Pedigree Analysis and Mendelian disorders Questions in English

(A) Marriages between close relatives (inbreeding) increase the probability that offspring will inherit two copies of a harmful recessive allele from a common ancestor.
Since close relatives share a significant portion of their genetic material, the likelihood of both parents being carriers for the same rare recessive genetic disorder is much higher than in the general population.
When both parents are carriers, there is a $25\%$ chance for each child to express the recessive trait (homozygous recessive condition).
Therefore, inbreeding increases the expression of deleterious recessive traits in the population.

(A) $1$. The woman has a color-blind son $(X^cY)$ from her first marriage,which means she must be a carrier $(X^CX^c)$ for color blindness.
$2$. In her second marriage,she marries a color-blind man $(X^cY)$.
$3$. The cross is: $X^CX^c \times X^cY$.
$4$. The possible genotypes of the offspring are: $X^CX^c$ (normal vision daughter,carrier),$X^cX^c$ (color-blind daughter),$X^CY$ (normal vision son),and $X^cY$ (color-blind son).
$5$. Thus,$50\%$ of the sons will be color-blind,and $50\%$ of the daughters will be color-blind (the other $50\%$ of daughters will be carriers).

(C) Colorblindness is an $X$-linked recessive trait. Let $X^c$ represent the allele for colorblindness and $X$ represent the normal allele.
$1$. The man is colorblind, so his genotype is $X^c Y$.
$2$. The woman is the daughter of a colorblind father, meaning she inherited an $X^c$ chromosome from her father. Since she is not stated to be colorblind, she must be a carrier, with genotype $X^c X$.
$3$. The cross is $X^c Y \times X^c X$.
$4$. The possible genotypes of the offspring are:
- $X^c X^c$ (Colorblind daughter)
- $X^c X$ (Carrier daughter)
- $X^c Y$ (Colorblind son)
- $XY$ (Normal son)
$5$. From the results, $50\%$ of the sons are colorblind and $50\%$ of the daughters are colorblind. However, looking at the options provided, the question asks for the outcome. Since $50\%$ of sons are colorblind, option $C$ is the correct description of the inheritance pattern.

(B) Hemophilia is an $X$-linked recessive disorder. Let $X^H$ be the normal allele and $X^h$ be the hemophilic allele.
Since the father is normal,his genotype is $X^H Y$.
If the mother is a carrier $(X^H X^h)$,the cross is $X^H Y \times X^H X^h$.
The offspring genotypes will be $X^H X^H$ (normal female),$X^H X^h$ (carrier female),$X^H Y$ (normal male),and $X^h Y$ (hemophilic male).
If the mother is normal $(X^H X^H)$,all children will be normal.
However,in the context of standard genetics problems where a trait is being analyzed,if we assume the mother is a carrier,the male children have a $50\%$ chance of inheriting the $X^h$ chromosome from the mother and the $Y$ chromosome from the father,resulting in hemophilia $(X^h Y)$.

(B) Color blindness is an $X$-linked recessive disorder.
Let $X^c$ represent the allele for color blindness and $X$ represent the normal allele.
The genotype of a color-blind woman is $X^cX^c$.
The genotype of a normal man is $XY$.
When these two cross:
$X^cX^c \times XY \rightarrow X^cX$ (carrier daughter),$X^cY$ (color-blind son).
Since the woman passes one $X^c$ to all her children,all sons will receive the $X^c$ chromosome from their mother and the $Y$ chromosome from their father,making them color-blind $(X^cY)$.
All daughters will receive one $X^c$ from their mother and one $X$ from their father,making them carriers (phenotypically normal) $(X^cX)$.
Therefore,all daughters are normal (carriers) and all sons are color-blind.

(B) Color blindness is an $X$-linked recessive disorder. Let $X^c$ represent the allele for color blindness and $X$ represent the normal allele.
The genotype of a color-blind woman is $X^c X^c$.
The genotype of a normal man is $XY$.
When these parents cross:
$X^c X^c \times XY$
Possible offspring genotypes:
$1$. $X^c X$ (Carrier daughter)
$2$. $X^c Y$ (Color-blind son)
Thus,all sons will be color-blind and all daughters will be carriers.

(D) Color blindness is an $X$-linked recessive disorder. Let $X^c$ represent the allele for color blindness and $X$ represent the normal allele.
For a daughter to be color-blind,her genotype must be $X^c X^c$.
She receives one $X^c$ chromosome from her mother and one $X^c$ chromosome from her father.
Therefore,the father must be color-blind $(X^c Y)$ and the mother must be either a carrier $(X X^c)$ or color-blind $(X^c X^c)$.
Among the given options,the condition where the mother is a carrier $(X X^c)$ and the father is color-blind $(X^c Y)$ allows for the birth of a color-blind daughter $(X^c X^c)$.

