Pedigree Analysis and Mendelian disorders Questions in English

(B) Albinism is an autosomal recessive disorder. Let the normal allele be '$A$' and the albino allele be '$a$'.
Since the first child is an albino $(aa)$,both parents must be carriers of the recessive allele $(Aa)$.
When two carrier parents $(Aa \times Aa)$ have a child,the possible genotypes are $AA, Aa, Aa, aa$.
The probability of an offspring being an albino $(aa)$ is $1/4$ or $25\%$.
Each child's birth is an independent event,so the probability for the second child remains $25\%$.

(C) Let $X^H$ be the gene for normal blood clotting and $X^h$ be the gene for hemophilia. Let $X^C$ be the gene for normal vision and $X^c$ be the gene for color blindness. The woman is a carrier for hemophilia $(X^H X^h)$ and also carries the color blindness gene on one $X$ chromosome. Assuming the genes are linked,her genotype is $X^{Hc} X^h$ (where $c$ is the color blindness allele). The normal man has the genotype $X^{HC} Y$.
When they cross:
$1$. $X^{Hc} X^h \times X^{HC} Y$
$2$. Offspring genotypes: $X^{Hc} X^{HC}$ (Carrier daughter),$X^{Hc} Y$ (Hemophilic and color-blind son),$X^h X^{HC}$ (Carrier daughter),$X^h Y$ (Hemophilic son).
Thus,the sons will be $50\%$ hemophilic and color-blind,and $50\%$ hemophilic.

(B) Hemophilia is an $X$-linked recessive disorder.
Let $X^H$ represent the normal allele and $X^h$ represent the hemophilic allele.
The genotype of the hemophilic man is $X^hY$ and the genotype of the normal woman is $X^HX^H$.
When they cross: $X^hY \times X^HX^H$.
The offspring genotypes will be $X^HX^h$ (carrier daughter) and $X^HY$ (normal son).
Since all daughters are carriers (phenotypically normal) and all sons are normal,all offspring are phenotypically normal.

(B) Colorblindness is an $X$-linked recessive disorder.
Let $X^C$ represent the allele for normal vision and $X^c$ represent the allele for colorblindness.
The woman's father was colorblind $(X^cY)$,so she must be a carrier $(X^CX^c)$.
The man is normal $(X^CY)$.
The cross is: $X^CX^c \times X^CY$.
The possible offspring are: $X^CX^C$ (normal daughter),$X^CX^Y$ (normal son),$X^CX^c$ (carrier daughter),and $X^cY$ (colorblind son).
Among the sons,$50\%$ are normal $(X^CY)$ and $50\%$ are colorblind $(X^cY)$.

(D) $1$. The parents are unaffected,but they have affected offspring,which indicates that the trait is recessive.
$2$. The disease appears only in sons ($5$ sons,$3$ affected) and not in daughters ($2$ daughters,$0$ affected).
$3$. This pattern of inheritance,where the trait skips a generation and affects primarily males,is characteristic of $X$-linked recessive inheritance.
$4$. In $X$-linked recessive inheritance,the mother is a carrier $(X^CX)$ and the father is unaffected $(XY)$. The sons inherit the $Y$ chromosome from the father and the $X$ chromosome from the mother. If the mother passes the affected $X$ chromosome to her son,he will express the disease.

(C) Color blindness is an $X$-linked recessive trait. Let $X^C$ be the allele for color blindness and $X$ be the allele for normal vision.
$1$. The woman's father was color-blind $(X^CY)$,so she must have inherited the $X^C$ allele from him. Since she has normal vision,her genotype is $XX^C$.
$2$. The man is color-blind,so his genotype is $X^CY$.
$3$. The cross is $XX^C \times X^CY$.
$4$. The possible genotypes for their children are: $XX^C$ (normal vision daughter),$X^CX^C$ (color-blind daughter),$XY$ (normal vision son),and $X^CY$ (color-blind son).
$5$. For any son,there is a $50\%$ chance of being color-blind and a $50\%$ chance of having normal vision. Therefore,the fourth boy can be either color-blind or have normal vision.

