Pedigree Analysis and Mendelian disorders Questions in English

(C) In pedigree analysis:
- $A$ square represents a male.
- $A$ blackened (shaded) square or circle represents an affected individual.
- $A$ horizontal line between two symbols represents mating (parents).
Pedigree analysis is the study of the inheritance of genetic traits in several generations of a human family,represented as a family tree diagram.
Advantages:
$(i)$ It helps in genetic counselling to avoid disorders.
$(ii)$ It shows the origin of a trait and the flow of a trait in a family.
$(iii)$ It is important to know the possibility of a recessive allele that can cause genetic disorders like colour blindness,haemophilia,etc.

(D) Haemophilia is a genetic disorder caused by a mutation in genes located on the $X$ chromosome.
It is inherited in an $X$-linked recessive pattern.
It is historically known as 'Royal disease' because it affected several royal families in Europe.
It is also known as 'Bleeder's disease' because the blood of affected individuals fails to clot properly due to a deficiency in clotting factors.
Therefore,it is not a $Y$-linked disorder.

(C) Colour blindness is an $X$-linked recessive disorder. Let the genotype of the colour blind man be $X^CY$ and the normal woman be $XX$.
The cross between them is as follows:
$X^CY \times XX \rightarrow X^CX, X^CX, XY, XY$.
The offspring produced are two carrier daughters $(X^CX)$ and two normal sons $(XY)$.
Since the sons receive their $X$ chromosome from their mother and $Y$ chromosome from their father,and the mother is normal $(XX)$,all sons will be normal.
Therefore,the probability of their sons having colour blindness is $0 \%$.

(B) Sex-linked diseases are genetic disorders caused by mutations in genes located on sex chromosomes ($X$ or $Y$).
Colour blindness is a classic example of an $X$-linked recessive disorder.
Haemophilia is another well-known example of an $X$-linked recessive disease.
$AIDS$,Syphilis,and Gonorrhoea are infectious diseases caused by pathogens,not genetic mutations.

(D) The study of inheritance of genetic traits in several generations of a human family in the form of a family tree diagram is called pedigree analysis.
Advantages:
$(i)$ It helps in genetic counseling to avoid disorders.
$(ii)$ It shows the origin of a trait and the flow of a trait in a family.
$(iii)$ It is important to know the possibility of a recessive allele that can cause genetic disorders like color blindness,hemophilia,etc.
The symbols used in pedigree analysis are as follows:
| Symbol | Meaning |
| :--- | :--- |
| Square | Male |
| Circle | Female |
| Diamond | Sex unspecified |
| Shaded symbols | Affected individuals |
| Horizontal line between square and circle | Mating |
| Double horizontal line | Mating between relatives (consanguineous mating) |
| Vertical line from horizontal line | Parents above and children below (in order of birth-left to right) |
| Square with number $5$ inside | Five unaffected offspring |

(A) Sickle-cell anaemia is a classic example of a point mutation.
It occurs due to a single base pair substitution in the gene coding for the $\beta$-globin chain of haemoglobin.
Specifically, the codon $GAG$ is replaced by $GUG$, which leads to the substitution of glutamic acid with valine at the sixth position of the $\beta$-globin chain.
Therefore, $A$ is point mutation and $B$ is $\beta$-chain.

(B) carrier organism is an individual who possesses a recessive gene for a specific trait or disease but does not show the phenotype because the dominant allele masks its expression. The individual is heterozygous for the trait.

(B) Colour blindness is a sex-linked recessive disorder caused by a defective gene located on the $X$-chromosome.
Since females have two $X$-chromosomes $(XX)$ and males have one $X$-chromosome $(XY)$,the inheritance pattern depends on the parents.
$A$ colour-blind male inherits the defective gene from his mother,while a colour-blind female inherits the defective gene from both her father and her mother.
Therefore,the gene for colour blindness is carried by both the father and the mother.

