following syndromes are associated with the chromosome complement of 47,+21? Prader Willi Syndrome Angelman Syndrome Klinefelter Syndrome Turner Syndrome Down Syndrome The majority of cases of DOWN syndrome occur as a result of Mosaicism of normal and trisomic cell lines Unbalanced 14:21 translocations End to end fusion of two chromosomes 21 Non-disjunction during paternal meiosis Non-disjunction during maternal meiosis Chromosomes 1, 3, 19 and 20 are Submetacentric Acrocentric Telocentric Sex-chromosomes Metacentric A man who is affected with phenylketonuria marries a woman who is heterozygous at that locus. What is the probability that their first child will have phenylketonuria? 1/8 1/4 3/4 1 1/2 Which of the following genotypes causes Klinefelter syndrome? 45/ XO 46/ XX 47/ XYY 46/ XY 47/ XXY Which of the following is an example of monosomy? 46,XX 47,XXX 69,XYY 47,+21 45,X Which of the following is an example of trisomy? 46,XX 69,XYY C. D. E. * 137. A. B. C. D. E. * 138. A. B. C. D. E. * 139. A. B. C. D. E. * 140. A. B. C. D. E. * 141. A. B. C. D. E. * 142. A. B. C. D. E. * 143. A. B. C. D. E. * 144. 45,X 46, XY 47,XXX Which of the following is an example of trisomy? 46,XX 69,XYY 45,X 46,XY 47,+21 Which of the following is an example of trisomy? 46,XX 69,XYY 45,X 46,XY 47,+18 Which of the following is an example of trisomy? 46,XX 69,XYY 45,X 46,XY 47,+13 Form of interaction between allelic genes is: epistasis continuous variation complementarity polimery complete dominance Form of interaction between allelic genes is: epistasis continuous variation complementarity polimery incomplete dominance Form of interaction between allelic genes is: epistasis continuous variation complementarity polimery superdominance Form of interaction between allelic genes is: epistasis continuous variation complementarity polimery codominance Form of interaction between non-allelic genes is: A. B. C. D. E. * 145. A. B. C. D. E. * 146. A. B. C. D. E. * 147. A. B. C. D. E. * 148. A. B. C. D. E. * 149. A. B. C. D. E. * 150. A. B. C. D. E. * 151. A. B. C. D. codominance complete dominance incomplete dominance super dominance complementarity Form of interaction between non-allelic genes is: codominance complete dominance incomplete dominance super dominance epistasis Form of interaction between non-allelic genes is: codominance complete dominance incomplete dominance superdominance continuous variation Form of interaction between non-allelic genes is: codominance complete dominance incomplete dominance superdominance polimery Genetics is the science about: the peculiarities of the hereditary the peculiarities of the variability karyotype chromosomes the peculiarities of the hereditary and variability Which of the following is not an example of a multifactorial trait? diabetes disorders intelligence obesity cystic fibrosis Cri-du-chat syndrome is associated with a missing chromosome number 5. an additional chromosome number 5. a deletion of the long arm of chromosome number 5. an invertion of chromosome number 5. a deletion of the short arm of chromosome number 5. Down syndrome is associated with a missing chromosome number 21. a deletion of the long arm of chromosome number 21. a deletion of the short arm of chromosome number 21. an invertion of chromosome number 21. E. * 152. A. B. C. D. E. * 153. A. B. C. D. E. * 154. A. * B. C. D. E. 155. A. B. C. D. E. * 156. A. B. C. D. E. * 157. A. B. C. D. E. * 158. A. B. C. D. E. * 159. A. B. an additional chromosome number 21. Patau syndrome is associated with a missing chromosome number 13. a deletion of the long arm of chromosome number 13. a deletion of the short arm of chromosome number 13. an invertion of chromosome number 13. an additional chromosome number 13. Edward’s syndrome is associated with a missing chromosome number 18. a deletion of the long arm of chromosome number 18. a deletion of the short arm of chromosome number 18. an invertion of chromosome number 18. an additional chromosome number 18. Turner syndrome is associated with a missing X-chromosome in fеmale. an additional X-chromosome in male. a deletion of the long arm of X-chromosome. a deletion of the short arm of X-chromosome. an additional X-chromosome in female. Klanfelter syndrome is associated with a missing X-chromosome in male. a deletion of the long arm of X-chromosome. a deletion of the short arm of X-chromosome. an additional X-chromosome in female. an additional X-chromosome in male. Approximately ____ percentage of spontaneous abortions result from extra or missing chromosomes. 10. 25. 75. 100. 