iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which of the following cannot be detected in a developing foetus by amniocentesis ?
A.
Sex of the foetus
B.
jaundice
C.
Klinefelter syndrome
D.
Down syndrome
Correct Answer: B
Explanation:
Amniocentesis is a fetal sex determination
test in which amniotic fluid containing fetal cells which
surrounds the developing embryo is extracted and cells
are tested for chromosomal pattern to identify genetic
disorders, if any. Jaundice is not a chromosomal
disorder thus cannot be tested by amniocentesis.
2013
Q102
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
If two persons with 'AB' blood group marry and have sufficiently large number of children, these children
could be classified as 'A' blood group : 'AB' blood group : 'B' blood group in 1 : 2 : 1 ratio. Modern technique
of protein electrophoresis reveals presence of both 'A' and 'B' type proteins in 'AB' blood group individuals.
This is an example of :
A.
Codominance
B.
Partial dominance
C.
Incomplete dominance
D.
Complete dominance
Correct Answer: A
Explanation:
ABO blood group system in human beings is an example of codominant, dominant
recessive and multiple alletes. Blood groups
are controlled by the gene I located on 9th
chromosome that has 3 multiple alleles, out
of which any two are found in a person.
In codominance both gene express it self
completely
2013
Q103
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Select the incorrect statement with regard to Haemophilia is :
A.
It is a dominant disease
B.
A single protein involved in the clotting of blood is affected
C.
It is a recessive disease
D.
It is a sex-linked disease
Correct Answer: A
Explanation:
Haemophilia is sex-linked disease which is
also known as bleeder’s disease as the patient will
continue to bleed even from a minor cut since he or
she does not possess the natural phenomenon of blood
clotting due to absence of antihaemophiliac globulin
or factor VIII (haemophilia – A) and plasma
thromboplastin factor IX (haemophilia–B, Christmas
disease) essential for it. As a result of continuous bleeding the patient may die of blood loss. It is
genetically due to the presence of a recessive sex linked
gene h, carried by X-chromosome. A female becomes
haemophiliac only when both of her X-chromosomes
carry the gene (XhXh).
However, such females
generally die before birth because the combination of
these two recessive alleles is lethal. A female having
only one allele for haemophilia (XXh) appears normal
because the allele for normal blood clotting present on
the other X-chromosome is dominant. Such females are
known as carriers. In case of males, a single gene for the
defect is able to express itself as the Y-chromosome is
devoid of any corresponding allele (XhY).
2012
Q104
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following is a wrong statement regarding mutations ?
A.
UV and Gamma rays are mutagens
B.
Deletion and insertion of base pairs cause frame-shift mutations
C.
Cancer cells commonly show chromosomal aberrations
D.
Change in a single base pair of DNA does not cause mutation
Correct Answer: D
Explanation:
Change in single base pair of DNA is also
a type of mutations called point mutations.
It is a type of mutation that causes the
replacement of a single base nucleotide with
another nucleotide of the genetic material,
DNA or RNA. For example, a point mutation
is the cause of sickle cell disease.
2012
Q105
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A test cross is carried out to :
A.
Determine whether two species or varieties will breed successfully
B.
Determine the genotype of a plant at F2
C.
Predict whether two traits are linked
D.
Assess the number of alleles of a gene
Correct Answer: B
Explanation:
Test cross is performed to determine the
genotype of F2
plant. In a typical test cross an organism
showing dominant phenotype and whose genotype is
to be determined is crossed with one that is homozygous
recessive for the allele being investigated, instead of
self-crossing. The progenies of such a cross can easily
be analysed to predict the genotype of the test organism.
2012
Q106
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Represented below is the inheritance pattern of the certain type of traits in humans. Which one of the
following conditions could be an example of this pattern ?
A.
Haemophilia
B.
Sickel cell anaemia
C.
Thalassemia
D.
Phenylketonuria
Correct Answer: A
Explanation:
The inheritance pattern of a particular
trait shown in the picture results in haemophilia.
