Sex determination Questions in English

(C) In humans,the diploid number of chromosomes is $46$ ($23$ pairs).
Out of these,$22$ pairs are known as autosomes,which are responsible for determining somatic (body) characteristics in both sexes.
The $23^{rd}$ pair is known as heterosomes or allosomes (sex chromosomes),which determine the sex of the individual.
Therefore,the correct complement is $22$ pairs of autosomes and $1$ pair of heterosomes.

(D) In humans,sex determination is based on the sex chromosomes.
Females have a pair of $XX$ chromosomes,while males have $XY$ chromosomes.
During fertilization,the mother always contributes an $X$ chromosome.
If the father contributes an $X$ chromosome,the zygote becomes $XX$ (female).
If the father contributes a $Y$ chromosome,the zygote becomes $XY$ (male).
Therefore,a male child is born when the genetic chromosomal constitution of the child is $XY$.

(A) In humans,the male sex chromosome constitution is $XY$.
During fertilization,the father contributes either an $X$ or a $Y$ chromosome to the offspring.
If the father contributes a $Y$ chromosome,the offspring is male.
The $X$ chromosome in a male $(XY)$ is always inherited from the mother,as the father contributes the $Y$ chromosome to determine the male sex.

(B) In grasshoppers,the sex determination mechanism is of the $XO$ type. In this type,males possess only one $X$ chromosome in addition to the autosomes,while females possess two $X$ chromosomes. Therefore,the males of grasshoppers possess two sets of autosomes and only one $X$ chromosome.

(D) normal human cell contains $46$ chromosomes,which consist of $23$ pairs.
These $46$ chromosomes are divided into $44$ autosomes ($22$ pairs) and $2$ sex chromosomes ($1$ pair).
In a normal human female,the sex chromosomes are $XX$.
Therefore,the number of autosomes is $44$.

(A) Humans have a total of $46$ chromosomes, which are organized into $23$ pairs.
Out of these $23$ pairs, $22$ pairs are known as autosomes, which are identical in both males and females.
The remaining $1$ pair consists of sex chromosomes ($XX$ in females and $XY$ in males).
Therefore, the number of autosomes in a human is $22$ pairs.

(A) human somatic cell or a fertilized egg (zygote) contains a total of $46$ chromosomes.
These $46$ chromosomes consist of $23$ pairs.
Out of these,$22$ pairs ($44$ chromosomes) are autosomes,which are responsible for determining somatic characteristics.
The remaining $1$ pair ($2$ chromosomes) consists of sex chromosomes ($XX$ in females or $XY$ in males).
Therefore,the total number of autosomes in a human zygote is $44$.

(C) The correct option is $(C)$.
In humans,the total number of chromosomes is $46$ ($23$ pairs).
Out of these,$44$ chromosomes are autosomes ($22$ pairs),which are identical in both males and females.
The remaining $2$ chromosomes are sex chromosomes.
In human males,the sex chromosomes are $XY$,resulting in a total chromosomal constitution of $44 AA + XY$.

(B) In humans,there are $23$ pairs of chromosomes. The first $22$ pairs are called autosomes,which are identical in both males and females. The $23$rd pair is known as the sex chromosome or heterosome because it determines the sex of the individual. In males,it consists of $XY$ chromosomes,and in females,it consists of $XX$ chromosomes. Therefore,the $23$rd pair is referred to as a heterosome.

(C) Chromosomes are classified into two types based on their function: $1.$ Autosomes and $2.$ Sex chromosomes.
Autosomes are the chromosomes that determine the somatic (body) characters of an organism.
Sex chromosomes are the chromosomes that determine the sex of an individual.
Therefore,the correct answer is Autosomes.

(C) Sex-linked traits are genes located on the sex chromosomes. In humans,the majority of sex-linked genes are located on the $X$-chromosome because it is much larger and contains more genes than the $Y$-chromosome. Since males have only one $X$-chromosome (hemizygous),any recessive allele present on their $X$-chromosome will express its phenotype,as there is no corresponding allele on the $Y$-chromosome to mask it.

