Pedigree Analysis and Mendelian disorders Questions in English

(A) The correct matches are:
$(1)$ Albinism is an autosomal recessive disorder caused by the absence of the enzyme tyrosinase,often represented by the genotype $cc$.
$(2)$ Phenylketonuria $(PKU)$ is an inborn error of metabolism caused by a recessive gene,often represented by $pp$.
$(3)$ Sickle cell anaemia is an autosomal recessive disorder caused by a mutation in the $\beta$-globin chain,represented by the genotype $Hb^sHb^s$.
$(4)$ Alkaptonuria is an autosomal recessive metabolic disorder caused by a deficiency of the enzyme homogentisate $1,2-$dioxygenase,often represented by the genotype $aa$.
Therefore,the correct matching is $(1-q), (2-s), (3-p), (4-r)$.

(A) The correct matches are as follows:
$1$. $(a)$ Sickle cell anaemia is caused by the production of $(w)$ Defective haemoglobin (specifically $HbS$ due to a point mutation).
$2$. $(b)$ Alkaptonuria is a metabolic disorder characterized by the accumulation of $(x)$ Homogentisic acid in the urine.
$3$. $(c)$ Albinism is a genetic condition where $(y)$ Melanin pigments are absent in the skin,hair,and eyes.
$4$. $(d)$ Phenylketonuria is a disorder caused by the inability to metabolize phenylalanine,leading to $(z)$ More amount of phenylalanine in the body.
Therefore,the correct sequence is $(a-w), (b-x), (c-y), (d-z)$.

(D) Phenylketonuria $(PKU)$ is an inborn error of metabolism inherited as an autosomal recessive trait.
In this disorder,the affected individual lacks the enzyme Phenylalanine hydroxylase,which is required to convert the amino acid Phenylalanine into Tyrosine.
Due to this deficiency,Phenylalanine accumulates and is converted into Phenylpyruvic acid and other derivatives,leading to mental retardation.
Accumulation of these derivatives in the brain results in mental retardation,and they are also excreted through urine because of poor absorption by the kidney.
Since the conversion of Phenylalanine to Tyrosine is blocked,the level of Tyrosine decreases,not increases.
Therefore,the statement 'Increase in the level of Tyrosine' is incorrect and not associated with $PKU$.

(D) Pedigree analysis is a tool used in human genetics to study the inheritance pattern of specific traits or diseases over several generations.
It is useful for:
$1$. Studying the inheritance of a specific trait or gene.
$2$. Identifying abnormalities or genetic disorders in offspring.
$3$. Diagnosing the mode of inheritance of a disease (e.g.,autosomal dominant,recessive,or sex-linked).
Since pedigree analysis is useful for all the purposes mentioned in options $A$,$B$,and $C$,the correct answer is $D$.

(D) Mendelian disorders are primarily determined by alteration or mutation in the single gene. Examples include Hemophilia,Cystic fibrosis,Sickle cell anemia,Color blindness,Phenylketonuria,and Thalassemia.
Down's syndrome is a chromosomal disorder caused by the presence of an additional copy of chromosome number $21$ (trisomy of $21$).
Therefore,Down's syndrome is not a Mendelian disorder but a chromosomal disorder.

(B) Hemophilia is a $X$-linked recessive genetic disorder.
In this condition,the gene responsible for the clotting factor is located on the $X$ chromosome.
Since it is recessive,a female must have the mutation on both $X$ chromosomes to express the disease,whereas a male only needs one mutated $X$ chromosome because he has only one $X$ chromosome ($XY$ genotype).
Therefore,it is classified as a sex-linked recessive disorder.

(C) Myotonic dystrophy is a genetic disorder characterized by the progressive weakness and wasting of muscles.
It is caused by a mutation in the $DMPK$ gene.
According to the principles of human genetics,myotonic dystrophy follows an autosomal dominant pattern of inheritance.
This means that an affected individual only needs to inherit one copy of the mutated gene from either parent to express the disorder.

