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Genetic Basis Of Inheritance Mcq
Quiz 2
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Q.1
What is true in case of Honey Bee? .... ... [ Bih PMT 1994]
0%
a) Male diploid, female haploid
0%
b) Male diploid, female diploid
0%
c) Male haploid, female haploid
0%
d) Male haploid, female diploid
Explanation
The sex determination in honey bee is based on the number of sets of chromosomes an individual receives. An offspring formed from the union of a sperm and an egg develops as a female (queen or worker), and an unfertilised egg develops as a male (drone) by means of parthenogenesis. This means that the males have half the number of chromosomes than that of a female. The females are diploid having 32 chromosomes and males are haploid, i.e., having 16 chromosomes. This is called as haplodiploid sex-determination system Answer : (d)
Q.2
Down's syndrome and Turner's syndrome are sue to respectively .... .. [ kerala 2007 ]
0%
a) Monosomic and nillisomic conditions
0%
b) Trisomic and monosomic conditions
0%
c) Monosomic and trisomic conditions
0%
d) Trisomic and tetrasomic conditions
Explanation
Aneuploidy is a chromosomal variation due to a loss or a gain of one or more chromosomes resulting in the deviation from the normal or the usual number of chromosomes. The different conditions of aneuploidy are nullisomy (2N-2), monosomy (2N-1), trisomy (2N+1), and tetrasomy (2N+2). Down’s syndrome results in the gain of extra copy of chromosome 21. This condition is trisomy. Turner’s syndrome results due to loss of an X chromosome in human females. This conition is monosomy. Answer : (b)
Q.3
If BB represents barr body and Y0 Y-body, XXY or Klinefelter's syndrome has .... ... [ CPMT 1997 ]
0%
a) BB - 1, Y0 - 0
0%
b) BB - 1, Y0 - 1
0%
c) BB - 0, Y0 - 1
0%
d) BB - 2, Y0 - 1
Explanation
A Barr body or X-chromatin is an inactive X chromosome in a cell with more than one X chromosome, rendered inactive in a process called lyonization, in species with XY sex-determination. Men with Klinefelter syndrome (47, XXY karyotype) have a single Barr body. Answer : (b)
Q.4
Which genotype and phenotype is a result of aneuploidy in sex chromosomes? ... [ CPMT 2004 ]
0%
a) 22 pairs + XXY male
0%
b) 22 + XX female
0%
c) 22+ pairs + XXXY female
0%
d) 22 pairs + Y female
Explanation
Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Aneuploidy in sex chromosomes means additional copy of a chromosome may be included in an individual or an individual may lack one of any one pair of chromosomes. example: Turner’s Syndrome : Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with X0. Klinefelter’s Syndrome : This genetic disorder is also caused due to the presence of an additional copy of Xchromosome resulting into a karyotype of 47, XXY. Answer : (a)
Q.5
A lady carries for haemophilia (Hh) marries a normal man (HO). Daughters of such a lady would be ... ... [ kerala 2001 ]
0%
a) 50% normal (HH) and 50% carrier (Hh)
0%
b) 50% normal (HH) and 50% haemophilic (hh)
0%
c) 25% carrier (Hh) and 75% haemophilic
0%
d) 75% carrier (Hh) and 25% haemophilic (hh)
Explanation
Gamete
X
X
h
X
XX
Normal daughter
X X
h
Carrier daughter
Y
XY
Normal boy
X
h
Y
Haemophilic boy
Haemophilia : This sex linked recessive disease, which shows its transmission from unaffected carrier female to some of the male progeny has been widely studied. Cross between haemophilic carrier woman and normal man will produce progeny as follows: 50% carrier daughter and 50% normal daughter. Answer : (a)
Q.6
Chromosomes other than sex chromosomes are called ... ... [ CMC 2003 ]