(A) In humans,males are hemizygous for the $X$-chromosome because they possess only one $X$-chromosome $(XY)$.
If a recessive gene is present on the $X$-chromosome of a male,there is no corresponding allele on the $Y$-chromosome to mask its effect.
Therefore,the recessive trait will always be expressed in males.
In contrast,females have two $X$-chromosomes $(XX)$,so a recessive gene on one $X$-chromosome is usually masked by the dominant allele on the other $X$-chromosome,unless the female is homozygous recessive.

(D) The $Y$ chromosome was discovered by $Netti \ Stevens$ in $1905$ while studying the mealworm $Tenebrio \ molitor$. She observed that the sex of the organism was determined by the presence or absence of specific chromosomes, which she identified as $X$ and $Y$ chromosomes.

(B) Holandric genes are those genes that are located on the non-homologous region of the $Y$ chromosome.
These genes are inherited directly from father to son.
Since they are present only on the $Y$ chromosome,they are expressed only in males and are never found in females.
Therefore,the correct option is $B$.

(C) Allosomes are the chromosomes that differ in size,form,and behavior from other chromosomes in an organism. They are responsible for the determination of sex in individuals and are commonly known as sex chromosomes or idiochromosomes. In humans,the $X$ and $Y$ chromosomes are examples of allosomes.

(C) The genes present on the differential (non-homologous) region of the $Y$ chromosome are exclusively inherited from father to son.
These genes are known as Holandric genes.
Since they are located only on the $Y$ chromosome,they do not have corresponding alleles on the $X$ chromosome and are expressed in all males who inherit the $Y$ chromosome.

(C) In $Honey \text{ } bees$, the process of development of an embryo from an unfertilized egg is known as $Parthenogenesis$.
Specifically, in the $Honey \text{ } bee$ colony, the unfertilized eggs develop into haploid males, which are known as $Drones$.
This is a natural form of asexual reproduction where fertilization does not occur to produce the male offspring.

(A) In honey bees, the queen lays both fertilized and unfertilized eggs.
Fertilized eggs develop into female bees $(queens$ or $workers)$, depending on the nutrition provided.
Unfertilized eggs develop into male bees, which are known as drones, through a process called arrhenotoky (a type of parthenogenesis).

(D) In honeybees,the sex determination is haplodiploid.
Queens and workers are diploid $(2n)$,while drones are haploid $(n)$.
Since drones are already haploid,they cannot undergo meiosis (reduction division) to produce sperm.
Instead,they produce sperm through mitosis.
This occurs because drones develop from unfertilized eggs through a process called parthenogenesis.

(B) In human populations,the sex ratio at birth is approximately $1:1$. This is due to the mechanism of sex determination,where the male produces two types of gametes ($X$ and $Y$ chromosomes) in equal proportions. During fertilization,there is an equal probability of a sperm carrying either an $X$ or a $Y$ chromosome fusing with the egg ($X$ chromosome). Therefore,statistically,the ratio of male to female offspring is $1:1$.

(A) The $Barr$ body is a condensed,inactive $X$ chromosome found in the nuclei of somatic cells of female mammals.
It is formed due to the process of $X$-inactivation (lyonization),where one of the two $X$ chromosomes in females is randomly inactivated to ensure dosage compensation.
Since males have only one $X$ chromosome $(XY)$,they do not typically possess a $Barr$ body.
Therefore,the $Barr$ body is specifically associated with the female sex chromosome complement.

(B) In grasshoppers,sex determination follows the $XO$ type mechanism.
In this system,the male grasshopper has only one $X$ chromosome and lacks a corresponding partner chromosome,which is represented as $XO$.
Therefore,the chromosomal constitution of a male grasshopper is $2A + XO$,where $2A$ represents the autosomes and $XO$ represents the sex chromosomes.
Females,on the other hand,possess two $X$ chromosomes $(2A + XX)$.

(D) In humans,sex determination is of the $XY$ type. The $Y$-chromosome carries a specific gene known as the $SRY$ (Sex-determining Region $Y$) gene. This gene acts as a master regulatory switch that triggers the development of testes in the embryo. In the absence of the $Y$-chromosome (and thus the $SRY$ gene),the embryo develops into a female. Therefore,the master regulatory gene is located on the $Y$-chromosome.