Haemophilia is a group of inherited blood
disorders in which the blood does not clot
properly. It is caused by a fault in one of the
genes that determine how the body makes
blood clotting factor VIII or IX. These genes
are located on the X chromosome.
2012
Q107
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2
: 1. It represents a case of
A.
Monohybrid cross with incomplete dominance
B.
Co - dominance
C.
Monohybrid cross with complete dominance
D.
Dihybrid cross
Correct Answer: A
Explanation:
The inheritance of flower colour in the dog
flower (snapdragon or Antirrhinum sp.) is a good
example which shows incomplete dominance. In a
cross between true-breeding red-flowered (RR) and
true-breeding white-flowered plants (rr), the F1
(Rr) was pink. When the F1 was self-pollinated the F2
resulted in the following ratio, 1 (RR) Red : 2 (Rr)
Pink : 1 (rr) White. Here the genotype ratios were 1 : 2
: 1 as in any Mendelian monohybrid cross, but the
phenotype ratios had changed from the 3 : 1 dominant
: recessive ratio to 1 : 2 : 1.
2012
Q108
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A normal - visioned man whose father was colour bilind, marries a woman whose father was also
colour - blind. They have their first child as a daughter. What are the chances that this child would be
colour blind
A.
50%
B.
25%
C.
100%
D.
Zero percent
Correct Answer: D
Explanation:
To determine the chances that the daughter would be color blind, we need to understand the genetics of color blindness. Color blindness is most commonly caused by defects in the X chromosome, and it is a recessive trait. Since males (XY) have only one X chromosome and females (XX) have two, a male will be color blind if his single X chromosome carries the defect. A female, on the other hand, would need to have the defect on both of her X chromosomes to express color blindness since it is recessive.
Given that the man has normal vision, we can infer that his X chromosome (which he got from his mother) does not have the color blindness defect; however, because his father was color blind, we know that he carries a Y chromosome without the color blindness trait (otherwise, the man would be color blind too). The man will pass on either his X or his Y chromosome to his offspring. If it's a daughter, he will pass on the X chromosome, which we know does not carry the color blindness defect.
The woman's father was color blind, meaning her father's only X chromosome carried the defect. Since women have two X chromosomes, the one she received from her mother could potentially not have the defect. Therefore, the woman can be heterozygous (one normal X chromosome and one with the color blindness defect) or homozygous normal (both X chromosomes without the defect). However, the problem does not provide information about the mother's vision or genotype to confirm whether she is a carrier or not. If the woman is not a carrier, none of her children would inherit color blindness. If she is a carrier, then there's a 50% chance she could pass on the X chromosome with the defect.
So, let's consider the two potential scenarios for the woman, represented by X (normal X chromosome) and X^c (X chromosome with color blindness defect):
1. The woman is a carrier (X X^c): There's a 50% chance she might pass on the X chromosome with the defect (X^c) since she is heterozygous.
2. The woman is not a carrier (X X): There's a 0% chance the child will be color blind since all her X chromosomes are without the color blindness defect.
In conclusion:
If the woman is a carrier, the daughter has a 50% chance of getting the X^c chromosome from her mother and would then be a carrier like her mother (heterozygous), but will not be color blind because the X chromosome from her father is normal.
If the woman is not a carrier, then the daughter has a 0% chance of being color blind because both of her parents would provide normal X chromosomes.
Because we don't know the mother's carrier status with the information given, we cannot say for certain what the probability is for the daughter to be color blind. However, we know it's either 0% (if the mother is not a carrier) or 50% (if the mother is a carrier) chance of being a carrier, but 0% chance of expressing color blindness since she receives one normal X chromosome from her father. Therefore, the probability that the child would be color blind (express the trait) is:
Option D: Zero percent.
2011
Q109
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Test cross in plants or in Drosophila involves
crossing :
A.
the F1 hybrid with a double recessive genotype
B.
between two genotypes with dominant trait
C.
between two F1 hybrids
D.
between two genotypes with recessive trait
Correct Answer: A
Explanation:
In test cross, genotype of an organism
showing dominant phenotype is determined
by crossing it with homozygous recessive
genotype.