(A) Thalassaemia is an autosome-linked recessive blood disorder.
$(i)$ It occurs due to either mutation or deletion,resulting in a reduced rate of synthesis of one of the globin chains of haemoglobin.
$(ii)$ Anaemia is the characteristic symptom of this disease.
$(iii)$ Thalassaemia is classified into two types:
$1.$ $\alpha$-thalassaemia: The production of the $\alpha$-globin chain is affected. It is controlled by the closely linked genes $HBA1$ and $HBA2$ on chromosome $16$. It occurs due to the mutation or deletion of one or more of the four genes.
$2.$ $\beta$-thalassaemia: The production of the $\beta$-globin chain is affected. It occurs due to the mutation of one or both $HBB$ genes on chromosome $11$.

(B) Phenylketonuria $(PKU)$ is an inborn error of metabolism inherited as an autosomal recessive trait. It is caused by a mutation in the gene that codes for the liver enzyme $Phenylalanine$ hydroxylase. This enzyme is responsible for converting the amino acid $Phenylalanine$ into $Tyrosine$. Due to the deficiency of this enzyme, $Phenylalanine$ accumulates in the body and is converted into phenylpyruvic acid and other derivatives, which leads to mental retardation and other symptoms.

(A) Genetic disorders are broadly grouped into two categories:
$(i)$ Mendelian Disorders: These genetic disorders are primarily caused by alterations or mutations in a single gene. They are transmitted to offspring following the principles of inheritance as described by Mendel. Mendelian disorders can be dominant or recessive. Examples include haemophilia,colour blindness,sickle-cell anaemia,cystic fibrosis,phenylketonuria,and thalassaemia.
$(ii)$ Chromosomal Disorders: These disorders are caused due to the excess,absence,or abnormal arrangement of one or more chromosomes. Examples include Turner's syndrome,Down's syndrome,etc.

(B) $\beta$-thalassaemia major is also known as Cooley's Anemia.
Thalassaemia is of two major kinds: $\alpha$-thalassaemia and $\beta$-thalassaemia,depending on the defective gene in the $\alpha$ or $\beta$-chain of haemoglobin.
Key characteristics of Thalassaemia:
$(i)$ It is an autosome-linked recessive disease.
$(ii)$ It occurs due to either mutation or deletion,resulting in a reduced rate of synthesis of one of the globin chains of haemoglobin.
$(iii)$ Anaemia is the characteristic symptom of this disease.
$(iv)$ $\alpha$-thalassaemia: Production of $\alpha$-globin chain is affected. It is controlled by the closely linked genes $HBA1$ and $HBA2$ on chromosome $16$.
$(v)$ $\beta$-thalassaemia: Production of $\beta$-globin chain is affected. It occurs due to mutation of one or both $HBB$ genes on chromosome $11$.

(C) Mendelian disorders are caused by mutations in a single gene. These disorders follow the principles of Mendelian inheritance and can be classified as either dominant or recessive,depending on whether the trait is expressed in the heterozygous or homozygous condition.

(A) Colour blindness is a sex-linked recessive disorder. It is caused by the presence of a recessive gene located on the $X$ chromosome. Since males have only one $X$ chromosome,they are more frequently affected than females,who require two copies of the recessive gene to express the trait.

(C) Phenylketonuria $(PKU)$ is an inborn error of metabolism that is inherited as an autosomal recessive trait.
It is caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase.
Due to the deficiency of this single enzyme,the amino acid phenylalanine cannot be converted into tyrosine,leading to its accumulation and subsequent conversion into phenylpyruvic acid and other derivatives,which causes mental retardation.

(A) Colour blindness 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.

(A) Haemophilia is also known as Bleeder's disease.
It is a sex-linked recessive disorder in which the blood fails to clot properly due to the absence of specific clotting factors.
This condition is typically transmitted from a carrier female to her male progeny.

(A) Colour blindness is a genetic disorder characterized by the inability to distinguish between certain colours,most commonly red and green.
$(i)$ It is a sex-linked recessive disorder.
$(ii)$ It results from a defect in either red or green cone cells of the eye,leading to a failure to discriminate between red and green colours.
$(iii)$ The gene responsible for colour blindness is located on the $X$-chromosome.
$(iv)$ It is observed more frequently in males $(X^{c}Y)$ because they possess only one $X$-chromosome,whereas females $(XX)$ require two copies of the recessive gene to express the trait.