(D) The diploid chromosome number $(2n)$ of $Drosophila \text{ melanogaster}$ is $8$.
This means there are $4$ pairs of chromosomes in total.
Out of these $4$ pairs, $3$ pairs are autosomes (non-sex chromosomes) and $1$ pair consists of sex chromosomes ($XX$ in females or $XY$ in males).
Therefore, the correct composition is $3$ pairs of autosomes and $1$ pair of sex chromosomes.

(A) In birds,sex determination follows the $ZZ - ZW$ type of mechanism.
In this system,the male is homogametic,possessing two similar sex chromosomes denoted as $ZZ$.
The female is heterogametic,possessing two dissimilar sex chromosomes denoted as $ZW$.
This is distinct from the $XX - XY$ system found in humans and many other organisms,where the male is heterogametic.

(A) Sex-linked traits are those traits whose genes are located on the sex chromosomes. In humans,the $X$-chromosome is significantly larger and contains a much higher number of genes compared to the $Y$-chromosome. Consequently,the vast majority of sex-linked traits (such as color blindness and hemophilia) are $X$-linked. While $Y$-linked traits (holandric traits) do exist,they are far less common than $X$-linked traits. Therefore,$X$-chromosomes are the primary carriers of sex-linked traits.

(A) In grasshoppers,sex determination is of the $XO$ type.
In this mechanism,males have only one $X$ chromosome along with autosomes $(A + XO)$,while females have a pair of $X$ chromosomes $(A + XX)$.
During spermatogenesis,males produce two types of gametes: $50\%$ with the $X$ chromosome and $50\%$ without it.
Fertilization of an egg by a sperm carrying an $X$ chromosome results in a female,while fertilization by a sperm lacking the $X$ chromosome results in a male.

(A) Hypertrichosis (excessive hair growth on the pinna of the ear) is a classic example of a holandric trait.
Holandric traits are determined by genes located on the non-homologous region of the $Y$ chromosome.
Since these genes are present only on the $Y$ chromosome,they are transmitted directly from father to son.
Therefore,hypertrichosis is inherited exclusively through the male line.

(C) In $Drosophila$,sex determination is based on the genic balance theory proposed by $C.B. Bridges$.
According to this theory,the sex of the individual is determined by the ratio of the number of $X$-chromosomes $(X)$ to the number of sets of autosomes $(A)$.
This ratio is denoted as $X/A$ ratio.
If the $X/A$ ratio is $1.0$,the individual is a female.
If the $X/A$ ratio is $0.5$,the individual is a male.
Therefore,the sex is determined by the ratio of the number of $X$-chromosomes to the sets of autosomes.

(A) $X$-linked recessive traits are genes located on the $X$ chromosome that express their phenotype only when the dominant allele is absent.
In males,who have only one $X$ chromosome $(XY)$,a single copy of the recessive gene is sufficient to express the trait because there is no corresponding allele on the $Y$ chromosome to mask it.
Therefore,$X$-linked recessive genes are always expressed in males if they inherit the affected $X$ chromosome from their mother.

(D) Genes located on the non-homologous region of the $Y$-chromosome are passed directly from father to son. These genes are known as $Y$-linked or holandric genes. Since the $Y$-chromosome is present only in males,these traits are expressed exclusively in males and are inherited from father to son.

(D) gynandromorph is an organism that contains both male and female tissues.
This condition arises due to an error during early embryonic development,such as the loss of an $X$ chromosome in one of the daughter cells during mitosis.
As a result,one part of the body develops with male characteristics (e.g.,$XY$ or $XO$) and another part develops with female characteristics (e.g.,$XX$).
Therefore,it possesses different sexes in different cells of the body.