2011
Q110
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following conditions of the
zygotic cell would lead to the birth of a normal
human female child ?
A.
only one Y chromosome
B.
only one X chromosome
C.
one X and one Y chromosome
D.
two X chromosomes
Correct Answer: D
Explanation:
Two X chromosomes would lead to the
birth of normal human female child.
2011
Q111
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following conditions correctly
describes the manner of determining the sex ?
A.
XO condition in humans as found in Turner
Syndrome, determines female sex
B.
XO type of sex chromosomes determine
male sex in grasshopper
C.
Homozygous sex chromosomes (ZZ)
determine female sex in Birds
D.
Homozygous sex chromosomes (XX)
produce male in Drosophila
Correct Answer: B
Explanation:
In grasshopper the males lack a Y-sex
chromosome and have only an X-chromosome.
They produce sperm cells that contain either
an X chromosome or no sex chromosome,
which is designated as O.
2011
Q112
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Mutations can be induced with :
A.
Gamma radiation
B.
Infra Red radiations
C.
IAA
D.
Ethylene
Correct Answer: A
Explanation:
Mutation can be induced by gamma
radiation.
2011
Q113
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
When two unrelated individuals or lines are
crossed, the per romance of F1 hybrid is often
superior to both parents. This phenomenon is
called :
A.
Sphcing
B.
Transformation
C.
Heterosis
D.
Metamorphosis
Correct Answer: C
Explanation:
The increased vigour displayed by the
offspring from a cross between genetically different
parents is called heterosis. Hybrids from crosses
between different crop varieties (F1
hybrids) are often
stronger and produce better yields than the original
varieties.
2010
Q114
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Study the pedigree chart of a certain family
given below and select the correct conculusion
which can be drawn for the character –
A.
The trait under study could not be colourblindness
B.
The female parent is heterozygous
C.
The parents could not have had a normal
daughter for this character
D.
The male parent is homozygous dominant
Correct Answer: B
Explanation:
The given pedigree chart shows that both the
daughters received the gene from the parents,
while son may be normal or affected. It shows
that the female parent is heterozygous.
2010
Q115
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
ABO blood grouping is controlled by gene I
which has three alleles and show co-dominance.
There are six genotypes. How many phenotypes
in all are possible -
A.
five
B.
four
C.
three
D.
six
Correct Answer: B
Explanation:
The three alleles IA, IB and i of gene I in
ABO blood group system can produce six different
genotypes and four different phenotypes as shown
below :
2010
Q116
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A cross in which an organism showing a
dominant phenotype in crossed with the
recessive parent in order to know its genotype is
called -
A.
Dihybrid cross
B.
Back cross
C.
Monohybrid cross
D.
Test cross
Correct Answer: D
Explanation:
Test cross is performed to determine the
genotype of F2
plant. In a typical test cross an organism
showing dominant phenotype and whose genotype is
to be determined is crossed with one that is homozygous
recessive for the allele being investigated, instead of
self-crossing. The progenies of such a cross can easily
be analysed to predict the genotype of the test organism.
2010
Q117
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
In antirrhinum two plants with pink flowers were
hybridized. The F1 plants produced red, pink and
white flowers in the proportion of 1 red, 2 pink
and 1 white. What could be the genotype of the
two plants used for hybridization ? Red flower
colour is determined by RR, and White by rr
genes -
A.
rrrr
B.
rr
C.
Rr
D.
RR
Correct Answer: C
Explanation:
The given situation is an example of
incomplete dominance where phenotype found in
F1
generation do not resemble either of the two
parents. The genotype of the two plants used for
cross will be
The incomplete dominance of dominant allele (here
‘R’) over recessive allele (here ‘r’) could be due to
mutations (insertion, deletion, substitution or inversion
of nucleotides). The mutant allele generally produces
a faulty or no product. This modification in the product
may lead to incomplete dominance of the (unmodified)
wild type dominant allele.