(D) Hemophilia and color blindness are $X$-linked recessive disorders.
In males,there is only one $X$ chromosome $(XY)$,so if the gene is present on the $X$ chromosome,the disorder is expressed.
In females,there are two $X$ chromosomes $(XX)$,so they usually act as carriers unless they are homozygous for the recessive trait.
Therefore,these disorders are significantly more common in males than in females.

(A) Hemophilia is a sex-linked recessive disorder that results in the failure of blood to clot. This occurs due to a defect in the proteins (clotting factors) required for the blood coagulation cascade. In affected individuals,a simple cut results in non-stop bleeding. Color blindness is a defect in color perception,while Thalassemia and Sickle cell anemia are disorders related to hemoglobin synthesis.

(D) Hemophilia is an $X$-linked recessive disorder.
Since a son receives his $Y$ chromosome from his father and his $X$ chromosome from his mother,the gene for hemophilia must be present on the $X$ chromosome inherited from the mother.
If the mother is a carrier $(X^HX^h)$,she can pass the $X^h$ chromosome to her son,resulting in a hemophilic son $(X^hY)$.
Therefore,the probability of the son inheriting the disease comes from a carrier mother.

(C) $1$. Color blindness and Hemophilia are $X$-linked recessive disorders.
$2$. Sickle cell anemia is an autosomal recessive disorder caused by a mutation in the hemoglobin gene on chromosome $11$.
$3$. Therefore,the correct option is $C$.

(D) Sickle cell anemia is an autosomal recessive disorder.
Both parents are carriers $(Hb^{A}Hb^{S} \times Hb^{A}Hb^{S})$.
The Punnett square for this cross results in:
- $Hb^{A}Hb^{A}$ (Normal): $25 \%$
- $Hb^{A}Hb^{S}$ (Carrier): $50 \%$
- $Hb^{S}Hb^{S}$ (Affected): $25 \%$
The probability of having an affected child $(Hb^{S}Hb^{S})$ is $1/4$ or $25 \%$.
Since the question asks for the probability of having an affected son,we must consider the probability of the child being male $(1/2)$ and the probability of the child being affected $(1/4)$.
Probability = $P(\text{Affected}) \times P(\text{Male}) = 1/4 \times 1/2 = 1/8$ or $12.5 \%$.

(D) Thalassemia is an autosomal recessive blood disorder caused by defects in the synthesis of globin chains of hemoglobin.
It is classified into two types: $\alpha$-thalassemia and $\beta$-thalassemia.
Unlike sickle-cell anemia,which is a qualitative defect (abnormal hemoglobin structure),thalassemia is a quantitative defect where there is a reduced synthesis of one of the globin chains ($ \alpha $ or $ \beta $),leading to the formation of abnormal hemoglobin molecules and anemia.

$(A)$ Sickle-cell anaemia is a qualitative defect where a single amino acid substitution (glutamic acid to valine) occurs in the globin chain, leading to abnormal haemoglobin structure.
Thalassemia is a quantitative defect where there is a reduced synthesis of one of the globin chains ($\alpha$ or $\beta$ chains).
However, the question specifically asks for a qualitative defect in the synthesis of globin molecules that impairs $O_2$ transport.
Sickle-cell anaemia is the classic example of a qualitative defect in the globin molecule structure.
Therefore, the correct option is $A$.

(B) Sickle cell anemia is an autosome-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene. The disease is controlled by a single pair of allele,$Hb^A$ and $Hb^S$. The defect is caused by the substitution of glutamic acid $(Glu)$ by valine $(Val)$ at the sixth position of the $\beta$-globin chain of the hemoglobin molecule. This substitution occurs due to a single base substitution at the sixth codon of the $\beta$-globin gene from $GAG$ to $GUG$. The mutant hemoglobin molecule undergoes polymerization under low oxygen tension causing the change in the shape of the $RBC$ from biconcave disc to elongated sickle-like structure.

(C) Phenylketonuria $(PKU)$ is an inborn error of metabolism.
It is caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase.
In affected individuals,the enzyme phenylalanine hydroxylase is deficient or absent.
This enzyme is responsible for converting the amino acid phenylalanine into tyrosine.
Due to the lack of this enzyme,phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives,which leads to mental retardation.
Since the gene responsible for this condition is located on an autosome and is expressed only in the homozygous recessive state $(pp)$,it is classified as an autosomal recessive disorder.

