Download In hemoglobin Tocucci there was a replacement of the amino acid

In hemoglobin Tocucci there was a replacement of the amino acid histidine to tyrosine. What kind
of mutation is this
A. Genomic mutation.
B. Aneuploidy.
C. Polyploidy.
D. Inversion.
E. Gene (point) mutation.
ANSWER E
In hemoglobin Tocucci there was a replacement of the amino acid histidine to tyrosine. What kind
of mutation is this
A. Genomic mutation.
B. Nonsense mutation.
C. Silent mutation.
D. Chromosomal mutation.
E. Missense mutation.
ANSWER E
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
A. Mistake.
B. Replication.
C. Chromosome aberration.
D. Mitosis.
E. Gene mutation.
ANSWER E
In protein there was a replacement of the amino acid histidine to glutamine. What kind of mutation
is this
A. Genomic mutation.
B. Nonsense mutation.
C. Silent mutation.
D. Chromosomal mutation.
E. Missense mutation.
ANSWER E
How many Barr bodies are present in the somatic cell of a female with karyotype 45, XO
A. 1
B. 2
C. 3
D. 4
E. 0
ANSWER E
How many Barr bodies are present in the somatic cell of a female with karyotype 47, XXX
A. 1.
B. 3
C. 0.
D. 4.
E. 2.
ANSWER E
How many Barr bodies are present in the somatic cell of a male with karyotype 47, XXY
A. 2
B. 3
C. 0
D. 4
E. 1
ANSWER E
Which methods of human genetics you have to use for diagnosis of Edwards syndrome
A. Genealogy method.
B. Twin method
C. Dermatoglyphics
D. Pedigree Analysis.
E. Cytogenetical method.
ANSWER E
Which methods of human genetics you have to use for diagnosis of Turner syndrome
A. Genealogy method.
B. Twin method
C. Dermatoglyphics
D. Pedigree Analysis.
E. Chromosomes Analysis.
ANSWER E
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
A. Down syndrome.
B. Turner syndrome.
C. Edwards syndrome.
D. Klinefelter syndrome
E. Patau syndrome.
ANSWER E
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
A. Down syndrome.
B. Patau syndrome.
C. Turner syndrome.
D. Klinefelter syndrome.
E. Edwards syndrome.
ANSWER E
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
A. Patau syndrome.
B. Turner syndrome.
C. Edwards syndrome.
D. Klinefelter syndrome.
E. Down syndrome.
ANSWER E
Give the karyotype formula of a female with cri-du-chat syndrome.
A. 46, XX.
B. 47, XXY.
C. 45, X0.
D. 47, XXX.
E. 46, XX, 5р-.
ANSWER E
Give the karyotype formula of a female with Turner syndrome.
A. 46, XX.
B. 47, XXY.
C. 47, XXX.
D. 46, XX, 5р-.
E. 45, X0.
ANSWER E
Give the karyotype formula of a male with Patau syndrome.
A. 46, XX, 9р+;
B. 46, XX, 5р-;
C. 47, XY, 18+;
D. 47, XY, 21+.
E. 47, XY, 13+.
ANSWER E
Give the karyotype formula of a male with Edwards syndrome.
A. 46, XX, 9р+;
B. 46, XX, 5р-;
C. 47, XY, 21+.
D. 47, XY, 13+.
E. 47, XY, 18+;
ANSWER E
Give the karyotype formula of a male with Down syndrome.
A. 46, XX, 9р+;
B. 46, XX, 5р-;
C. 47, XY, 18+;
D. 47, XY, 13+.
E. 47, XY, 21+.
ANSWER E
Allelic genes ...
A. Have the different loci in the homologous chromosomes.
B. Are situated in the non-homologous chromosomes
C. Are situated in the homologous chromosomes.
D. Have the different loci in the heterochromosomes.
E. Have the same loci in the homologous chromosomes.
ANSWER E
Which form of gene interactions is characterized that one gene masks the phenotypic effect of the