2010
Q118
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
ABO blood groups in humans are controlled by
the gene I. It has three alleles – IA, IB
and i.
Since there are three different alleles, six
different genotypes are possible. How many
phenotypes can occur –
A.
One
B.
Two
C.
Three
D.
Four
Correct Answer: D
Explanation:
The three alleles in ABO blood groups in
humans can produce six different genotypes
and four different phenotypes.
2010
Q119
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following cannot be explained
on the basis of Mendel's Law of Dominance?
A.
Out of one pair of factors one is dominant,
and the other recessive
B.
Alleles do not show any blending and both
the characters recover as such in F2
generation
C.
The discrete unit controlling a particular
character is called a factor
D.
Factors occur in pairs
Correct Answer: B
Explanation:
According to Mendel’s law of Dominance,
out of two contrasting allelomorphic factors
only one expresses itself in an individual. The
factor that expresses itself is called dominant
while the other which has not shown its effect
in the heterozygous individual is termed as
recessive. The option (c) in the given question
cannot be explained on the basis of law of
dominance. It can only be explained on
the basis of Mendel’s Law of independent
assortment, according to which in a dihybrid
cross, the two alleles of each character assort
independently of the alleles of other character
and separate at the time of gamete formation.
2010
Q120
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
The genotype of a plant showing the dominant
phenotype can be determined
by –
A.
Pedigree analysis
B.
Back cross
C.
Test cross
D.
Dihybrid cross
Correct Answer: C
Explanation:
Test cross is the cross of an individual
with an individual having recessive phenotype.
It is used to determine the genotype of a plant
showing the dominant phenotype, that means
to determine whether the individual exhibiting
dominating characters are homozygous or
heterozygous.
2010
Q121
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Select the correct statement from the ones given
below with respect to dihybrid cross –
A.
Genes far apart on the same chromosome
show very few recombinations
B.
Tightly linked genes on the same
chromosome show very few recombination
C.
Tightly linked genes on the same
chromosome show higher recombinations
D.
Genes loosely linked on the same
chromosome show similar recombinations
as the tightly linked ones
Correct Answer: B
Explanation:
Linkage is the phenomenon of certain genes
staying together during inheritance through
generations without any change or separation due to
their being present on the same chromosome. Linked
genes occur in the same chromosome. Strength of the
linkage between two genes is inversely proportional
to the distance between the two i.e., two linked genes
show higher frequency of crossing over
(recombination) if the distance between them is higher
and lower frequency if the distance is small.
2010
Q122
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following symbols and its
representation, used in human pedigree
analysis is correct –
A.
B.
C.
D.
Correct Answer: A
Explanation:
A record of inheritance of certain genetic
traits for two or more generations presented in the form
of a diagram or family tree is called pedigree. In a
pedigree a square represents the male, a circle the
female, solid (blackened) symbol shows the trait under
study or affected individual; unaffected or normal
individual by an open or clear symbol and a cross or
shade (of any type) in the symbol signifies the carrier
of a recessive allele. Words can also be used in place
of symbols. Parents are shown by horizontal line while
their offsprings are connected to it by a vertical line.
The offsprings are then shown in the form of a
horizontal line below the parents and numbered with
arabic numerals.
2009
Q123
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Select the incorrect statement from the
following :
A.
Baldness is a sex limited trait
B.
Galactosemia is an inborn error of
metabolism
C.
Linkage is an exception to the principle of
independent assortment in heredity.
D.
Small population size results in random
genetic drift in a population
Correct Answer: A
Explanation:
Baldness is a sex influenced trait.
The dominance of alleles may differ in
heterozygotes of the two sexes.
2009
Q124
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
The most popularly known blood grouping is
the ABO grouping. It is named ABO and not
ABC, because "O" in it refers to having :
A.
Other antigens besides A and B on RBCs
B.
No antigens A and B on RBCs
C.
Overdominance of this type on the genes for
A and B types
D.