0%
a) Allosomes
0%
b) Autosomes
0%
c) Lampbrush chromosomes
0%
d) Heterosomes
Explanation
Answer : (b)
Q.7
A monosomic (2N-1) abnormality in human is ... ... [ JIPMER 2002 ]
0%
a) Klinefelter's syndrome
0%
b) Turner's syndrome
0%
c) Edward's syndrome
0%
d) Down's syndrome
Explanation
Answer : (b)
Q.8
If father is normal while mother is carrier of haemophilia .... .... [ KCET 2012 ] Explanation is provided, please click on
0%
a) All female offspring ill be carriers
0%
b) A male offspring has 50% chance of active disease
0%
c) A female offspring has 50% chance of active disease
0%
d) All female offspring will be normal
Explanation
The father's sex chromosomes are labeled XY. The father only passes half of his sex chromosomes to the baby, either the X or the Y. If the baby gets the Y chromosome from the father it will be a boy. The son can get from the mother either her X chromosome with the hemophilia gene or her X chromosome with the normal blood clotting gene. If the son gets his mother's X chromosome with the hemophilia gene he will have hemophilia. If he inherits his mother's other X chromosome, he will have normal blood clotting. So a carrier's son has a 50% chance of having hemophilia. A baby girl will inherit an X chromosome with a dominant gene for normal blood clotting from her father. So the daughter will not have hemophilia. A daughter will get either her mother's X chromosome with the hemophilia gene or her mother's X chromosome with the normal gene for clotting. If she gets the X chromosome with the hemophilia gene she will be a carrier. So a carrier's daughter has a 50% chance of being a carrier. Answer : (b)
Q.9
Frequency of A allele is 0.6 and that of a allele is 0.What would be frequency of hetero zygotes in random mating population? ...... [CBSE 2005]
0%
a) 0.36
0%
b) 0.16
0%
c) 0.24
0%
d) 0.48
Explanation
Hardy-Weinberg equation is p2 + 2pq + q2 = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Hence, frequency of heterozygous genotype= 2pq= 2 x 0.6 x 0.4= 0.48. Answer : (d)
Q.10
Sickle cell anaemia has not been eliminated from African population as ... ... [ CBSE 2006 ]
0%
a) It is controlled by dominant genes
0%
b) it is controlled by recessive genes
0%
c) It is not a fata disease
0%
d) It provides immunity against malaria
Explanation
Sickle cell anaemia provides immunity against malaria. The sickle cells have membranes, stretched by their unusual shape, that become porous and leak nutrients that the parasites need to survive and the faulty cells eventually get eliminated quite fast by the organisms, destroying the parasite along the way. Answer : (d)
Q.11
Wilson detected colour blindness in ... .. [ kerala 2003]
0%
a) Colour blind
0%
b) Normal male
0%
c) Normal female
0%
d) Haemophilic
Explanation
Answer : (c)
Q.12
In a chromosome the protein content is .... .. [ DPMT 1996]
0%
a) Nil
0%
b) trace
0%
c) Half of DNA
0%
d) Same as DNA
Explanation
Chromosomes are made up of a DNA-protein complex called chromatin that is organized into subunits called nucleosomes. Protein content is same as in DNA. Answer : (d)
Q.13
Marriage between colour blind man and normal woman shall result in .... .. [ EAMCET 1999 ]
0%
a) Colour blind female progeny
0%
b) Colour blind male progeny
0%
c) Normal vision female progeny
0%
d) Normal vision male and female progeny
Explanation
Gamete
X
X
X
C
X
C
X
Carrier normal vision girl
X
C
X
Carrier, normal vision girl
Y
XY
Normal boy
X Y
Normal boy
A colour blind man has XCY and normal female has XX chromosomes. Progeny will be as follows: All normal vision progeny. Answer : (d)