(A) In pedigree analysis,standard symbols are used to represent individuals.
Symbol $A$ represents a carrier female for an $X$-linked recessive trait (often shown as a circle with a central dot).
Symbol $B$ represents an affected female (a shaded circle).
Symbol $C$ represents a carrier female for an autosomal recessive trait (a circle with half-shaded area).
Therefore,symbol $C$ indicates a carrier female.

(B) $1$. In the given pedigree,both parents are unaffected (unshaded),but they have affected (shaded) offspring (both male and female).
$2$. This pattern indicates that the trait is recessive because it skips generations and appears in the offspring of unaffected parents.
$3$. Since both male and female offspring are affected,it cannot be $Y$-linked. If it were $X$-linked recessive,an affected daughter would require her father to be affected,which is not the case here.
$4$. Therefore,the trait must be autosomal recessive. Phenylketonuria is a classic example of an autosomal recessive metabolic disorder.
$5$. Thus,the pedigree represents the inheritance of a condition like phenylketonuria as an autosomal recessive trait.

(A) Brachydactyly is a genetic condition characterized by abnormally short fingers or toes. It is inherited as an autosomal dominant trait. This means that an individual only needs to inherit one copy of the mutated gene from one parent to express the condition.

(D) Colour blindness is an $X$-linked recessive disorder.
Let the normal vision allele be $X^C$ and the colourblind allele be $X^c$.
Since the husband has normal vision,his genotype is $X^CY$.
Since his father was colourblind,he inherited his $Y$ chromosome from his father and his $X^C$ chromosome from his mother.
Since the wife has normal vision but her father was colourblind,she must be a carrier of the trait. Her genotype is $X^CX^c$.
When we cross the husband $(X^CY)$ and the wife $(X^CX^c)$:
Possible genotypes for their children are $X^CX^C$ (normal daughter),$X^CX^c$ (carrier daughter),$X^CY$ (normal son),and $X^cY$ (colourblind son).
For a daughter to be colourblind,she must have the genotype $X^cX^c$.
Since the father is $X^CY$,he can only pass the $X^C$ allele to his daughters.
Therefore,the probability of their daughter being colourblind is $0\, \%$.

(D) Colour blindness is a sex-linked recessive disorder. The genetic basis of colour blindness was first described by Wilson in $1910$.

(C) Haemophilia is a sex-linked recessive disorder,which is caused due to the deficiency of specific blood clotting factors (such as Factor-$VIII$ in Haemophilia-$A$ or Factor-$IX$ in Haemophilia-$B$).
It is transmitted from an unaffected carrier female to some of her male progeny.
It is characterized by the inability of the blood to clot properly,leading to excessive bleeding even from minor injuries.

(D) The man has normal vision,so his genotype is $X^CY$. Since his father was colourblind,the man received his $Y$ chromosome from his father and his $X^C$ chromosome from his mother,but he is phenotypically normal,meaning he carries the dominant allele. Wait,if the man has normal vision,his genotype is $X^CY$. The woman's father was colourblind,so she must be a carrier $(X^CX^c)$.
When a normal man $(X^CY)$ marries a carrier woman $(X^CX^c)$,the possible genotypes for their daughters are $X^CX^C$ (normal) and $X^CX^c$ (carrier).
Neither of these genotypes results in colourblindness for the daughter,as colourblindness is a recessive $X$-linked trait requiring the daughter to inherit the recessive allele from both parents. Since the father is normal $(X^CY)$,he cannot pass the recessive allele to his daughter. Therefore,the chance of the daughter being colourblind is $0 \%$.

(A) $1$. In the given pedigree,the trait appears in the offspring of unaffected parents (first generation). This indicates that the trait is recessive.
$2$. The trait is observed in both males and females,and it does not show a specific bias towards one sex,which is characteristic of an autosomal trait.
$3$. Therefore,the pedigree chart represents an autosomal recessive trait.