One antibody only-either anti-A and anti-B
on the RBCs
Correct Answer: B
Explanation:
In ABO blood group O refers to O blood
group. It has no antigen (A and B) on RBCs.
2009
Q125
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Study the pedigree chart given below :
What does it show ?
A.
Inheritance of a recessive sex-linked disease
like haemophilia
B.
Inheritance of a sex-linked inborn error of
metabolism like phenylketonuria
C.
The pedigree chart is wrong as this is not
possible
D.
Inheritance of a condition like
phenylketonuria as an autosomal recessive
trait
Correct Answer: D
Explanation:
The chart shows the inheritance of
a condition like phenylketonuria as an
autosomal recessive trait. Parents’ needs to
be heterozygous as two of their children are
known to be sufferer of the disease. It cannot
be recessive sex linked inheritance because
then the male parent would also be sufferer.
2009
Q126
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
The genetic defect-adenosine deaminase (ADA)
deficiency may be cured permanently by
A.
Introducing bone marrow cells producing
ADA into cells at early embryonic stages
B.
Periodic infusion of genetically engineered
lymphocytes having functional ADA cDNA
C.
Administering adenosine deaminase
activators
D.
Enzyme replacement therapy
Correct Answer: A
Explanation:
ADA deficiency can be permanently cured
if the isolated gene from bone marrow cells producing
ADA is introduced into cells at early embryonic stages.
2009
Q127
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Sickle-cell anemia is :
A.
Caused by substitute of valine by glutamic
acid in the beta globin chain of haemoglobin
B.
An autosomal linked dominant trait
C.
Caused by a change in a single base pair of
DNA
D.
Characterized by elongated sickle like RBCs
with a nucleus
Correct Answer: C
Explanation:
Sickle-cell anaemia is an autosomal hereditary
disorder in which erythrocytes become sickle shaped.
It is caused by the formation of abnormal haemoglobin
called haemoglobin-S. Haemoglobin-S is formed when
6th amino acid of $\beta $-chain, i.e., glutamic acid is replaced by valine due to substitution. It occurs due to
a single nucleotide change (A $ \to $ T) in the $\beta $-globin
gene of coding strand. In the normal $\beta $-globin gene
the DNA sequence is CCTGAGGAG, while in sicklecell anaemia, the sequence is CCTGTGGAG.
2008
Q128
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following conditions in
humans is correctly matched with its
chromosomal abnormality/linkage?
A.
Klinfelter's syndrome — 44 autosomes + XXY
B.
Colour-blindness — Y-linked
C.
Down syndrome — 44 autosomes + XO
D.
Erythroblastosis foetalis — X-linked
Correct Answer: A
Explanation:
Klinefelter's syndrome is a genetic
disorder affecting men in which an individual
gains an extra X chromosome, so that the
usual Karyotype of XY is replaced by one of
XXY. Symptoms of Klinefelter's syndrome
named after us physician H.P. Klinefelter,
include female characteristics (such as breast
enlargement).
So, the corresponding genotype will be AaBb.
2008
Q129
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Haploids are more suitable for mutation studies
than the diploids. This is because:
A.
all mutations, whether dominant or recessive
are expressed in haploids
B.
mutagens penetrate in haploids more
effectively than in diploids
C.
haploids are reproductively more stable than
diploids
D.
haploids are more abundant in nature than
diploids
Correct Answer: A
Explanation:
Haploid plants, are always pure because they
possess only one set of chromosomes. So, the
mutations are expressed very easily in haploid plants
as compared to diploid plants.
2007
Q130
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A common test to find the genotype of a hybrid
is by:-
A.
Crossing of one F1 progeny with male parent
B.
Crossing of one F2 progeny with female
parent
C.
Crossing of one F2 progeny with male parent
D.
Studying the sexual behaviour of F1
progenies
Correct Answer: A
Explanation:
crossing of one F1 progeny with male
parent. e.g
To find the genotype of hybrid test cross is
done.
2007
Q131
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A human male produces sperms with the
genotypes AB, Ab, AB, and ab pertaining to two
diallelic characters in equal proportions. What is
the corresponding genotype of this person ?