Q.14
Crossing over produces ... .. [ CPMT 1991 ]
0%
a) Recombination of linked genes
0%
b) Synapsis of linked genes
0%
c) Expression of recessive genes
0%
d) Linkage of dominant genes
Explanation
Crossing over is a process that happens between homologous chromosomes in order to increase genetic diversity. During crossing over, part of one chromosome is exchanged with another. The result is a hybrid chromosome with a unique pattern of genetic material. This generation of non-parental gene combinations are known as recombinant. Answer : (a)
Q.15
Cause of chromosome laggards in meiosis is .... .. { Chd. CET 2012 ]
0%
a) Inversion
0%
b) Di centric chromosome
0%
c) A centric chromosome
0%
d) Duplication of a gene
Explanation
A laggard was defined as a chromosome that did not overlap along the long axis of the spindle with any of the properly segregating chromosomes. Dicentric chromosome is the cause of chromosome laggards in meiosis as it is an abnormal chromosome with two centromeres. Answer : (b)
Q.16
Genes for cytoplasmic male sterility in plants are located in ... .. [ AMU 2011 ]
0%
a) Chloroplast genome
0%
b) Mitochondrial genome
0%
c) Nuclear genome
0%
d) Cytosol
Explanation
Cytoplasmic male sterility can be defined as total or partial male sterility in plants due to specific nuclear and mitochondrial interactions. Genes responsible for this are present in mitochondrial genome. Answer : (b)
Q.17
Which represents correct hexaploid nature of Wheat? [ AIIMS 2003 ]
0%
a) Monosomic - 21 ; Haploid - 28; Nullisomic - 42; Trisomic -43
0%
b) Monosomic - 7 ; Haploid - 28; Nullisomic -40; Trisomic -42
0%
c) Monosomic - 21 ; Haploid - 7; Nullisomic -42; Trisomic -43
0%
d) Monosomic - 41 ; Haploid - 21; Nullisomic -40; Trisomic -43
Explanation
Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and, this phenomenon is known as polyploidy. Wheat, Triticum, has basic chromosome number is n= 7; thus the hexaploid species (Triticum aestivum) have 6n= 42. Monosomic =6n-1 = 41 Nullisomic = 6n-2 = 40 Trisomic = 6n+1 = 43 Haploid = 6n/2 = 21 Answer : (d)
Q.18
Mutations can be induced in bacteria by ... ... [ AFMC 1999]
0%
a) Growing different strains in same culture
0%
b) Starvation
0%
c) Providing growth substances
0%
d) High energy radiation
Explanation
Mutations can result from errors during DNA replication or induced by exposure to mutagens (like chemicals and radiation). Answer : (d)
Q.19
Failure of separation of sister chromatids is ... ... [ kerala 2004 ]
0%
a) Fusion
0%
b) Non disjunction
0%
c) Complementation
0%
d) Interference
Explanation
During anaphase, sister chromatids (or homologous chromosomes for meiosis I), will separate and move to opposite poles of the cell, pulled by microtubules. In nondisjunction, the separation fails to occur causing both sister chromatids or homologous chromosomes to be pulled to one pole of the cell. Answer : (b)
Q.20
Number of chromosomes in male grasshopper is .... .. [ MHTCET 2007]
0%
a) 8
0%
b) 45
0%
c) 46
0%
d) 23
Explanation
In grasshoppers, males have 23 chromosomes and females have 24 chromosomes Answer : (d)
Q.21
Linkage decreases frequency of .. ... [ CPMT 1998 ]
0%
a) recessive allele
0%
b) Dominant allele
0%
c) Hybridization
0%
d) Both B and C
Explanation
The term linkage is used to describe physical association of genes on a chromosome. The genes that are close to each other have a tendency to be more linked than those that are farther. The more the linkage lesser will be the recombination frequency. Answer : (c)
Q.22
Huntington's chorea appears at the age of .... ....