(B) $1$. The pedigree shows an $X$-linked recessive trait because the affected mother $(A)$ passes the trait to her sons.
$2$. Since $A$ is affected,her genotype is $X^cX^c$. Her husband is normal,so his genotype is $XY$.
$3$. Individual $C$ is a normal female daughter of $A$. She must have inherited one $X^c$ from her mother and one $X$ from her father. Thus,the genotype of $C$ is $XX^c$.
$4$. Individual $D$ is a normal male son of $C$. Since he is normal,he must have inherited the $X$ chromosome from his mother $(C)$ and the $Y$ chromosome from his father. Thus,the genotype of $D$ is $XY$.

(D) Haemophilia is an $X$-linked recessive disorder.
Since the woman has a haemophilic son $(X^hY)$,the son must have inherited the $X^h$ chromosome from his mother.
Therefore,the mother must be a carrier of the haemophilia gene,meaning her genotype is $X^hX$.
Since the woman also has normal children and the husband is not mentioned as haemophilic,we assume the husband is normal $(XY)$.
Thus,the genotype of the woman is $X^hX$ and the husband is $XY$.

(B) Haemophilia is an $X$-linked recessive disorder. $A$ haemophilic woman has the genotype $X^hX^h$,and a normal man has the genotype $XY$.
When they cross $(X^hX^h \times XY)$,the gametes produced are $X^h$ from the mother and $X$ or $Y$ from the father.
The resulting progeny will have genotypes $X^hX$ (carrier daughter) and $X^hY$ (haemophilic son).
Therefore,all sons will be haemophilic and all daughters will be carriers.

(A) The pedigree chart shows that the trait is $X$-linked dominant.
$1$. The trait is present in every generation,which is a characteristic of a dominant trait.
$2$. An affected father passes the trait to all his daughters but none of his sons,which indicates $X$-linkage.
$3$. Since the trait appears in females (who have two $X$ chromosomes) and is passed from an affected father to his daughters,it must be $X$-linked dominant.

(B) -Dizygotic (fraternal) twins result from the fertilization of two separate ova by two separate sperm. They are genetically as similar as any other siblings.
$B$-Monozygotic (identical) twins result from the fertilization of a single ovum by a single sperm,forming one zygote. This zygote subsequently divides into two or more embryos. These individuals share an identical genome.

(B) Sickle cell anaemia is an autosome-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for the gene.
This disease is controlled by a single pair of allele,$Hb^A$ and $Hb^S$.
It is caused by a point mutation in the $\beta$-globin chain of the haemoglobin molecule.
Specifically,there is a substitution of Glutamic acid $(Glu)$ by Valine $(Val)$ at the sixth position of the $\beta$-globin chain.
This substitution occurs due to a single base pair change in the $DNA$ sequence,where the codon $GAG$ is mutated to $GUG$.

(A) In pedigree analysis,standard symbols are used to represent individuals in a family tree.
- $A$ square $(A)$ represents a male.
- $A$ circle represents a female.
- $A$ diamond shape represents an individual whose sex is unspecified.
- $A$ filled circle or square represents an affected individual.
- $A$ square connected to a circle represents a mating between a male and a female.
Therefore,the symbol for sex unspecified is a diamond shape.

(C) Cystic fibrosis,Myotonic dystrophy,and Thalassemia are classified as Mendelian disorders.
These disorders are primarily determined by alteration or mutation in a single gene.
They are transmitted to the offspring through the same principles of inheritance as described by Mendel,hence they are called Mendelian disorders.

(A) Phenylketonuria $(PKU)$ is an inborn error of metabolism caused by a mutation in the gene encoding the enzyme phenylalanine hydroxylase.
This gene is located on chromosome $12$.
Since the disorder only manifests when an individual inherits two copies of the mutated gene (one from each parent),it is classified as an autosomal recessive trait.

(C) $1$. Colour blindness is an $X$-linked recessive disorder.
$2$. Thalassemia is an autosomal recessive blood disorder.
$3$. Myotonic dystrophy is an autosomal dominant disorder caused by a mutation in the $DMPK$ gene.
$4$. Haemophilia is an $X$-linked recessive disorder.
Therefore,the correct answer is Myotonic dystrophy.