(C) Hemophilia is a sex-linked recessive disorder.
It is caused by a mutation in the genes located on the $X$ chromosome.
Since males have only one $X$ chromosome $(XY)$,a single recessive gene is sufficient to express the disease.
In contrast,females have two $X$ chromosomes $(XX)$,so they require two copies of the recessive gene to express the disease,making them less likely to be affected compared to males.

(B) $1$. In the given pedigree,the parents are unaffected (unshaded),but they have affected (shaded) offspring of both sexes (a male and a female).
$2$. This pattern of inheritance,where unaffected parents produce affected offspring,is characteristic of an autosomal recessive trait.
$3$. Phenylketonuria is a well-known example of an autosomal recessive disorder.
$4$. Sex-linked recessive traits (like Haemophilia) typically show a different pattern,often affecting males more frequently and skipping generations in a specific way related to the $X$-chromosome.
$5$. Therefore,the pedigree represents an autosomal recessive trait like Phenylketonuria.

(B) Color blindness is an $X$-linked recessive disorder.
Let $X^C$ be the allele for color blindness and $X$ be the normal allele.
The man is normal-visioned,so his genotype is $XY$. Since his father was color-blind $(X^CY)$,he inherited his $Y$ chromosome from his father and his $X$ chromosome from his mother. Thus,his genotype is $XY$.
The woman's father was color-blind $(X^CY)$,so she must have inherited the $X^C$ allele from him. Since she is not mentioned as color-blind,she is a carrier with the genotype $X^CX$.
Cross: $XY$ (man) $\times$ $X^CX$ (woman).
Possible offspring genotypes: $XX, X^CX, XY, X^CY$.
The female offspring genotypes are $XX$ (normal) and $X^CX$ (carrier).
Neither of the female offspring genotypes is color-blind $(X^CX^C)$.
Therefore,the probability of their daughter being color-blind is $0\%$.

(C) $Hemophilia$ is a sex-linked recessive disorder.
It is caused by a mutation in the genes that code for proteins involved in the blood clotting cascade.
Since it is a recessive trait,it is expressed only when the individual is homozygous for the recessive allele in females or hemizygous in males.
Therefore,the statement that it is a dominant disease is incorrect.

(C) Thalassemia is an autosomal recessive disorder. Let the normal allele be represented by '$A$' and the recessive disease-causing allele by '$a$'.
Since both parents are carriers,their genotypes are '$Aa$'.
When we perform a cross between two carriers $(Aa \times Aa)$:
The possible offspring genotypes are:
$AA$ (Normal - $25\%$)
$Aa$ (Carrier - $50\%$)
$aa$ (Affected - $25\%$)
Therefore,the probability of having an affected child $(aa)$ is $25\%$.

(A) Colorblindness is an $X$-linked recessive trait.
$1$. The man's father was colorblind,so the man must have received his $Y$ chromosome from his father and his $X$ chromosome from his mother. Since he is a male,his genotype is $X^CY$ (where $X^C$ is the colorblind allele). However,the man himself is normal,so his genotype is $XY$.
$2$. The woman has a colorblind mother $(X^CX^C)$ and a normal father $(XY)$. She must have inherited one $X^C$ from her mother and one $X$ from her father. Thus,her genotype is $X^CX$.
$3$. The cross is $XY$ (man) $\times$ $X^CX$ (woman).
$4$. The possible genotypes of the offspring are: $XX^C$ (carrier daughter),$XX$ (normal daughter),$X^CY$ (colorblind son),and $XY$ (normal son).
$5$. Among the male children,the genotypes are $X^CY$ and $XY$. The probability of a male child being colorblind is $1/2$ or $50\%$.