0%
a) 25 -55 years
0%
b) 15 -25 years
0%
c) 50 -60 years
0%
d) 10 -15 years
Explanation
Huntington's disease is caused by an inherited defect in a single gene. Huntington's disease is an autosomal dominant disorder, which means that a person needs only one copy of the defective gene to develop the disorder. An inherited condition in which nerve cells in the brain break down over time. It typically starts in a person's 30s or 40s. Usually, Huntington's disease results in progressive movement, thinking (cognitive) and psychiatric symptoms. Answer : (a)
Q.23
The word chromosome was called by .... ... [ MPPMT 1997 ]
0%
a) Benda
0%
b) Waldeyer
0%
c) Robert Hooke
0%
d) T.H. Morgan
Explanation
Chromosomes were discovered by Hofmeister (1848), studied by Strasburger (1875) and given the present name by Waldeyer (1888) after their staining by dyes like Janus Green. Answer : (b)
Q.24
Which is a base analogue .... ... [MPPMT 1990 ]
0%
a) 5-Bromouracil
0%
b) Caffeine
0%
c) Colchicine
0%
d) Nitrous acid
Explanation
Base analogues are molecules that can substitute for normal bases in nucleic acids. Usually, substitution of a base analogue will result in altered base pairings and structural changes that affect DNA replication and transcription of genes. Examples of base analogues include 5-bromouracil, 2-aminopurine, 6-mercaptopurine, and acycloguanosine. Since 5-bromouracil can pair with either adenine or guanine, it also affects base pairing during DNA replication, which leads to mutations. Answer : (a)
Q.25
Two linked genes a and b show 20% recombination. The individuals of dihybrid cross between + +/+ + × ab/ab shall show gametes ... .. [ BHU 1997 ]
0%
a) + + 80 : ab : 20
0%
b) + + 50 : ab : 50
0%
c) + + 40ab40: +a10 : +b:10
0%
d) + + 30ab30: +a20 : +b:20
Explanation
Answer : (c)
Q.26
Which one is found in males only ? ..... ... [ RPMT 1996]
0%
a) X-chromosome
0%
b) Y-chromosome
0%
c) 2X chromosomes
0%
d) X+X - chromosomes
Explanation
Answer : (b)
Q.27
Which one of the following symbols and is representation, used in human pedigree analysis is correct? [ CBSE 2010 ]
0%
a) ◯ = unaffected male
0%
b) ◹ = unaffected female
0%
c) ♦ = male affected
0%
d) □=◯ = mating between relatives
Explanation
In human pedigree analysis full-shaded symbols represent the affected individuals while the half-shaded stands for carriers. The square shape stands for a male individual while the circle stands for females. A diamond shape represents the unidentified sex. A square connected by two horizontal lines with circle represents the mating between relatives. Answer : (d)
Q.28
Exchange of one part of a chromosome with a part of the same or another chromosome is . ..... [ AFMC 2002 ]
0%
a) inversion
0%
b) crossing over
0%
c) translocation
0%
d) linkage
Explanation
A translocation occurs when a piece of one chromosome breaks off and attaches to another chromosome. Answer : (c)
Q.29
A disease found only in males is .... ... [ AMU 2002 ]
0%
a) Gaucher's disease
0%
b) Lesch-Nyhan disease
0%
c) Hunter's disease
0%
d) Fabry's disease
Explanation
Lesch-Nyhan syndrome is a condition that occurs almost exclusively in males. It is characterized by neurological and behavioral abnormalities and the overproduction of uric acid. Uric acid is a waste product of normal chemical processes and is found in blood and urine. Answer : (b)
Q.30
If haemophilic female survives and marries a normal male, the theoretical ratio of their offspring regarding haemophilia will be ... ... [ CPMT 2005 ]
0%
a) All offspring haemophilic
0%
b) All girls haemophilic
0%
c) All sons haemophilic
0%
d) 50% of sons and 50% daughters haemophilic
Explanation
Gamete
X
h
X
h
X
X
h
X
Carrier daughter
X X
h
Carrier daughter
Y
X
h
Y
Haemophilic boy
X
h
Y
Haemophilic boy
Haemophilia : This sex linked recessive disease. Cross between haemophilic woman and normal man will produce progeny as follows: All daughter will be carrier and all the sons are haemophilic. Answer : (c)
Q.31
Drosophila is metamale with chromosomal formula ... ... [ kerala 2007 ]
0%
a) 2A + 3X
0%
b) 3A + 3X
0%
c) 4A +3X
0%
d) 3A + XY
Explanation
According to this theory of sex determination, if the ratio of the X chromosome to a total number of sets of autosomes X/A ratio < 0.5 gives supermale or metamale. Drosophila with 1 X chromosome and 3 set of autosomes have the (X/A = 1/3) ratio of 0.33 and is metamale. Answer : (d)
Q.32
Albinism and phenylketonuria are disorders due to ... ... [ AIIMS 2000 ]
0%
a) recessive autosomal genes
0%
b) Dominant autosomal genes
0%
c) Dominant sex genes
0%
d) Recessive sex genes
Explanation
Phenylketonuria : This inborn error of metabolism is also inherited as the autosomal recessive trait. albinism is passed on in an autosomal recessive inheritance pattern. Answer : (a)
Q.33
Haemophilia does not occur in woman .... .. [ BHU 2003 ]
0%
a) It is autosomal recessive
0%
b) Women have to be homozygous which is fatal
0%
c) They have only one X-chromosoe
0%
d) They are more resistant to this disorder
Explanation
Haemophilia : This sex linked recessive disease. Female can be haemophilic when woman has hozygous pattern i.e. XhXh A homozygous recessive female express the disease but they do not survive due to impaired ability or inability of blood clotting which causes heavy blood loss even in the case of small injury. Answer : (b)
Q.34
Epicanthus skin fold above the eyes and transverse palmer crease are typical symptoms of ... ...