(A) In $Drosophila$,sex is determined by the ratio of $X$ chromosomes to the number of sets of autosomes ($X/A$ ratio). The presence of the $Y$ chromosome does not determine maleness in $Drosophila$; it is only required for male fertility. Thus,$XXY$ is a female.
In humans,sex is determined by the presence of the $SRY$ gene located on the $Y$ chromosome. The presence of even a single $Y$ chromosome triggers the development of testes,leading to a male phenotype. Therefore,$XXY$ results in a male with $Klinefelter's$ syndrome. This confirms that the $Y$ chromosome is essential for sex determination in humans.

(D) In humans,sex determination is based on the $XY$ type of sex chromosomes.
Females are homogametic,producing gametes with $22 + X$ chromosomes.
Males are heterogametic,producing two types of sperms: $50\%$ with $22 + X$ and $50\%$ with $22 + Y$ chromosomes.
If a sperm carrying the $X$ chromosome fertilizes the ovum,the zygote develops into a female $(XX)$.
If a sperm carrying the $Y$ chromosome fertilizes the ovum,the zygote develops into a male $(XY)$.
Therefore,the sex of the offspring is determined by the sex chromosome contributed by the father.

(B) In the marine worm $Bonellia$ $viridis$,sex determination is environmentally controlled.
$1$. Larvae that settle on the proboscis of an adult female are exposed to secretions that cause them to develop into small,parasitic males.
$2$. Larvae that do not encounter a female and develop freely in the open water grow into much larger,independent females.
Therefore,if the larva develops freely in the water,it develops into a female.

(B) In humans,the sex determination mechanism is of the $XX-XY$ type.
- The female is homogametic,producing ova with $22 + X$ chromosomes.
- The male is heterogametic,producing sperms with either $22 + X$ or $22 + Y$ chromosomes.
- When a sperm carrying an $X$ chromosome fertilizes an ovum $(X)$,the resulting zygote has $XX$ chromosomes,which develops into a female child.
- When a sperm carrying a $Y$ chromosome fertilizes an ovum $(X)$,the resulting zygote has $XY$ chromosomes,which develops into a male child.
- Therefore,the combination of two $X$ chromosomes (one from the mother and one from the father) results in a female child.

(B) The Genic Balance Theory of sex determination was proposed by $C.B. Bridges$ in $1921$ based on his studies on $Drosophila$ $melanogaster$.
According to this theory,the sex of an individual is determined by the ratio of the number of $X$-chromosomes to the number of sets of autosomes $(A)$.
This ratio is expressed as $Ratio = X/A$.
If the ratio is $1.0$,the individual is a female; if the ratio is $0.5$,the individual is a male; and if the ratio is between $0.5$ and $1.0$,it results in an intersex individual.

(D) Holandric inheritance refers to the transmission of genes located on the $Y$ chromosome. Since only males possess the $Y$ chromosome,these traits are passed directly from father to son. Therefore,the inheritance is transmitted exclusively through the male line.

(B) $1$. In birds,sex determination is of the $ZW-ZZ$ type,where $ZZ$ represents the male and $ZW$ represents the female. Thus,option $A$ is incorrect.
$2$. In grasshoppers,sex determination is of the $XO$ type. Males have only one $X$ chromosome $(XO)$,while females have two $(XX)$. Therefore,the presence of a single $X$ chromosome determines the male sex. Thus,option $B$ is correct.
$3$. In humans,the $XO$ condition (Turner syndrome) results in a female phenotype,but these individuals are sterile. However,the standard mechanism for female sex determination in humans is the presence of $XX$ chromosomes. Option $C$ is partially correct in description but $B$ is the standard textbook example for $XO$ sex determination.
$4$. In Drosophila,sex is determined by the ratio of $X$ chromosomes to autosomes. $XX$ individuals are female,not male. Thus,option $D$ is incorrect.

(C) According to Bridges' genic balance theory of sex determination in Drosophila,the sex of an individual is determined by the ratio of the number of $X$ chromosomes to the number of sets of autosomes $(A)$.
- If $X/A = 1.0$,the organism is a normal female.
- If $X/A = 0.5$,the organism is a normal male.
- If $X/A > 1.0$ (e.g.,$1.5$),the organism is a metafemale or super female.
- If $0.5 < X/A < 1.0$,the organism is an intersex.
- If $X/A < 0.5$,the organism is a metamale.
Since the given ratio is $1.5$,which is greater than $1.0$,the organism is a super female.