(C) Color blindness is an $X$-linked recessive trait. Let $X^c$ represent the color-blind allele and $X$ represent the normal allele.
$1$. The man is color-blind $(X^c Y)$. The woman is normal and has no family history,so she is homozygous normal $(XX)$.
$2$. Their offspring: The daughter receives $X^c$ from her father and $X$ from her mother,making her a carrier $(X X^c)$. The son receives $Y$ from his father and $X$ from his mother,making him normal $(XY)$.
$3$. The daughter (carrier $X X^c$) marries a normal man $(XY)$.
$4$. Their children's genotypes: $XX$ (normal daughter),$X X^c$ (carrier daughter),$XY$ (normal son),$X^c Y$ (color-blind son).
$5$. The probability of a grandson being color-blind is $0.5$ (since $X^c Y$ is one of two possibilities for a son),and the probability of a granddaughter being color-blind is $0$ (since she must inherit $X^c$ from her father,who is normal).
$6$. The question asks for the probability of the daughter's children being color-blind. Since the daughter has a $1/4$ chance of having a color-blind son and $0$ chance of a color-blind daughter,the overall probability for any child is $0.25$.

(B) $1$. Analyze the pedigree: The trait appears in every generation (vertical transmission),which is characteristic of dominant traits.
$2$. Check for Autosomal Dominant inheritance: In autosomal dominant traits,an affected individual must have at least one affected parent. Here,the affected individuals in generation $II$ and $III$ have affected parents.
$3$. Check for $X$-linked dominant: If it were $X$-linked dominant,an affected father would pass the trait to all his daughters. In generation $I$,the affected father has an unaffected daughter in generation $II$,which rules out $X$-linked dominant.
$4$. Conclusion: The pattern is consistent with Autosomal Dominant inheritance,as it does not skip generations and affected individuals have affected parents.

(C) Color blindness is an $X$-linked recessive trait.
Let $X^C$ be the allele for color blindness and $X$ be the normal allele.
The genotype of the colorblind man is $X^C Y$.
The genotype of the woman with normal vision is $X X$.
When they cross: $X^C Y \times X X \rightarrow X^C X, X^C X, XY, XY$.
The offspring are: two carrier daughters $(X^C X)$ and two normal sons $(XY)$.
Since the sons receive the $Y$ chromosome from the father and the $X$ chromosome from the mother,and the mother is normal $(XX)$,all sons will have normal vision.
Therefore,the probability of their son being colorblind is $0$.

(C) The correct answer is $C$.
$1$. Phenylketonuria is an autosomal recessive disorder caused by a mutation in the gene for the enzyme phenylalanine hydroxylase.
$2$. Down syndrome is a chromosomal disorder caused by the presence of an extra copy of chromosome $21$ (trisomy $21$),which is a type of aneuploidy.
$3$. Sickle cell anemia is an autosomal recessive disorder,not an $X$-linked disorder. It is caused by a mutation in the hemoglobin beta gene on chromosome $11$.
$4$. Hemophilia is a well-known $X$-linked recessive disorder where blood fails to clot properly.

(A) Hemophilia is a sex-linked genetic disorder where the gene responsible for the clotting factor is located on the $X$ chromosome.
Since it is a recessive trait,it is expressed more frequently in males $(XY)$ because they have only one $X$ chromosome,whereas females $(XX)$ require two copies of the recessive allele to express the disease.
Therefore,it is specifically classified as an $X$-linked recessive gene disorder.

(C) Thalassemia is a quantitative disorder where there is a defect in the synthesis of globin chains,leading to the production of fewer globin molecules.
Sickle-cell anemia is a qualitative disorder caused by a point mutation in the gene coding for the $\beta$-globin chain,resulting in the synthesis of abnormal hemoglobin $(HbS)$.

(D) Sickle cell anemia is a genetic disorder caused by a mutation in the hemoglobin gene.
Individuals who are heterozygous for the sickle cell trait $(Hb^A Hb^S)$ possess red blood cells that are less hospitable to the malaria parasite,$Plasmodium$.
This provides a selective advantage in regions where malaria is endemic,as these individuals are less likely to develop severe or fatal cases of malaria compared to individuals with normal hemoglobin $(Hb^A Hb^A)$.