A.
AaBb
B.
AABB
C.
AaBB
D.
AABb
Correct Answer: A
Explanation:
As sperms produced are with genotypes AB,
Ab, aB, ab (two diallelic character) the person must
be heterozygous for both genes. So his genotype will
be AaBb.
2007
Q132
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
In the hexaploid wheat, the haploid(n) and
basic(x) numbers of chromosomes are :
A.
n = 21 and x = 14
B.
n = 21 and x = 21
C.
n = 7 and x = 21
D.
n = 21 and x = 7
Correct Answer: D
Explanation:
Hexaploid wheat has six sets of homologous chromosomes. In botanical nomenclature, the term "haploid" (n) refers to the number of unique chromosomes in a gamete, which in the case of hexaploid wheat is 21. The "basic" number (x) is the number of chromosomes in a single set within the organism, which in hexaploid wheat is 7, as there are six such sets (6x = 42 total chromosomes in a somatic cell).
Therefore, the correct answer is :
Option D : n = 21 and x = 7.
2007
Q133
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Inheritance of skin colour in humans is an
example of:-
A.
Chromosomal aberration
B.
Polygenic inheritance
C.
Codominance
D.
Point mutation
Correct Answer: B
Explanation:
Inheritance of skin colour in human is
controlled by three genes, A B and C which
is polygenic inheritance.
2007
Q134
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
In pea plants, yellow seeds are dominant to
green. If a heterozygous yellow seeded plant is
crossed with a green seeded plant, what ratio of
yellow and green seeded plants would you
expect in F1 generation ?
A.
9 : 1
B.
3 : 1
C.
50 : 50
D.
1 : 3
Correct Answer: C
Explanation:
Let GG ⇒ homozygous yellow seed
plant.
Gg ⇒ heterozygous green
gg ⇒ homozygous green
According to the question :
So, the ratio will be 50 : 50
2006
Q135
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Sickle cell anaemia has not been eliminated from
the African population because-
A.
It is not a fatal disease
B.
It is controlled by dominant genes
C.
It is controlled by recessive genes
D.
It provides immunity against malaria
Correct Answer: D
Explanation:
In sickle cell anaemia RBCs become
sickle shaped which are not supportive for the
growth of malarial parasite Plasmodium so it
provides immunity against malaria disease.
2006
Q136
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Test cross involves-
A.
Crossing between two F1 hybrids
B.
Crossing the F1 hybrid with a double
recessive genotype
C.
Crossing between two genotypes with
recessive trait
D.
Crossing between two genotypes with
dominant trait
Correct Answer: B
Explanation:
A cross of F1 hybrid with its recessive
homozygous parent is called the test cross. It
is done to determine the genotype of a given
plant. If the given plant has homozygous
dominant traits then on test cross it gives all
dominant trait plants but if it is heterozygous dominant than it gives dominant and recessive
phenotypes in 1 : 1 ratio.
2006
Q137
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
In Mendel's experiments with garden pea, round
seed shape (RR) was dominant over wrinkled
seeds (rr), yellow cotyledon (YY) was dominant
over green cotyledon (yy). What are the
expected phenotypes in the F2 generation of the
cross RRYY × rryy ?
A.
Round seeds with yellow cotyledons, and
wrinkled seeds with yellow cotyledons
B.
Only wrinkled seeds with green cotyledons
C.
Only wrinkled seeds with yellow cotyledons
D.
Only round seeds with green cotyledons
Correct Answer: A
Explanation:
Since round seed shape is dominant over
wrinkled seed shape and yellow cotyledon is dominant
over green cotyledon so RRYY individuals is round
yellow and rryy is wrinkled green.
2006
Q138
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
How many different kinds of gametes will be
produced by a plant having the genotype
AABbCC ?
A.
Two
B.
Nine
C.
Four
D.
Three
Correct Answer: A
Explanation:
It would make only two types of gametes,
these are ABC & AbC.