(A) The sex determination mechanism in the plant $Melandrium$ $album$ (also known as $Silene$ $latifolia$) was discovered by $H. E. Warmke$ in $1946$.
He demonstrated that the $Y$ chromosome in this species carries male-determining factors,making it a classic model for studying sex determination in dioecious plants.

(D) In males,there is only one $X$-chromosome (hemizygous condition). Therefore,any recessive trait present on the $X$-chromosome will express its effect because there is no corresponding allele on the $Y$-chromosome to mask it. This is why $X$-linked recessive disorders are more common in males.

(D) In humans,males are hemizygous for the $X$-chromosome because they possess only one $X$-chromosome $(XY)$.
If a recessive gene is present on the $X$-chromosome of a male,there is no corresponding allele on the $Y$-chromosome to mask its effect.
Therefore,the recessive trait will always be expressed in males.
In contrast,females have two $X$-chromosomes $(XX)$,so a recessive gene on one $X$-chromosome is usually masked by the dominant allele on the other $X$-chromosome,unless the female is homozygous recessive.

(C) In Drosophila,sex determination is based on the ratio of the number of $X$ chromosomes to the number of sets of autosomes ($X/A$ ratio).
$1$. If the $X/A$ ratio is $1.0$,the individual is a female.
$2$. If the $X/A$ ratio is $0.5$,the individual is a male.
$3$. If the $X/A$ ratio is between $0.5$ and $1.0$ (e.g.,$0.67$),the individual is an intersex.
For option $C$ $(3A + XXY)$: The number of $X$ chromosomes is $2$ and the number of sets of autosomes is $3$. The ratio is $2/3 \approx 0.67$. Since this value lies between $0.5$ and $1.0$,it represents an intersex individual.

(D) In grasshoppers,sex determination follows the $XO$ type.
In this mechanism,the male has only one $X$ chromosome and lacks a $Y$ chromosome,making it monosomic $(XO)$.
The female has two $X$ chromosomes $(XX)$.
Therefore,the male is $XO$ and the female is $XX$.

(A) In many insects,including grasshoppers and butterflies,the mechanism of sex determination is of the $XO$ type.
In this type,males have only one $X$ chromosome along with autosomes,while females have a pair of $X$ chromosomes $(XX)$.
Therefore,the male grasshopper possesses autosomes and only one $X$ chromosome ($XO$ condition).

(C) According to Bridges' Genic Balance Theory of sex determination in Drosophila:
$1$. If the $X/A$ ratio is $1.0$,the individual is a normal female.
$2$. If the $X/A$ ratio is $0.5$,the individual is a normal male.
$3$. If the $X/A$ ratio is between $0.5$ and $1.0$ (e.g.,$0.67$),the individual is an intersex.
$4$. If the $X/A$ ratio is less than $0.5$ (e.g.,$0.33$),the individual is a supermale.
$5$. If the $X/A$ ratio is greater than $1.0$ (e.g.,$1.5$),the individual is a superfemale.
Given values:
- For $X/A = 0.6$,the value lies between $0.5$ and $1.0$,which corresponds to an intersex.
- For $X/A = 0.33$,the value is less than $0.5$,which corresponds to a supermale.
Therefore,the correct answer is Intersex and Supermale.

(B) In chickens,sex determination follows the $ZW-ZZ$ type mechanism.
In this system,the female is heterogametic $(ZW)$ and the male is homogametic $(ZZ)$.
Since the female produces two types of gametes ($Z$ and $W$),the sex of the offspring is determined by the egg (ovum) provided by the female.
Therefore,the ovum is responsible for sex determination in chickens.

(A) The sex of a child is determined by the combination of sex chromosomes ($XX$ or $XY$) at the time of fertilization.
Each pregnancy is an independent event.
The probability of having a boy or a girl in any single pregnancy is always $50\%$ or $1/2$,regardless of the gender of previous children.
Therefore,the probability of having a boy as the sixth child remains $1/2$.