(B) Sickle cell anemia is an autosomal recessive genetic disorder caused by a mutation in the hemoglobin gene.
Huntington's chorea is an autosomal dominant genetic disorder caused by a mutation in the huntingtin gene.
Since both of these conditions are inherited from parents to offspring through genes,they are classified as congenital disorders (genetic disorders present at birth).
Therefore,the correct option is $B$.

(C) Sickle cell anemia is a genetic disorder caused by a mutation in the hemoglobin gene.
Individuals who are heterozygous for the sickle cell trait $(Hb^A Hb^S)$ possess a survival advantage in regions where malaria is endemic.
The presence of sickle-shaped red blood cells inhibits the growth and reproduction of the malaria parasite $(Plasmodium)$.
Therefore,natural selection maintains the sickle cell gene in the population because it confers resistance to malaria,preventing its complete elimination despite the disease being harmful in the homozygous recessive $(Hb^S Hb^S)$ state.

(C) Haemophilia is a sex-linked recessive disorder caused by the deficiency of blood clotting factor $VIII$ (Haemophilia $A$) or factor $IX$ (Haemophilia $B$).
In patients with haemophilia,the blood clotting mechanism is impaired because the cascade of reactions required for fibrin formation is interrupted due to the absence of these specific clotting factors.
The Assertion is correct because haemophilia is indeed characterized by the failure to produce functional clotting factor $VIII$.
The Reason is incorrect because the primary defect in haemophilia is not a low concentration of platelets or a deficiency in prothrombin production; rather,it is a genetic deficiency of specific clotting factors. Platelet counts in haemophiliacs are typically normal.

(B) Colour blindness is an $X$-linked recessive trait.
$1$. The woman is normal but her father was colour blind,so she must be a carrier for the trait. Her genotype is $X^hX$.
$2$. The man is normal,so his genotype is $XY$.
$3$. The cross between the carrier woman $(X^hX)$ and the normal man $(XY)$ results in the following offspring:
- $X^hX$ (Carrier daughter)
- $XX$ (Normal daughter)
- $X^hY$ (Colour blind son)
- $XY$ (Normal son)
$4$. Among the sons,one is colour blind $(X^hY)$ and one is normal $(XY)$. Therefore,$50\%$ of the sons would be colour blind.

(B) Thalassemia is an autosomal recessive disorder. Let '$A$' represent the dominant normal allele and '$a$' represent the recessive mutant allele.
Carrier parents have the genotype '$Aa$'.
When two carrier parents $(Aa \times Aa)$ reproduce,the possible genotypes of the offspring are determined by a Punnett square:
- $AA$ (Normal child): $25\%$
- $Aa$ (Carrier child): $50\%$
- $aa$ (Affected child): $25\%$
Therefore,the probability of having an affected child $(aa)$ is $25\%$.

(B) Huntington's disease is an autosomal dominant disorder. In an autosomal dominant pedigree,an affected individual must have at least one affected parent.
In the given pedigree,individual $11$ is affected (represented by a filled circle). Since the disease is autosomal dominant,at least one of her parents ($6$ or $7$) must also be affected (represented by a filled symbol).
However,in the pedigree,both individuals $6$ and $7$ are unaffected (represented by empty symbols). Therefore,this pedigree is erroneous because it violates the rule of autosomal dominant inheritance.

(A) Phenylketonuria $(PKU)$ is an autosomal recessive genetic disorder.
It is caused by the deficiency of the enzyme phenylalanine hydroxylase,which is responsible for converting the amino acid phenylalanine into tyrosine.
Due to the lack of this enzyme,phenylalanine accumulates in the body and is converted into phenylpyruvic acid and other derivatives.
These derivatives are excreted in the urine,hence the name Phenylketonuria. Therefore,both the Assertion and the Reason are correct,and the Reason explains why the condition is named as such and why the substance appears in the urine.