(D) Phenylketonuria $(PKU)$ is an inborn error of metabolism that is inherited as an autosomal recessive trait.
In this disorder,the affected individual lacks the enzyme phenylalanine hydroxylase,which converts the amino acid phenylalanine into tyrosine.
As a result,phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives.
Accumulation of these substances in the brain results in mental retardation.
Other symptoms include reduction in hair and skin pigmentation.

(C) Thalassemia is an autosomal recessive blood disorder.
In $\alpha$-Thalassemia,the production of the $\alpha$-globin chain is affected,which is controlled by two closely linked genes,$HBA1$ and $HBA2$,located on chromosome $16$.
Defects in these genes lead to reduced or absent production of $\alpha$-globin chains,which impairs the formation of hemoglobin,thereby affecting the transport of respiratory gases like oxygen.
In contrast,$\beta$-Thalassemia is caused by mutations in the $HBB$ gene located on chromosome $11$.

(C) The disorder described is $\beta$-Thalassemia.
$1$. $\beta$-Thalassemia is a quantitative disorder where the production of the $\beta$-globin chain of hemoglobin is reduced or absent.
$2$. This is caused by mutations in the $HBB$ gene located on chromosome $11$.
$3$. In contrast,Sickle cell anemia is a qualitative disorder caused by a mutation in the $HBB$ gene that results in the production of abnormal hemoglobin $(HbS)$ rather than a reduction in the quantity of the chain.
$4$. Hemophilia is a sex-linked recessive disorder related to blood clotting factors,not hemoglobin synthesis.

(D) The disorders listed,$Sickle$ $cell$ $anemia$,$Thalassemia$,and $Phenylketonuria$,are all examples of autosomal recessive disorders.
Autosomal recessive disorders occur when an individual inherits two copies of the mutated gene (one from each parent).
$Sickle$ $cell$ $anemia$ is caused by a mutation in the $HbS$ gene.
$Thalassemia$ is a group of inherited blood disorders characterized by less hemoglobin and fewer red blood cells in the body.
$Phenylketonuria$ $(PKU)$ is an inborn error of metabolism caused by a deficiency of the enzyme phenylalanine hydroxylase.
Since none of these are autosomal dominant,the correct answer is $D$.

(C) Sickle cell anemia is an autosome-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene.
In this disease,the hemoglobin molecule undergoes polymerization under low oxygen tension,causing the change in the shape of the red blood cell from a biconcave disc to an elongated sickle-like structure.
This distortion occurs due to a single base substitution at the sixth codon of the $\beta -$globin gene from $GAG$ to $GUG$,resulting in the substitution of glutamic acid by valine.

(A) In humans,a male offspring has $XY$ sex chromosomes. The $Y$ chromosome is inherited from the father,while the $X$ chromosome is inherited from the mother. If the defect is $X$-linked recessive (the most common type of genetic defect in males),the gene is located on the $X$ chromosome. Since the male receives his only $X$ chromosome from his mother,the defect is inherited from the mother.

(A) Colorblindness is an $X$-linked recessive disorder.
For a daughter to be colorblind,she must inherit the recessive allele $(X^c)$ from both parents.
This means her genotype must be $X^cX^c$.
$1$. The father must be colorblind $(X^cY)$,so he contributes an $X^c$ chromosome to his daughter.
$2$. The mother must be either a carrier $(X^cX)$ or colorblind $(X^cX^c)$ to contribute an $X^c$ chromosome.
Among the given options,if the father is colorblind $(X^cY)$ and the mother is a carrier $(X^cX)$,there is a $25\%$ chance of having a colorblind daughter $(X^cX^c)$.
Option $A$ is the most appropriate condition where the father is colorblind and the mother is a carrier,ensuring the daughter receives the recessive allele from both.

(C) Color blindness is an $X$-linked recessive genetic disorder.
The genotype of the color-blind man is $X^cY$ (where $X^c$ is the gene for color blindness).
The woman's father was color-blind,so the woman is a carrier: $X^cX$.
When a cross occurs between $X^cY$ and $X^cX$:
Offspring: $X^cX^c$ (color-blind daughter),$X^cX$ (carrier daughter),$X^cY$ (color-blind son),$XY$ (normal son).
Thus,there is a $50 \%$ probability that the sons will be color-blind.