gene from different pair
A. Complete dominance.
B. Incomplete dominance.
C. Superdominance
D. Codominance
E. Epistasis
ANSWER E
Which type of the human ABO blood group is example of codominance
A. I (O)
B. II (A)
C. III (B)
D. All of the above.
E. IV (AB)
ANSWER E
How many types of gametes does the person with NnffPp genotype produce
A. 1
B. 2
C. 6
D. 8
E. 4
ANSWER E
How many types of gametes does the person with AaBbCC genotype produce
A. 1
B. 2
C. 6
D. 8
E. 4
ANSWER E
How many types of gametes does the person with NnFfPp genotype produce
A. 1
B. 2
C. 4
D. 6
E. 8
ANSWER E
How many types of gametes does the person with AaBbCc genotype produce
A. 1
B. 2
C. 4
D. 6
E. 8
ANSWER E
Linkage between two genes was interrupted by crossing-over when
A. Genes stay together at a very short distance on the same chromosome.
B. Genes stay on the two non-homologous chromosomes.
C. Genes stay together at distance 60 centimorgans.
D. Genes stay together at distance 70 centimorgans.
E. Genes stay together at distance 20 centimorgans.
ANSWER E
Linkage between two genes was interrupted by crossing-over when
A. Genes stay together at a very short distance on the same chromosome.
B. Genes stay on the two non-homologous chromosomes.
C. Genes stay together at distance 60 centimorgans.
D. Genes stay together at distance 70 centimorgans.
E. Genes stay together at distance 30 centimorgans.
ANSWER E
How many types of gametes does the person with AABBCC genotype produce
A. 2
B. 4
C. 6
D. 8
E. 1
ANSWER E
How many types of gametes does the person with AaBBCC genotype produce
A. 1
B. 4
C. 6
D. 8
E. 2
ANSWER E
Which sex-limited trait do you know
A. Baldness
B. Hemophilia
C. Polydactyly
D. Albinism
E. Beard growth.
ANSWER E
Which sex-limited trait do you know
A. Baldness
B. Hemophilia
C. Polydactyly
D. Albinism
E. Breast size.
ANSWER E
Which sex-influenced trait do you know
A. Breast size.
B. Hemophilia
C. Polydactyly
D. Albinism
E. Baldness
ANSWER E
Ichthiosis is inherited as Y-linked trait (d- allele of the ichthiosis). Which types of gametes does a
man with ichtiosis produce
A. D, d.
B. D.
C. d.
D. Xd, Y.
E. X, Yd
ANSWER E
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
A. Father
B. Aunt
C. Uncle
D. Grandfather
E. Mother
ANSWER E
Molecular disorders can be caused by
A. Changes in chromosome structure.
B. Changes in chromosome number.
C. Genomic mutations.
D. All of the above.
E. Gene mutations.
ANSWER E
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
A. Deletion
B. Duplication
C. Insertion
D. Translocation
E. Inversion
ANSWER E
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
A. Duplication
B. Inversion
C. Insertion
D. All of the above.
E. Deletion
ANSWER E
What is the cause of phenylketonuria
A. The lack of an enzyme tyrosinase.
B. The lack of an enzyme hexosaminidase.
C. The lack of an enzyme halactose.
D. All of the above.
E. The lack of an enzyme phenylalaninehydroxilase.
ANSWER E
How many Barr bodies are present in the somatic cell of a female with triplo-X syndrome
A. 1
B. 3
C. 0
D. 4
E. 2
ANSWER E
How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome
(karyotype is 47, XXY)
A. 2
B. 3
C. 0
D. 4
E. 1
ANSWER E
How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome
(karyotype is 48, XXXY)
A. 1
B. 3
C. 0
D. 4
E. 2
ANSWER E
How many Barr bodies are present in the somatic cell of a male with Klinfelter syndrome
(karyotype is 49, XXXXY)
A. 1
B. 2
C. 0
D. 4
E. 3
ANSWER E
How many Barr bodies are present in the somatic cell of a female with karyotype 48, XXXX
A. 1
B. 2
C. 0
D. 4
E. 3
ANSWER E
How many Barr bodies are present in the somatic cell of a female with karyotype 49, XXXXX
A. 1
B. 2
C. 0
D. 3
E. 4
ANSWER E
How many Barr bodies are present in the somatic cell of a female with Down syndrome