(C) Haemophilia,also known as bleeder's disease,is a classic example of recessive sex-linked inheritance in humans.
It is masked in the heterozygous condition.
The person suffering from this disease lacks clotting factors $VIII$ and $IX$,which are essential for blood coagulation.
$A$ small cut may lead to continuous bleeding.
Men are primarily affected by this disease,while women typically act as carriers.
The mutation of a structural gene on chromosome number $15$ is associated with Marfan syndrome,not haemophilia.
Marfan syndrome results in the formation of abnormal connective tissues and extreme looseness of joints.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) Haemophilia is a group of hereditary genetic disorders that impair the body's ability to control blood clotting or coagulation.
In its most common form,Haemophilia $A$,the blood clotting factor $VIII$ is absent or deficient.
In Haemophilia $B$,factor $IX$ is deficient.
Factor $VIII$ acts as a cofactor for factor $IXa$,which in the presence of $Ca^{+2}$ and phospholipids,forms a complex that converts factor $X$ to its activated form,$Xa$.
Defects in the gene encoding factor $VIII$ result in Haemophilia $A$,which is a recessive $X$-linked coagulation disorder.
The Reason statement is incorrect because the concentration of platelets in haemophilia patients is typically normal; the disorder is caused by a deficiency in specific clotting factors,not a deficiency in platelet count or prothrombin production.

(N/A) Pedigree analysis is a record of the occurrence of a trait in several generations of a family. It is based on the fact that certain characteristic features are heritable in a family,for example,eye colour,skin colour,hair form and colour,and other facial characteristics. Along with these features,there are other genetic disorders such as Mendelian disorders that are inherited in a family,generation after generation. Hence,by using pedigree analysis for the study of specific traits or disorders,generation after generation,it is possible to trace the pattern of inheritance. In this analysis,the inheritance of a trait is represented as a tree,called a family tree. Genetic counselors use pedigree charts for the analysis of various traits and diseases in a family and predict their inheritance patterns. It is useful in preventing hemophilia,sickle cell anemia,and other genetic disorders in future generations.

(N/A) Two autosomal genetic disorders are as follows:
$1$. Sickle cell anaemia: It is an autosomal-linked recessive disorder caused by a point mutation in the beta-globin chain of the haemoglobin molecule. The disease is controlled by $Hb^A$ and $Hb^S$ alleles. Homozygous individuals $(Hb^S Hb^S)$ show symptoms,while heterozygous individuals $(Hb^A Hb^S)$ are carriers.
Symptoms: Rapid heart rate,breathlessness,delayed growth,jaundice,weakness,fever,excessive thirst,chest pain,and decreased fertility.
$2$. Down's syndrome: It is an autosomal disorder caused by the trisomy of chromosome $21$.
Symptoms: Short stature,round head,open mouth,protruding tongue,short neck,slanting eyes,broad short hands,and retarded mental and physical growth.

(N/A) Pedigree analysis is the study of the inheritance of a particular trait,abnormality,or disease in a family over several generations. It is represented as a family tree.
In human genetics,pedigree analysis is a powerful tool because controlled crosses,which are possible in organisms like pea plants,cannot be performed in humans due to reasons such as: $(i)$ The progeny produced is very small (usually one) and takes a long time to mature. $(ii)$ Controlled crosses are not ethically or practically possible.
The standard symbols used in pedigree analysis are as follows:
| Symbol | Meaning |
| :--- | :--- |
| Square | Male |
| Circle | Female |
| Diamond | Sex unspecified |
| Shaded symbols | Affected individuals |
| Square-Circle connected by horizontal line | Mating |
| Square-Circle connected by double horizontal line | Mating between relatives (consanguineous mating) |
| Parents connected by horizontal line with offspring below | Parents with children (in order of birth from left to right) |
| Parents with a shaded square below | Parents with an affected male child |
| Diamond with number $5$ inside | Five unaffected offspring |