(C) Color blindness is an $X$-linked recessive disorder.
Let the normal vision allele be $X^C$ and the color-blindness allele be $X^c$.
The son is color-blind,so his genotype must be $X^c Y$.
$A$ son inherits his $Y$ chromosome from his father and his $X$ chromosome from his mother.
Therefore,the son must have inherited the $X^c$ chromosome from his mother.
Since the mother has normal vision,she must possess at least one dominant $X^C$ allele.
Thus,the mother's genotype must be $X^C X^c$ (a carrier).

(D) In genetics,many Mendelian disorders are inherited in a recessive manner.
$1$. Autosomal recessive disorders (e.g.,Sickle-cell anaemia) require two copies of the mutated gene to manifest.
$2$. Sex-linked recessive disorders (e.g.,Haemophilia,Colour blindness) are more common in males because they have only one $X$ chromosome.
$3$. Mutation-induced disorders can also be recessive if the mutation results in a loss-of-function allele.
Therefore,all the listed categories frequently exhibit recessive inheritance patterns.

(D) Hemophilia is a sex-linked recessive disorder caused by a mutation in genes located on the $X$ chromosome.
In females,there are two $X$ chromosomes $(XX)$. For a female to be affected,she must inherit the recessive gene on both $X$ chromosomes $(X^hX^h)$.
In males,there is only one $X$ chromosome $(XY)$. If a male inherits the recessive gene on his single $X$ chromosome $(X^hY)$,he will express the disorder.
Therefore,males are more frequently affected because they only require one copy of the defective gene to manifest the disease,whereas females require two.

(C) In pedigree analysis, a circle with a dot in the center $(\odot)$ is used to represent a carrier female for an $X$-linked recessive trait. This symbol indicates that the individual is phenotypically normal but carries the recessive allele for the disorder.

(B) $1$. In the pedigree,the mother is a carrier (indicated by a dot in the circle) and the father is unaffected. They have an affected son.
$2$. The affected son marries an unaffected woman and they have two carrier daughters.
$3$. This pattern is characteristic of an Autosomal Recessive disorder. If it were $X$-linked recessive,the affected father would pass the trait to all his daughters,making them carriers,but he would not pass it to his sons. Here,the father is affected and the daughters are carriers,which is consistent with an autosomal recessive trait where the mother is a carrier and the father is affected (homozygous recessive). However,looking at the first generation,the mother is a carrier and the father is normal,resulting in an affected son,which is a classic indicator of an Autosomal Recessive trait.

(B) In the given pedigree chart,the disorder is autosomal recessive.
Individuals in generation $II$ are affected (shaded),meaning they must have the genotype $aa$.
Since the parents in generation $I$ are unaffected but have affected children in generation $II$,both parents must be carriers of the recessive allele.
Therefore,the genotype of the parents in generation $I$ must be $Aa$ and $Aa$.

(A) $1$. In the given pedigree,the parents are unaffected (unshaded),but they have an affected (shaded) offspring. This indicates that the trait is recessive.
$2$. Since the trait appears in both generations and can skip generations,it is likely autosomal recessive.
$3$. Phenylketonuria is a well-known example of an autosomal recessive metabolic disorder.
$4$. Sex-linked recessive disorders like Haemophilia typically show a criss-cross pattern of inheritance,which is not exclusively depicted here,and the autosomal recessive pattern fits the data perfectly.

(A) Albinism is an autosomal recessive disorder. Let the normal allele be '$A$' and the recessive allele for albinism be '$a$'.
An affected individual must have the genotype '$aa$'.
In the pedigree,the female parent of individuals $5$ and $6$ is affected,so her genotype is '$aa$'.
Individual $4$ is given as homozygous dominant,so his genotype is '$AA$'.
The offspring of '$aa$' and '$AA$' will all have the genotype '$Aa$'.
Therefore,individuals $5$ and $6$ are carriers ('$Aa$').
Since individuals $5$ and $6$ received one '$a$' allele from their affected mother,and their father $(4)$ is '$AA$',they must be carriers.
For the first generation,since the affected female's parents ($1$ and $2$) produced an affected child $(aa)$,both parents $1$ and $2$ must be carriers ('$Aa$').
Individual $3$ is a normal male,but his genotype could be '$AA$' or '$Aa$',so he is not necessarily a carrier.
Thus,the carriers are $1, 2, 5,$ and $6$.