(C) In the provided pedigree,we observe that the trait is expressed in every generation (vertical transmission).
Furthermore,an affected individual has at least one affected parent.
This pattern of inheritance,where the trait does not skip generations and affected individuals appear in every generation,is characteristic of a dominant trait.
Therefore,the shaded symbols represent the dominant allele.

(C) Haemophilia is a sex-linked recessive disorder in which a single protein that is a part of the cascade of proteins involved in blood clotting is affected.
In Haemophilia $A$,the blood clotting factor $VIII$ is absent or defective.
In Haemophilia $B$,the blood clotting factor $IX$ is absent or defective.
Due to this,a simple cut results in non-stop bleeding.

(B) Sickle-cell anaemia is an autosome-linked recessive trait.
Individuals with the genotype $Hb^{A}Hb^{A}$ are normal.
Individuals with the genotype $Hb^{A}Hb^{S}$ are carriers and exhibit the sickle-cell trait,meaning they are generally healthy but can pass the gene to their offspring.
Individuals with the genotype $Hb^{S}Hb^{S}$ are affected by the disease.
Therefore,the correct statement is that heterozygous individuals $(Hb^{A}Hb^{S})$ exhibit the sickle-cell trait.

(A) $1$. Colour blindness is an $X$-linked recessive trait.
$2$. The woman is normal but her father was colour blind,so she must be a carrier $(XX^c)$.
$3$. The man is normal $(XY)$.
$4$. The cross is $XX^c \times XY$.
$5$. The Punnett square results are:
- $XX$ (Normal daughter)
- $XX^c$ (Carrier daughter)
- $XY$ (Normal son)
- $X^cY$ (Colour blind son)
$6$. Out of four possible outcomes,one $(X^cY)$ is colour blind.
$7$. Therefore,the probability of colour blindness in the progeny is $1/4$ or $25\;\%$.

(C) Haemophilia is an $X$-linked recessive disorder,and cystic fibrosis is an autosomal recessive disorder.
Mr. Stevan is suffering from both,so his genotype is $X^hY$ for haemophilia and $cc$ for cystic fibrosis.
Since he is a male,his sperm cells will carry either the $X$ chromosome or the $Y$ chromosome.
For the $X$-linked trait,the sperm will carry the recessive allele $X^h$ with a probability of $1$.
For the autosomal trait,since he is $cc$,all his gametes will carry the recessive allele $c$ with a probability of $1$.
Therefore,the probability of a sperm carrying both the recessive $X$-linked allele $(X^h)$ and the recessive autosomal allele $(c)$ is $1 \times 1 = 1$. However,considering the segregation of chromosomes,the sperm carrying the $X^h$ allele is $1/2$ of the total sperm produced.
Thus,the probability is $1/2$.

(C) In pedigree analysis, standard symbols are used to represent genetic relationships and traits:
$a.$ Solid symbol represents an affected individual (Trait to be studied) - $(iii)$.
$b.$ $A$ horizontal line between two symbols represents mating between parents - $(iv)$.
$c.$ $A$ horizontal line above symbols represents offspring - $(ii)$.
$d.$ $A$ dot in the centre of a symbol represents a carrier of a sex-linked trait - $(i)$.
Therefore, the correct matching is $a(iii), b(iv), c(ii), d(i)$.

(B) Phenylketonuria is an inborn error of metabolism inherited as an autosomal recessive trait. It is caused by a deficiency of the enzyme phenylalanine hydroxylase,which converts the amino acid phenylalanine into tyrosine. Due to the absence of this enzyme,phenylalanine accumulates in the body and is converted into phenylpyruvic acid and other derivatives,which leads to mental retardation.

(A) Haemophilia is an $X$-linked recessive disorder.
For a female to be haemophilic (genotype $X^hX^h$),she must inherit one recessive allele $(X^h)$ from her mother and one recessive allele $(X^h)$ from her father.
This means the father must be haemophilic $(X^hY)$ and the mother must be either a carrier $(X^HX^h)$ or haemophilic $(X^hX^h)$.
Since the condition is rare,the probability of both parents carrying the gene is very low,making the occurrence of a haemophilic female extremely rare.
Thus,both the Assertion and the Reason are correct,and the Reason explains why the condition is rare in females.