(N/A) Color blindness is a sex-linked recessive disorder caused by a mutation in certain genes present on the $X$ chromosome.
This defect results in the failure to discriminate between red and green colors due to the defect in either red or green cone cells of the eye.
It occurs in about $8\%$ of males and only about $0.4\%$ of females because the genes for red-green color blindness are located on the $X$ chromosome.
Males have only one $X$ chromosome and a $Y$ chromosome,whereas females have two $X$ chromosomes.
The son of a woman who carries the gene has a $50\%$ chance of being color blind.
The mother is not herself color blind because the gene is recessive and its effect is suppressed by her other normal dominant gene.
$A$ daughter will be color blind only when her mother is a carrier and her father is color blind.

(N/A) Haemophilia is a sex-linked recessive disorder that is transmitted from an unaffected carrier female to some of the male progeny.
$1$. Genetic Basis: The gene for haemophilia is located on the $X$ chromosome. Since it is recessive,females $(XX)$ require two copies of the defective gene to express the disease,whereas males $(XY)$ only require one copy.
$2$. Physiological Effect: The disorder results in a defect in the blood clotting mechanism. $A$ single protein that is part of the cascade of proteins involved in the clotting of blood is affected.
$3$. Clinical Manifestation: Due to the defect,a simple cut results in non-stop bleeding in an affected individual.
$4$. Inheritance Pattern: The possibility of a female becoming a haemophilic is extremely rare because the mother of such a female has to be at least a carrier and the father should be haemophilic (unviable in the later stage of life).

(N/A) Sickle-cell anemia: This is an autosome-linked recessive trait that can be transmitted from parents to offspring when both partners are carriers for the gene (or heterozygous).
The disease is controlled by a pair of alleles,$Hb^A$ and $Hb^S$. Out of the three possible genotypes,only individuals homozygous for $Hb^S$ $(Hb^SHb^S)$ show the disease symptoms.
Heterozygous $(Hb^AHb^S)$ individuals appear unaffected but are carriers of the disease. There is a $50\%$ probability of transmission of the mutant gene to the progeny.
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 of amino acid in the globin protein results due to the single base pair substitution at the sixth codon of the $\beta$-globin gene from $GAG$ to $GUG$.
Under low oxygen tension,the mutant hemoglobin molecule undergoes polymerization,causing the shape of the red blood cell to change from biconcave disc to an elongated sickle-like structure.
Phenylketonuria $(PKU)$: This is an inborn error of metabolism that is also inherited as an autosomal recessive trait.
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.
Accumulation of these in the brain results in mental retardation. These are also excreted through urine because of poor absorption by the kidney.

(N/A) Thalassemia is an autosomal recessive blood disorder characterized by abnormal hemoglobin production.
$1$. It is caused by mutations in the genes that code for the globin chains of hemoglobin.
$2$. The disorder is classified into two main types: $\alpha$-thalassemia and $\beta$-thalassemia.
$3$. In $\alpha$-thalassemia,the production of the $\alpha$-globin chain is affected,which is controlled by two closely linked genes,$HBA1$ and $HBA2$,on chromosome $16$.
$4$. In $\beta$-thalassemia,the production of the $\beta$-globin chain is affected,which is controlled by a single gene,$HBB$,on chromosome $11$.
$5$. The condition leads to the formation of abnormal hemoglobin molecules,resulting in the destruction of red blood cells (anemia).

(B) Color blindness is a sex-linked recessive disorder caused by a mutation in genes located on the $X$ chromosome.
Since males have only one $X$ chromosome $(XY)$,a single recessive allele is sufficient to express the trait.
Females have two $X$ chromosomes $(XX)$,so they only express the trait if they are homozygous recessive $(X^cX^c)$.
If a female has one recessive allele,she acts as a carrier $(X^cX)$ and does not show the symptoms.