A. 2
B. 3
C. 0
D. 4
E. 1
ANSWER E
What is the genotype of man with the rhesus-positive II group of blood
A. IAirhrh.
B. IBirhrh.
C. IAIB Rhrh.
D. IAIB rhrh.
E. IAIARhRh.
ANSWER E
Non-allelic genes
A. Are situated in the sex chromosomes.
B. Are situated in the homologous chromosomes.
C. Have the same loci in the homologous chromosomes.
D. Have the same loci in the heterochromosomes.
E. Are situated in the non-homologous chromosomes.
ANSWER E
How many types of gametes does the person with mmBBCc genotype produce
A. 4
B. 8
C. 3
D. 7
E. 2
ANSWER E
Which holandric trait do you know
A. Baldness
B. Beard growth.
C. Polydactyly
D. Hemophilia
E. Hairy pinnae.
ANSWER E
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
A. D, d.
B. D.
C. d.
D. Xd, Y.
E. XD, Y.
ANSWER E
What is the cause of albinism
A. The lack of an enzyme phenylalaninehydroxilase.
B. The lack of an enzyme hexosaminidase.
C. The lack of an enzyme halactose.
D. The lack of an enzyme helicase.
E. The lack of an enzyme tyrosinase.
ANSWER E
Which type of chromosomal aberration is characterized by separation and loss of a chromosome
segment
A. Duplication
B. Translocation
C. Inversion
D. Polyploidy
E. Deletion
ANSWER E
A segment of DNA of normal diploid cell has the sequence of the nucleotides
A. After ionizing radiation mutant DNA have the sequence of the nucleotides AATGCTA.What
kind of mutation is this
B. Translocation
C. Deletion
D. Polyploidy
E. Inversion
ANSWER D
A permanent transmissible change in the genetic material (modification in chromosomes and genes)
is
A. Modification
B. Gametogenesis
C. Fertilization
D. Phenotypic variation.
E. Genetic variation
ANSWER E
A nineteen year old female with short stature, wide spaced nipples, and primary amenorrhea most
likely has the karyotype of
A. 47,XX,+18
B. 46,XY
C. 47,XXY
D. 46,XX
E. 45,X0
ANSWER E
A boy, 9 years of age, is admitted to the pediatric ward with hemophilia A. He inherited this
condition through a ___________ trait.
A. X-linked dominant
B. Y-linked
C. Autosomal dominant
D. Autosomal recessive
E. X-linked recessive
ANSWER E
How many chromosomes does a triploid human karyotype have
A. 23
B. 46
C. 92
D. 44
E. 69
ANSWER E
How many chromosomes does a person with Edward syndrome have
A. 45
B. 46
C. 48
D. 44
E. 47
ANSWER E
How many chromosomes does a person with Down syndrome have
A. 45
B. 46
C. 48
D. 44
E. 47
ANSWER E
How many chromosomes does a person with Turner syndrome have
A. 46
B. 47
C. 48
D. 44
E. 45
ANSWER E
How many chromosomes does a person with Klanfelter syndrome have
A. 45
B. 46
C. 48
D. 44
E. 47
ANSWER E
In human ichthyosis is holandric disorder. A couple have a single child with ichthyosis. Who passed
the gene of ichthyosis to the child
A. Mother
B. Aunt
C. Uncle
D. Grandfather
E. Father
ANSWER E
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
A. 5 centimorgans.
B. 3 centimorgans.
C. 10 centimorgans.
D. 13 centimorgans.
E. 8 centimorgans.
ANSWER E
Sicle-cell anaemia is caused by an _____ gene.
A. X-linked dominant
B. X-linked recessive
C. Y-linked
D. Autosomal recessive
E. Autosomal dominant
ANSWER E
Newborn screening for _______is example of a screening test
A. Down syndrome.
B. Patau syndrome.
C. Albinism.
D. Sicle-cell anaemia.
E. Phenylketonuria.
ANSWER E
Color blindness is an X-linked trait in humans. If a color-blind woman marries a man with normal
vision, the children will be
A. all color-blind daughters, but normal sons.
B. all normal sons, but carrier daughters.
C. all color-blind children
D. all normal children.
E. all color-blind sons, but carrier daughters
ANSWER E
Match the disease with the chromosomal abnormality single X chromosome with no homologous X
or Y chromosome
A. Down syndrome.
B. Patau syndrome.
C. Edwards syndrome.
D. Klinefelter syndrome.
E. Turner syndrome.
ANSWER E
What is the most clinically useful technique for prenatal diagnosis of chromosomal abnormalities at
2 months’ gestation
A. Gene mapping
B. Linkage analysis
C. Amniocentesis
D. Ultrasonography
E. Chorionic villi biopsy
ANSWER E
People who have neurofibromatosis have a varying degree of the condition because of the genetic
principle of
A. Penetrance
B. Dominance
C. Recessiveness
D. Pleiotropy
E. Expressivity
ANSWER E
Cystic fibrosis is caused by an _____ gene
A. X-linked dominant
B. X-linked recessive
C. Y-linked
D. Autosomal dominant
E. Autosomal recessive
ANSWER E
Match the disease with the chromosomal abnormality two X chromosomes and one Y chromosome