(C) Polydactyly is a condition in which a person has more than the normal number of fingers or toes.
It is a genetic disorder that follows an autosomal dominant pattern of inheritance.
This means that an individual only needs to inherit one copy of the mutated gene from either parent to express the trait.
Since it is autosomal,it is not linked to the sex chromosomes ($X$ or $Y$).

(D) Sickle-cell anemia is an autosome-linked recessive trait.
In this condition,the gene $Hb^A$ represents the normal hemoglobin,while $Hb^S$ represents the mutant hemoglobin.
- $Hb^A Hb^A$: Normal individual.
- $Hb^S Hb^S$: Affected individual (suffers from the disease).
- $Hb^A Hb^S$: Carrier individual (heterozygous,phenotypically normal but carries the trait).
Therefore,the genotype $Hb^A Hb^S$ is responsible for the carrier state.

(C) Hemophilia is a sex-linked recessive disorder,which shows its transmission from unaffected carrier female to some of the male progeny.
In this disease,a single protein that is a part of the cascade of proteins involved in the clotting of blood is affected.
Due to this,in an affected individual,a simple cut will result in non-stop bleeding.
The gene for this disease is located on the $X$ chromosome.
Therefore,it is known as an $X$-linked recessive disorder and is commonly referred to as 'bleeder's disease'.

(C) Hemophilia is a group of genetic disorders that impair the body's ability to make blood clots.
- Hemophilia-$A$ is caused by a deficiency of clotting factor $VIII$ (Anti-hemophilic factor).
- Hemophilia-$B$ (also known as Christmas disease) is caused by a deficiency of clotting factor $IX$ (Christmas factor).
- Hemophilia-$C$ is caused by a deficiency of clotting factor $XI$ (Plasma thromboplastin antecedent).
Therefore,the correct sequence of deficient factors is $VIII, IX, XI$.

(A) In the given pedigree,both parents are affected (shaded),yet they have unaffected offspring (unshaded).
This indicates that the trait is dominant,as two affected parents can produce an unaffected child only if the trait is dominant and both parents are heterozygous $(Aa \times Aa \rightarrow aa)$.
If the trait were recessive,two affected parents would only produce affected offspring.
Therefore,the pedigree represents dominant inheritance.

(D) Based on standard pedigree analysis symbols:
$(i)$ $A$ filled circle represents an affected female.
(ii) $A$ circle with a dot represents a carrier female.
(iii) $A$ diamond with a number inside represents unspecified sex with that number of offspring.
(iv) $A$ filled circle connected to a filled square represents mating between an affected female and an affected male.
Matching the columns:
$(i)$ Affected female -> $(q)$
(ii) Carrier female -> $(r)$
(iii) Four normal offspring (unspecified sex) -> $(s)$
(iv) Mating -> $(p)$
Therefore,the correct matching is $(i - q), (ii - r), (iii - s), (iv - p)$.

(D) The correct matches are as follows:
$(P)$ Haemophilia is an $X$-linked recessive disorder,represented by the genotype $X^{h}X^{h}$ in females.
$(Q)$ Colour blindness is an $X$-linked recessive disorder,represented by the genotype $X^{c}X^{c}$ in females.
$(R)$ Phenylketonuria is an autosomal recessive metabolic disorder,represented by the genotype $pp$.
$(S)$ Sickle-cell anaemia is an autosomal recessive disorder,represented by the genotype $Hb^{s}Hb^{s}$.
Therefore,the correct matching is $P-i, Q-iv, R-ii, S-iii$.

(A) In a pedigree analysis,a diamond shape with a number inside represents a group of offspring of the specified number whose sex is unspecified. The number inside the diamond indicates the total count of such offspring. Therefore,a diamond with the number $5$ inside represents $5$ unaffected offspring.