(A) Thalassemia is a blood-related genetic disorder characterized by the production of insufficient hemoglobin.
In $\beta$-thalassemia,the production of the $\beta$-globin chain of hemoglobin is affected.
It is caused by mutations or deletions in the $HBB$ gene located on chromosome $11$.
Since it is an autosomal recessive disorder,if both parents are carriers,there is a $25\%$ chance of the offspring being affected.
Therefore,genetic counseling and testing before marriage are recommended to prevent the transmission of the disease to the next generation.

(A) Pedigree analysis is a powerful tool used to study human genetics and understand the inheritance patterns of genetic disorders.
After the rediscovery of Mendel's work,the study of the inheritance patterns of traits in humans began.
Pedigree analysis is the record-keeping of a specific trait across several generations in a human family.
In this analysis,data regarding the history of a specific trait is collected,and then the expression of that trait is represented through a chart or family tree.

(N/A) The trait is an autosomal recessive trait.
Since the trait is absent in the parents but appears in the offspring,the parents must be heterozygous (carriers) for the trait.
Because the trait appears in both male and female offspring,it is not sex-linked.
Therefore,the inheritance pattern is autosomal recessive,where the genotype of both parents is $Aa$ and the affected offspring have the genotype $aa$.

(B) The harmful alleles are typically eliminated from a population over time,yet sickle cell anemia persists in the human population. This is because while the homozygous condition $(Hb^S Hb^S)$ is harmful,the heterozygous condition $(Hb^A Hb^S)$ provides a survival advantage against malaria.
The malaria-causing parasite,$Plasmodium$,spends part of its life cycle in red blood cells and triggers an abnormal drop in oxygen levels within the cell.
In carriers $(Hb^A Hb^S)$,this drop in oxygen is sufficient to trigger the sickling of red blood cells. These sickled cells are then rapidly removed from circulation by the spleen,which significantly limits the progression of the malarial infection.
These individuals possess a high resistance to malaria and have a greater chance of surviving outbreaks.
This resistance to infection is the primary reason the $Hb^S$ allele persists in the population. It is found at the highest frequency in regions where malaria is or was a serious health problem.
In contrast,normal individuals $(Hb^A Hb^A)$ are fully susceptible to $Plasmodium$ infection.

(C) Red-green colour blindness is an $X$-linked recessive disorder.
Females have two $X$ chromosomes $(XX)$,so for a female to be colourblind,both $X$ chromosomes must carry the recessive allele $(X^cX^c)$.
In contrast,males have only one $X$ chromosome $(XY)$. If an $X$ chromosome carries the recessive allele $(X^cY)$,the male will express the trait because there is no second $X$ chromosome to mask the effect of the recessive allele.
Therefore,the probability of males being affected is significantly higher than that of females.

(N/A) The gene for red-green colour blindness is $X$-linked recessive.
In humans,a son inherits his $Y$-chromosome from his father and his $X$-chromosome from his mother.
Since the trait is $X$-linked,it cannot be passed directly from father to son.
Therefore,the son did not inherit the trait from his father.
In this case,the mother must be a carrier (heterozygous) for the colour blindness gene,and she passed the affected $X$-chromosome to her son.

(D) The woman is a carrier for colour blindness because her father was colour-blind (genotype $X^c Y$). She inherited the $X^c$ chromosome from her father, so her genotype is $X^C X^c$. The man is normal-visioned, so his genotype is $X^C Y$.
Cross: $X^C X^c$ (woman) $\times$ $X^C Y$ (man)
Offspring genotypes:
$1$. $X^C X^C$ (Normal-visioned daughter)
$2$. $X^C X^c$ (Carrier daughter, normal vision)
$3$. $X^C Y$ (Normal-visioned son)
$4$. $X^c Y$ (Colour-blind son)
Probability:
- Daughters: $0 \%$ chance of being colour-blind ($50 \%$ are carriers).
- Sons: $50 \%$ chance of being colour-blind.