A. Down syndrome.
B. Patau syndrome.
C. Turner syndrome.
D. Edwards syndrome.
E. Klinefelter syndrome.
ANSWER E
Which of the following is an example of monosomy
A. 46,XX
B. 47,XXX
C. 69,XYY
D. 47,+21
E. 45,X
ANSWER E
Which of the following is an example of trisomy
A. 46,XX
B. 69,XYY
C. 45,X
D. 46, XY
E. 47,XXX
ANSWER E
Which of the following is an example of trisomy
A. 46,XX
B. 69,XYY
C. 45,X
D. 46,XY
E. 47,+21
ANSWER E
Which of the following is an example of trisomy
A. 46,XX
B. 69,XYY
C. 45,X
D. 46,XY
E. 47,+18
ANSWER E
Which of the following is an example of trisomy
A. 46,XX
B. 69,XYY
C. 45,X
D. 46,XY
E. 47,+13
ANSWER E
Form of interaction between allelic genes is
A. epistasis
B. continuous variation
C. complementarity
D. polimery
E. complete dominance
ANSWER E
Form of interaction between allelic genes is
A. epistasis
B. continuous variation
C. complementarity
D. polimery
E. incomplete dominance
ANSWER E
Form of interaction between allelic genes is
A. epistasis
B. continuous variation
C. complementarity
D. polimery
E. superdominance
ANSWER E
Form of interaction between allelic genes is
A. epistasis
B. continuous variation
C. complementarity
D. polimery
E. codominance
ANSWER E
Form of interaction between non-allelic genes is
A. codominance
B. complete dominance
C. incomplete dominance
D. super dominance
E. complementarity
ANSWER E
Form of interaction between non-allelic genes is
A. codominance
B. complete dominance
C. incomplete dominance
D. super dominance
E. epistasis
ANSWER E
Form of interaction between non-allelic genes is
A. codominance
B. complete dominance
C. incomplete dominance
D. superdominance
E. continuous variation
ANSWER E
Form of interaction between non-allelic genes is
A. codominance
B. complete dominance
C. incomplete dominance
D. superdominance
E. polimery
ANSWER E
Genetics is the science about
A. the peculiarities of the hereditary
B. the peculiarities of the variability
C. karyotype
D. chromosomes
E. the peculiarities of the hereditary and variability
ANSWER E
The most common autosomal aneuploid is
A. trisomy 13.
B. trisomy 18.
C. trisomy X.
D. trisomy Y.
E. trisomy 21.
ANSWER E
An individual with two normal sets of autosomes and a single X chromosome has
A. Down syndrome.
B. Patau syndrome.
C. Klinefelter syndrome.
D. Angelman syndrome.
E. Turner syndrome.
ANSWER E
Which parent does nondisjunction lead to the sex chromosome aneuploid XYY in
A. mother.
B. either parent.
C. both parents.
D. Non of above
E. father.
ANSWER E
An individual with the chromosomal description 45, X would be a
A. normal female.
B. normal male.
C. male with Klinefelter syndrome.
D. cannot be determined.
E. female with Turner syndrome.
ANSWER E
How many chromosomes does a triploid human karyotype have
A. 23
B. 46
C. 92
D. 44
E. 69
ANSWER E
What process can lead to break of genes linkage
A. mitosis
B. pleiotropy
C. meiosis
D. mutation
E. crossing-over
ANSWER E
All of the following aneuploids are resulted of female nondisjunction except
A. triplo-X.
B. Jacobs syndrome.
C. Down syndrome.
D. Patau syndrome.
E. Klinefelter syndrome.
ANSWER E
Which of the following cell types is not used to examine chromosomes
A. white blood cells.
B. bone marrow cells.
C. cells of skin.
D. all of the above can be used.
E. red blood cells.
ANSWER E
About 90% of trisomy 13 are due to nondisjunction during
A. meiosis I in the female.
B. meiosis I in the male.
C. meiosis II in the male.
D. mitosis in the male.
E. meiosis II in the female.
ANSWER E
About 90% of trisomy 18 are due to nondisjunction during
A. meiosis I in the female.
B. meiosis I in the male.
C. meiosis II in the male.
D. mitosis in the male.
E. meiosis II in the female.
ANSWER E
Which of the following genetic disorders is expressed in individuals after the prenatal period
A. Polydactyly
B. Cystic fibrosis
C. Albinism
D. Phenilketonuria
E. Huntington disease
ANSWER E
Which of the following is a rapid-aging disorder
A. Cystic fibrosis
B. Marfan syndrome
C. Patau syndrom
D. Phenilketonuria
E. Progeria
ANSWER E
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.
A. heritability
B. coefficient of relationship.
C. empiric risk.
D. none of the above
E. concordance.
ANSWER E
Which of the following cell types is used to examine chromosomes during cytological method
A. bone marrow cells.
B. red blood cells.
C. cells of skin.
D. all of the above can be used.
E. white blood cells.
ANSWER E
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.
A. Lithium
B. Diethylstilbestrol
C. Penicillamine
D. Kolchicyn
E. Thalidomide
ANSWER E
Genes are arranged in a chromosome
A. irregulary
B. orderly
C. in conglomerations
D. in a definite locus
E. linearly
ANSWER E