(A) Haemophilia is an $X$-linked recessive disorder. $A$ heterozygotic female (carrier) has one normal $X$ chromosome and one $X$ chromosome carrying the haemophilia gene $(X^hX)$.
During gametogenesis,she produces two types of eggs: $X$ and $X^h$. If an egg with the $X^h$ chromosome is fertilized by a $Y$-bearing sperm,the resulting son will be haemophilic $(X^hY)$. Thus,the assertion is correct.
Haemophilia exhibits criss-cross inheritance,where the trait is passed from an affected father to his daughter (who becomes a carrier) and then to his grandsons. Since $X$-linked recessive traits are transmitted from mother to son and father to daughter,they follow a criss-cross pattern. Thus,the reason is correct and explains why the disease is transmitted in this specific manner.

(C) Sickle cell anaemia is an autosomal recessive disorder.
Let the normal allele be $Hb^A$ and the sickle cell allele be $Hb^S$.
Both parents are heterozygous, meaning their genotypes are $Hb^A Hb^S$.
The cross is: $Hb^A Hb^S \times Hb^A Hb^S$.
The Punnett square results are:
$1$ $Hb^A Hb^A$ (Normal)
$2$ $Hb^A Hb^S$ (Carrier/Heterozygous)
$1$ $Hb^S Hb^S$ (Diseased)
Thus, $1$ out of $4$ offspring will be diseased.
Percentage = $(1/4) \times 100 = 25\%$.

(N/A) Thalassemia and haemophilia are categorised as Mendelian disorders because these disorders are caused by an alteration or mutation in a single gene. They are transmitted to offspring following Mendelian principles of inheritance.
Symptoms and patterns of inheritance are as follows:
$(a)$ Thalassemia: It is an autosomal-linked recessive blood disorder characterized by defects in the $\alpha$,$\beta$,or $\delta$ globin chains,resulting in abnormal $Hb$ molecules.
Symptom: Severe anaemia.
Inheritance: It follows an autosomal recessive pattern. Two mutant alleles (one from each parent) must be inherited for an individual to be affected (homozygous). Heterozygous individuals are carriers and may pass the mutant allele to their children.
$(b)$ Haemophilia: It is a sex-linked recessive disorder where the defective gene is located on the $X$-chromosome.
Symptom: Prolonged blood clotting time and excessive bleeding even from minor injuries.
Inheritance: Since the gene is on the $X$-chromosome,it is more common in males as they possess only a single $X$-chromosome (hemizygous). Females have two $X$-chromosomes,so they are rarely affected unless the father is haemophilic and the mother is at least a carrier.

(A) $1$. Myotonic dystrophy is an example of an autosome-linked dominant trait.
$2$. Haemophilia is an $X$-linked recessive disorder.
$3$. Thalassemia is an autosome-linked recessive blood disorder.
$4$. Sickle cell anaemia is an autosome-linked recessive disorder.
Therefore,the correct option is $A$.

(C) Colour blindness is an $X$-linked recessive disorder.
Let $X^c$ represent the allele for colour blindness and $X$ represent the normal allele.
$A$ colour blind female has the genotype $X^c X^c$.
$A$ man whose mother was colour blind must have inherited the $X^c$ allele from her,so his genotype is $X^c Y$.
When they marry $(X^c X^c \times X^c Y)$:
- Offspring genotypes: $X^c X^c$ (colour blind daughter),$X^c Y$ (colour blind son),$X^c X^c$ (colour blind daughter),$X^c Y$ (colour blind son).
All offspring $(100 \%)$ will be colour blind.

(B) Factor $VIII$ (Anti-hemophilic factor) is a crucial protein involved in the blood clotting cascade.
Its deficiency leads to Hemophilia $A$,a genetic disorder where the blood fails to clot properly,resulting in prolonged bleeding even from minor injuries.
Color blindness is a vision-related disorder,Jaundice is related to liver function,and Sickle cell anemia is a hemoglobin-related genetic disorder.