0%
a) Cri-du-chat
0%
b) Klinefelter's syndrome
0%
c) Down's syndrome
0%
d) Turner's syndrome
Explanation
Down’s Syndrome : The cause of this genetic disorder is the presence of an additional copy of the chromosome number 21 (trisomy of 21). This disorder was first described by Langdon Down (1866). The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded. The eyes of an individual with Down syndrome might slant upwards a little bit and are almond shaped. They might have small folds of skin at the inner corners, which are called Epicanthal Folds. Answer : (c)
Q.35
Nobel Prize for jumping gene/transposable DNA elements was given to ... ... [AMU 1996 ]
0%
a) Muller
0%
b) Mc Clintock
0%
c) Morgan
0%
d) Kornberg
Explanation
Barbara McClintock,American scientist whose discovery in the 1940s and '50s of mobile genetic elements, or “jumping genes, won her the Nobel Prize for Physiology or Medicine in 1983. Answer : (b)
Q.36
More men suffer from colour blindness than women because .... ... [ KCET 2011 ]
0%
a) Women are more resistant to diseases
0%
b) Male sex hormone testosterone causes the disease
0%
c) Colour blindness gene occurs on Y-chromosome
0%
d) Men are hemizygous and are defective allele is enough to cause the disease
Explanation
Colour Blindness : It is a sex-linked recessive disorder due to defect in either red or green cone of eye resulting in failure to discriminate between red and green colour. This defect is due to mutation in certain genes present in the X chromosome. For female to colour blindess, she should have homozygous defective allele, heterzygous makes a carrier. But for male, defect in X chromosomes can make then colour blind. Answer : (d)
Q.37
Percentage of similarity of β-chain of Hb in humans and Rhesus monkey is ... [ Odisha 2004 ]
0%
a) 2%
0%
b) 4%
0%
c) 8%
0%
d) 40%
Explanation
40% Answer : (d)
Q.38
Chromosomal doubling for producing polyploid plants is carried out by ... .. [ CMC 2003 ]
0%
a) PEG
0%
b) NAA
0%
c) EMS
0%
d) Colchicine
Explanation
Colchicine is an important mutagen that works by preventing the microtubules formation and doubles the number of chromosomes. It is commonly used to develop polyploid plants and functions as a mitotic poison by producing many mutagenic effects on plants Answer : (d)
Q.39
Rearrangement of genes occurs due to .... ... [ CPMT 1997 ]
0%
a) translocation and duplication
0%
b) translocation and deficiency
0%
c) deletion and deficiency
0%
d) translocation and inversion
Explanation
A translocation occurs when a piece of one chromosome breaks off and attaches to another chromosome. An inversion is a chromosome rearrangement in which a segment of a chromosome is reversed end-to-end. Answer : (d)
Q.40
Datura is a classical example of study of ..... ...