2006
Q139
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Phenotype of an organism is the result of-
A.
Environmental changes and sexual dimorphism
B.
Cytoplasmic effects and nutrition
C.
Genotype and environment interactions
D.
Mutations and linkages
Correct Answer: C
Explanation:
Phenotype is the appearance of one
organism while genotype is the gene
complement it has from its ancestors. These
genes only show their effect in phenotype but
environment also plays an important role in
this. Hence phenotype is a result of genotype
and environmental interaction.
2006
Q140
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
If a colour blind woman marries a normal visioned
man, their sons will be
A.
All normal visioned
B.
Three-fourths colourblind and one-fourth normal
C.
One-half colourblind and one-half normal
D.
All colourblind
Correct Answer: D
Explanation:
Colour blindness in a X-chromosome
linked character. So they’ll be having all
colour blind sons and carrier daughters.
2006
Q141
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which one of the following is an example of
polygenic inheritance ?
A.
Skin colour in humans
B.
Pod shape in garden pea
C.
Production of male honey bee
D.
Flower colour in Mirabilis jalapa
Correct Answer: A
Explanation:
Polygenic inheritance is the inheritance
of traits which are dependent on the no.
of genes such as the skin colour of human
beings. eg. AABB is black AaBB in neither
dark nor black. AaBb is again wheatish Aabb
is light and aabb is white colour.
2006
Q142
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Both sickle cell anemia and Huntington's chorea
are-
A.
Congenital disorders
B.
Bacteria-related diseases
C.
Virus-related diseases
D.
Pollutant-induced disorders
Correct Answer: A
Explanation:
A congenital disorder is a medical condition
that is present at birth. Congenital disorders can be a
result of genetic abnormalities, the intrauterine
environment, or unknown factors. Sickle cell disease
[a group of genetic disorders caused by sickle
haemoglobin (HbS). HbS
molecules tend to clump
together, making red blood cells sticky, stiff and more
fragile and causing them to form into a curved, sickle
shape] and Huntington’s chorea (an inherited disorder
characterised by degenerative changes in the basal
ganglia structures, which ultimately result in a severely
shrunken brain and enlarged ventricles, abnormal body
movements called chorea and loss of memory) are
congenital disorders.
2006
Q143
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Cri-du-chat syndrome in humans is caused by
the-
A.
Loss of half of the long arm of chromosome 5
B.
Loss of half of the short arm of chromosome 5
C.
Trisomy of 21st chromosome
D.
Fertilization of an XX egg by a normal
Y-bearing sperm
Correct Answer: B
Explanation:
Cri-du-chat syndrome, also called deletion
5p syndrome, (or 5p minus), is a rare genetic disorder.
Cri-du-chat syndrome is due to a partial deletion of
the short arm of chromosome number 5. The name of
this syndrome is French for “cry of the cat,” referring
to the distinctive cry of children with this disorder.
The cry is caused by abnormal larynx development,
which becomes normal within a few weeks of birth.
Infants with cri-du-chat have low birth weight and may
have respiratory problems. Some people with this
disorder have a shortened lifespan, but most have a
normal life expectancy.
2005
Q144
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
G-6-P dehydrogenase deficiency is associated
with haemolysis of -
A.
RBCs
B.
Lymphocytes
C.
Platelets
D.
Leucocytes
Correct Answer: A
Explanation:
Glucose-6-P dehydrogenase is the
first enzyme of glucose oxidation during
Pentose Phosphate Pathway. RBC contain
haemoglobin which combines with oxygen
to form oxyhaemoglobin which gives its
oxygen for oxidation of food. In haemolysis
there is destruction of RBCs with release
of haemoglobin into plasma resulting in
jaundice. So, now RBCs cannot provide
oxygen for oxidation of food thereby causing
deficiency of G-6-P dehydrogenase.
2005
Q145
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Which of the following is not a hereditary
disease ?
A.
Cretinism
B.
Cystic fibrosis
C.
Haemophilia
D.