(D) $1$. Thalassemia is an autosomal recessive blood disorder.
$2$. Haemophilia is an $X$-linked recessive disorder.
$3$. Phenylketonuria is an autosomal recessive metabolic disorder.
$4$. Sickle cell anaemia is an autosomal recessive trait caused by a mutation in the gene located on chromosome $11$.

(C) $1$. Hemophilia is an $X$-linked recessive disorder.
$2$. Phenylketonuria is an autosomal recessive disorder.
$3$. Sickle cell anemia is an autosomal recessive disorder caused by a mutation in the gene on chromosome $-11$.
$4$. Thalassemia is an autosomal recessive blood disorder.
Therefore,the correct pair is $C$.

(N/A) Mendelian disorders are primarily determined by alterations or mutations in a single gene.
These disorders are transmitted to the offspring following the principles of inheritance.
The pattern of inheritance of Mendelian disorders can be traced in a family through pedigree analysis.
Some common and prevalent Mendelian disorders include Haemophilia,Cystic fibrosis,Sickle cell anaemia,Colour blindness,Phenylketonuria,and Thalassemia.
Such Mendelian disorders may be dominant or recessive.
Through pedigree analysis,one can easily determine whether the trait in question is dominant or recessive.
Similarly,the trait may also be linked to the sex chromosome,as in the case of haemophilia.
It is evident that this $X$-linked recessive trait shows transmission from a carrier female to male progeny.
$A$ representative pedigree is shown in the figure for dominant and recessive traits.
Colour blindness is a sex-linked recessive disorder caused by a defect in either the red or green cones of the eye,resulting in a failure to discriminate between red and green colours.

(N/A)

Haemophilia	Sickle-cell Anaemia
$(1)$ It occurs due to a defective recessive allele present on the $X$-chromosome.	$(1)$ It occurs due to point mutation,i.e.,a single base pair change leads to a change in amino acid.
$(2)$ The defective allele produces a non-functional protein involved in the blood clotting cascade.	$(2)$ In this,glutamic acid $(Glu)$ is replaced by valine $(Val)$ at the sixth position of the beta-globin chain of the haemoglobin molecule.
$(3)$ It is a sex-linked recessive disorder.	$(3)$ It is an autosomal-linked recessive disorder.
$(4)$ More males are affected than females.	$(4)$ Both males and females are affected equally.

(N/A)

Mendelian disorders	Chromosomal disorders
$(1)$ These are caused by an alteration or mutation in a single gene.	$(1)$ These are caused by the absence,excess,or abnormal arrangement of one or more chromosomes.
$(2)$ They are transmitted to subsequent generations following Mendelian principles of inheritance.	$(2)$ They are generally not transmitted as the affected individuals are often sterile.
$(3)$ They can be recessive or dominant in nature.	$(3)$ These are typically associated with large-scale genomic changes and are often dominant in their phenotypic expression.
$(4)$ Examples: Colour blindness,Phenylketonuria,Haemophilia.	$(4)$ Examples: Down's syndrome,Turner's syndrome,Klinefelter's syndrome.

(N/A) Phenylketonuria $(PKU)$ is an inborn error of metabolism caused by the deficiency of the enzyme $Phenylalanine$ hydroxylase.
This enzyme is essential for converting the amino acid $Phenylalanine$ into $Tyrosine$ in the liver.
Due to the lack of this enzyme, $Phenylalanine$ accumulates in the body and is diverted into an alternative metabolic pathway.
It gets converted into $Phenylpyruvic$ acid and other related derivatives.
These metabolites accumulate in the brain, leading to mental retardation, and are also excreted in the urine due to poor reabsorption by the kidneys.

(N/A) The gene for haemophilia is present on the $X$-chromosome. $A$ male has only one $X$-chromosome,which he receives from his mother,and a $Y$-chromosome,which he receives from his father. During reproduction,the human male passes his $X$-chromosome to his daughters and his $Y$-chromosome to his sons. Since the haemophilia gene is located on the $X$-chromosome,a father cannot pass this gene to his son because the son inherits the $Y$-chromosome from the father,not the $X$-chromosome.