(B) Sickle cell anemia is an autosome-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene.
It is caused by a point mutation in the gene coding for the $\beta$-globin chain of hemoglobin.
This mutation results in the substitution of glutamic acid (Glu) by valine (Val) at the sixth position of the $\beta$-globin chain of the hemoglobin molecule.
Due to this amino acid substitution,the mutant hemoglobin molecule undergoes polymerization under low oxygen tension,causing the change in the shape of the $RBC$ from biconcave disc to an elongated sickle-like structure.

(B) $PKU$ stands for Phenylketonuria.
It is an inborn error of metabolism that is inherited as an autosomal recessive trait.
In this condition,the affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine.
As a result,phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives,which leads to mental retardation.

(A) The woman is a carrier for colourblindness because her father was colourblind $(X^{C}X)$. The man is normal $(XY)$.
The cross is as follows:
Parents: $X^{C}X \times XY$
Gametes: $X^{C}, X$ and $X, Y$
Offspring genotypes:
$X^{C}X$ (Carrier daughter,phenotypically normal)
$XX$ (Normal daughter)
$X^{C}Y$ (Colourblind son)
$XY$ (Normal son)
Thus,all daughters are phenotypically normal (though $50 \%$ are carriers),and $50 \%$ of the sons are colourblind.

(B) Red-green color blindness is a well-known $X$-linked recessive disorder.
In this condition,the gene responsible for the trait is located on the $X$ chromosome.
Females have two $X$ chromosomes $(XX)$,so they only express the trait if they are homozygous recessive $(X^cX^c)$.
Males have only one $X$ chromosome $(XY)$,so if they inherit the defective allele $(X^c)$,they will express the condition.
Therefore,red-green color blindness is a classic example of a sex-linked trait.

(A) . Cystic fibrosis is a genetic disorder characterized by the production of thick,sticky mucus that clogs the airways and digestive system. This is the correct match.
$B$. Sickle cell anaemia is caused by defective haemoglobin in $RBC$s,leading to abnormal cell shape and impaired oxygen transport,not brain deterioration.
$C$. Achondroplasia is a form of short-limbed dwarfism,not the presence of extra fingers or toes (which is polydactyly).
$D$. Huntington's disease is a progressive neurodegenerative disorder characterized by the breakdown of nerve cells in the brain,not skeletal or cardiovascular defects.

(D) Alkaptonuria is a rare genetic metabolic disorder. In this condition,the body cannot produce enough of the enzyme homogentisate $1,2-$dioxygenase. As a result,homogentisic acid accumulates in the body and is excreted in large amounts in the urine. When this urine is exposed to air,the homogentisic acid oxidizes,causing the urine to turn black.

(C) Alkaptonuria is a rare genetic metabolic disorder caused by a deficiency of the enzyme homogentisate $1,2-$dioxygenase.
This enzyme deficiency leads to the accumulation of homogentisic acid in the body.
Because the body cannot break down this acid,it is excreted through the kidneys into the urine.
When exposed to air,the homogentisic acid in the urine oxidizes,turning the urine dark or black in color.

(B) Sickle cell anaemia is an autosomal recessive genetic disorder.
It is caused by a point mutation in the gene coding for the $\beta$-globin chain of haemoglobin.
Specifically,the amino acid glutamic acid at the $6^{th}$ position is replaced by valine due to a single base pair substitution ($GAG$ to $GUG$ in $mRNA$).
This change leads to the formation of abnormal haemoglobin,causing $RBCs$ to become sickle-shaped under low oxygen tension.

(C) Sickle-cell anaemia is a genetic disorder caused by a mutation in the haemoglobin gene.
In regions where malaria is endemic,such as parts of Africa,individuals who are heterozygous for the sickle-cell trait (carriers) possess a survival advantage.
This is because the presence of some abnormal haemoglobin interferes with the life cycle of the malaria parasite $(Plasmodium)$,thereby providing partial immunity or resistance against malaria.
Consequently,the gene for sickle-cell anaemia is maintained in the population through natural selection.