Thalasasemia
Correct Answer: A
Explanation:
Cretinism occurs due to hyposecretion of
thyroid hormones. Haemophilia is a sex linked
recessive trait. Cystic fibrosis is also a recessive
autosomal disorder resulting in mucus clogging in
lungs. Thalassemia involves a gene mutation in the
polypeptide chains of haemoglobin.
2005
Q146
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A man and a woman, who do not show any
apparent signs of a certain inherited disease,
have seven children (2 daughter and 5 sons).
Three of the sons suffer from the given disease
but none of the daughters are affected. Which of
the following mode of inheritance do you
suggest for this disease
A.
Autosomal dominant
B.
Sex-linked recessive
C.
Sex-linked dominant
D.
Sex-limited recessive
Correct Answer: B
Explanation:
The daughters receive one X chromosome
from the father and one X chromosome from
the mother. Since all the daughters suffer from
their father’s disease, the X chromosome
from the father must be carrying a dominant
trait.
2005
Q147
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A women with 47 chromosomes due to three
copies of chromosome 21 is characterized by -
A.
Superfemaleness
B.
Turner's syndrome
C.
Triploidy
D.
Down's syndrome
Correct Answer: D
Explanation:
A woman with 47 chromosomes due to three copies of chromosome 21 is characterized by Down's syndrome, which is Option D.
Here's a brief explanation of each option :
Option A, Superfemaleness: This term is sometimes colloquially used for a condition known as Triple X syndrome, where a female has an extra X chromosome (47,XXX instead of the typical 46,XX). It's not associated with three copies of chromosome 21.
Option B, Turner's syndrome: This is a condition where a female partially or completely lacks one of the two X chromosomes (usually represented as 45,X instead of 46,XX). It's unrelated to having an extra copy of chromosome 21.
Option C, Triploidy: This is a rare chromosomal disorder where individuals have three copies of every chromosome (69 chromosomes in total), not just an extra copy of chromosome 21.
Option D, Down's syndrome: This is the correct answer. Down syndrome, also known as trisomy 21, is a genetic disorder caused by the presence of all or part of a third copy of chromosome 21. It's the most common chromosome abnormality in humans.
2005
Q148
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A women with 47 chromosomes due to three
copies of chromosome 21 is characterized by
A.
triploidy
B.
Down’s syndrome
C.
superfemaleness
D.
Turner’s syndrome
Correct Answer: B
Explanation:
Down's syndrome is caused by the
presence of an extra chromosome number
21 and the off spring has 47 chromosomes.
2005
Q149
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
A woman with normal vision, but whose father
was colour bind, marries a colour blind man.
Suppose that the fourth child of this couple was
a boy. This boy -
A.
May be colour blind or may be normal
vision
B.
Must have normal colour vision
C.
Must be colour blind
D.
Will be partially colour blind since he is
heterozygous for the colour blind mutant
allele.
Correct Answer: A
Explanation:
Since the woman’s father was colour
blind. She would be a carrier of the colour
blindness gene. When she marries a colour
blind man. Their progeny could be
2005
Q150
NEET
MCQ
iCON Education HYD, 79930 92826, 73309 7282610 Mar 2026
Haemophilia is more commonly seen in human
males than in human females because -
A.
This disease is due to an X-linked dominant
mutation
B.
This disease is due to a Y-linked recessive
mutation
C.
This disease is due to an X-linked recessive
mutation
D.
A greater proportion of girls die in infancy
Correct Answer: C
Explanation:
This disease is due to an X-linked
recessive mutation. Males suffer this disorder
since they have only one X chromosome and
hence express any trait on this chromosome.
The incomplete dominance of dominant allele (here
‘R’) over recessive allele (here ‘r’) could be due to
mutations (insertion, deletion, substitution or inversion
of nucleotides). The mutant allele generally produces
a faulty or no product. This modification in the product
may lead to incomplete dominance of the (unmodified)
wild type dominant allele.
What does it show ?
To find the genotype of hybrid test cross is
done.
So, the ratio will be 50 : 50
