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Transcript
1.
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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.