0%
a) Monosomics
0%
b) Trisomics
0%
c) Triploids
0%
d) Nullisomics
Explanation
Each chromosome may produce two iso-chromosome, one for each arm; thus for each primary trisomic, two types of secondary trisomics are possible. Thus in Datura (2n – 24), 12 primary and 24 secondary trisomics are possible. Answer : (b)
Q.41
An auxotroph is .... ... [ DPMT 2001]
0%
a) Plant capable of synthesizing own carbohydrates
0%
b) Plant showing quick bending response to sunlight
0%
c) A mutant having lost the ability to synthesis one or more nutrients
0%
d) An organism dependent on another for nutritional requirements
Explanation
Auxotrophy is the inability of an organism to synthesize a particular organic compound required for its growth. Auxotroph = mutant that cannot grow on minimal medium, requires certain supplement(s). Answer : (c)
Q.42
Nucleoprotein structures found at the end of chromosome are ... ...[ JKCMEE 2007 ]
0%
a) Lilium
0%
b) Zea mays
0%
c) Allium
0%
d) Trillium
Explanation
Telomeres are nucleoprotein structures that protect and maintain chromosome ends. Answer : (d)
Q.43
requency of recessive allele is 0.What is the frequency of homozygous dominant? [ EAMCET 2002 ]
0%
a) 0.64
0%
b) 0.32
0%
c) 0.8
0%
d) 0.064
Explanation
(p + q)2 = p2 + q2 + 2pq =1 where; p2 is the frequency of homozygous dominants. q2 is the frequency of homozygous recessive individuals. The 2pq in equation shows frequency of heterozygotes in the population. p+q=1. q=0.2 Then p = 0.8 p2 = (0.8)2 = 0.64 Answer : (a)
Q.44
Exchange of segments between non homologous chromosomes is ... ... [ DPMT 1996]
0%
a) Translocation
0%
b) Inversion
0%
c) Crossing over
0%
d) Tetrasomy
Explanation
A translocation occurs when a piece of one chromosome breaks off and attaches to another chromosome. Answer : (a)
Q.45
A fruit fly exhibiting both male and female traits is .... ... [ CBSE 1994]
0%
a) Heterozygous
0%
b) Gynandromorph
0%
c) Hemizygous
0%
d) Gynander
Explanation
A gynandromorph is an organism that contains both male and female characteristics. Gynandromorph is an animal that's half male and half female when splitting at the midline. Meanwhile, a hermaphrodite is an animal that appears as male or female but has both male and female reproductive organs. Answer : (b)
Q.46
A man with enlarged breasts, sparse body hair and XXY chromosme complement is suffering from ? .... .. [ AIIMS 2000 ]
0%
a) Down's syndrome
0%
b) Turner's syndrome
0%
c) Klinefelter's syndrome
0%
d) Super female
Explanation
Klinefelter’s Syndrome : This genetic disorder is also caused due to the presence of an additional copy of X chromosome resulting into a karyotype of 47, XXY. Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile. Answer : (c)
Q.47
Which one contains haploid set of chromosomes ... ....[ kerala 2000 ]
0%
a) Spermatogonium
0%
b) Primary spermatocyte
0%
c) Secondary spermatocyte
0%
d) Primordial germ cell
Explanation
Secondary spermatocytes are haploid (N) cells that contain half the number of chromosomes. Answer : (c)
Q.48
Which statement about colour blindness is correct .... ... [ AMU 2010 ]
0%
a) 6% men are red colour blind, 2% are green colour blind
0%
b) 2% men are red colour blind, 6% are green colour blind
0%
c) 10% men are red colour blind, 5% are green colour blind
0%
d) 5% men are colour blind, 10% green colour blind.
Explanation
Deuteranomaly (most common—6% of males, 0.4% of females): These individuals have a mutated form of the medium-wavelength (green) pigment. Protanomaly (2% of males, 0.01% of females): Having a mutated form of the long-wavelength (red) pigment Answer : (b)
Q.49
Ultraviolet radiations cause mutations due to ... ... [ Odisha 2003 ]
0%
a) Formation of thymine dimers/ thymidine
0%
b) Deletion of base pairs
0%
c) Addition of base pairs
0%
d) Methylation of base pairs
Explanation
A hallmark of UVC and UVB mutagenesis is the high frequency of transition mutations at dipyrimidine sequences containing cytosine. Because of these mutations, the connection between ultraviolet damage to DNA and cancer is quite clear. Answer : (a)
Q.50
The plant on which Hugo de Vries based his evolution theory is ... ... [ CPMT 1992 ]
0%
a) Antirrhinum majus
0%
b) lathyrus odoratus
0%
c) Oenothera lamarckiana / Evening Primrose
0%
d) Pisum sativum
Explanation
Hugo De Vries conducted his experiments on the plant Oenothera lamarckiana, the common name of this plant is “Evening Primrose”. Answer : (c)
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