50. The most common autosomal aneuploid is trisomy 13. trisomy 18. trisomy X. trisomy Y. trisomy 21. An individual with two normal sets of autosomes and a single X chromosome has Down syndrome. Patau syndrome. Klinefelter syndrome. Angelman syndrome. Turner syndrome. Which parent does nondisjunction lead to the sex chromosome aneuploid XYY in? mother. either parent. C. D. E. * 160. A. B. C. D. E. * 161. A. B. C. D. E. * 162. A. B. C. D. E. * 163. A. B. C. D. E. * 164. A. B. C. D. E. * 165. A. B. C. D. E. * 166. A. B. C. D. E. * 167. both parents. Non of above father. An individual with the chromosomal description 45, X would be a normal female. normal male. male with Klinefelter syndrome. cannot be determined. female with Turner syndrome. How many chromosomes does a triploid human karyotype have? 23 46 92 44 69 What process can lead to break of genes linkage? mitosis pleiotropy meiosis mutation crossing-over All of the following aneuploids are resulted of female nondisjunction except triplo-X. Jacobs syndrome. Down syndrome. Patau syndrome. Klinefelter syndrome. Which of the following cell types is not used to examine chromosomes? white blood cells. bone marrow cells. cells of skin. all of the above can be used. red blood cells. About 90% of trisomy 13 are due to nondisjunction during meiosis I in the female. meiosis I in the male. meiosis II in the male. mitosis in the male. meiosis II in the female. About 90% of trisomy 18 are due to nondisjunction during meiosis I in the female. meiosis I in the male. meiosis II in the male. mitosis in the male. meiosis II in the female. Which of the following genetic disorders is expressed in individuals after the prenatal period? A. B. C. D. E. * 168. A. B. C. D. E. * 169. A. B. C. D. E. * 170. A. B. C. D. E. * 171. A. B. C. D. E. * 172. A. B. C. D. E. * 173. A. B. C. D. E. * 174. A. B. Polydactyly Cystic fibrosis Albinism Phenilketonuria Huntington disease Which of the following is a rapid-aging disorder? Cystic fibrosis Marfan syndrome Patau syndrom Phenilketonuria Progeria Geneticists calculate the ____ of a trait, or the degree to which it is inherited, as the percentage of pairs in which both twins express the trait. heritability coefficient of relationship. empiric risk. none of the above concordance. Which of the following cell types is used to examine chromosomes during cytological method? bone marrow cells. red blood cells. cells of skin. all of the above can be used. white blood cells. The drug ____ was used to treat morning sickness between 1957 and 1961. Unfortunately, babies born to mothers who used this drug were born with incomplete or missing legs and arms. Lithium Diethylstilbestrol Penicillamine Kolchicyn Thalidomide Genes are arranged in a chromosome: irregulary orderly in conglomerations in a definite locus linearly A full diploid chromosome number of a somatic cell is called: genotype phenotype genome hemizygote karyotype Human height is an example of a (an) ____ trait. autosomal. X-linked. C. D. E. * 175. A. B. C. D. E. * 176. A. B. C. D. E. * 177. A. B. C. D. E. * 178. A. B. C. D. E. * 179. A. B. C. D. E. * 180. A. B. C. D. E. * 181. A. B. Y-linked. epistatic. polygenic. Human color of skin is an example of a (an) ____ trait. autosomal. X-linked. Y-linked. epistatic. polygenic. Which genotype would result in an individual with the most black-colored skin? AabbCC. AABBcc. AaBbcc. aabbcc. A1A1A2A2A3A3. Which genotype would result in an individual with medium-colored skin? AABBcc. AaBbcc. aabbcc. AABBCC. A1a1A2a2A3a3. Which genotype would result in an individual with the most white-colored skin? AabbCC. AABBcc. AaBbcc. AABBCC. a1a1а2а2а3а3. A geneticist crosses a plant with red flowers to a plant with white flowers. The offspring include plants with red flowers (1/4), pink flowers (1/2), and white flowers (1/4). Color of flowers expression is an example of complete dominance. codominance. epistasis. pleiotropy incomplete dominance. Consider the cross Aa x Aa. If the alleles for both genes exhibit incomplete dominance, what phenotypic ratio is expected in the resulting offspring? 1:1:1:1. 9:3:3:1. 3:6:3:1:2:1. 1:2:1:2:4:2:1:2:1. 1:2:1 Consider the cross AaBb x AaBb. If the alleles for both genes exhibit complete dominance, what phenotypic ratio is expected in the resulting offspring? 1:1:1:1. 3:6:3:1:2:1. C. D. E. * 182. A. B. C. D. E. * 15:1. 1:2:1 9:3:3:1. Consider the cross AaBb x AaBb. If the alleles for both genes exhibit complementary, what phenotypic ratio is expected in the resulting offspring? 15:1. 3:6:3:1:2:1. 1:2:1:2:4:2:1:2:1. 1:2:1 9:7.