(A) Polydactyly is a condition where an individual has extra fingers or toes.
It is a genetic disorder inherited in an autosomal dominant pattern.
This means that only one copy of the mutated gene from either parent is sufficient to cause the condition.
Therefore,the correct option is $A$.

(C) Sickle-cell anemia is an autosomal-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for the gene.
It is caused by a point mutation in the $DNA$ sequence,specifically 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 the single base pair change in the $DNA$ sequence ($GAG$ to $GTG$).
While option $D$ mentions a change in a single base pair,option $C$ provides the most specific and accurate biological description of the molecular defect associated with the disease.

(B) Many genetic disorders are caused by recessive alleles. In a heterozygous state,the individual is a carrier and does not express the disease phenotype,thus the harmful allele is not exposed to natural selection. Because these alleles remain hidden in the gene pool through carriers,they are not eliminated even if they are harmful in the homozygous recessive condition.

(D) Hemophilia is a sex-linked recessive disorder in humans.
Color blindness is also a sex-linked recessive disorder caused by a mutation in certain genes on the $X$ chromosome.
Both conditions follow the same pattern of inheritance,where the gene is located on the $X$ chromosome and typically affects males more frequently than females.
Therefore,color blindness belongs to the same category as hemophilia.

(B) Color blindness in humans is a sex-linked recessive disorder.
It is caused by a mutation in certain genes present on the $X$ chromosome.
Because males have only one $X$ chromosome $(XY)$,they are more frequently affected than females,who have two $X$ chromosomes $(XX)$ and can be carriers.
Therefore,it is a classic example of sex-linked inheritance.

(D) Color blindness is a sex-linked recessive disorder caused by mutations in genes on the $X$ chromosome. It is a genetic condition and is not caused by the deficiency of any vitamin. Deficiency of Vitamin $A$ typically leads to night blindness (nyctalopia),not color blindness. Therefore,the correct answer is $D$.

(B) $Sickle$ $cell$ $anaemia$ is caused due to a point mutation in which glutamic acid is replaced by valine at the $6th$ position of the $\beta$-globin chain. Thus,it is a qualitative defect in the functioning of globin molecules.
$Thalassemia$ is caused due to either mutation or deletion,which ultimately results in a reduced rate of synthesis of one of the globin chains that make up haemoglobin. Hence,it is a quantitative defect in the functioning of globin molecules.

(A) The genotype of a colour-blind man is $X^c Y$,where $X^c$ represents the $X$-linked recessive allele for colour blindness.
The genotype of a woman homozygous for normal colour vision is $XX$.
When these parents produce offspring,the possible genotypes are:
- Daughters: $X^c X$ (carriers with normal vision)
- Sons: $XY$ (normal vision)
Since the son receives his $Y$ chromosome from the father and his $X$ chromosome from the mother (who is homozygous normal),all sons will have normal vision.
Therefore,the probability of their son being colour-blind is $0$.

(B) The correct statements are $(1), (2),$ and $(3)$.
$(1)$ Haemophilia is a sex-linked recessive disease,which shows its transmission from unaffected carrier female to some of the male progeny.
$(2)$ Down's syndrome is a chromosomal disorder caused by the presence of an additional copy of chromosome number $21$ (trisomy of $21$),which is a type of aneuploidy.
$(3)$ Phenylketonuria is an inborn error of metabolism that is inherited as an autosomal recessive trait.
$(4)$ Sickle cell anaemia is an autosomal recessive trait that can be transmitted from parents to the offspring when both the partners are carriers of the gene (or heterozygous). Therefore,statement $(4)$ is incorrect.

(D) Haemophilia is a sex-linked genetic disorder.
It is caused by a mutation in the genes responsible for the production of blood clotting factors.
Specifically,it is an $X$-linked recessive disorder,meaning the defective gene is located on the $X$ chromosome.
Since it is recessive,males (who have only one $X$ chromosome) are more frequently affected than females,as they only need one copy of the defective gene to express the disease.

(B) colour blind man has the genotype $X^cY$,and a normal woman has the genotype $XX$.
When they marry,the offspring are: $X^cX$ (carrier daughter) and $XY$ (normal son).
All daughters are carriers $(X^cX)$ and all sons are normal $(XY)$.
When a carrier daughter $(X^cX)$ marries a normal man $(XY)$,the possible genotypes for their children are: $XX$ (normal daughter),$X^cX$ (carrier daughter),$XY$ (normal son),and $X^cY$ (colour blind son).
The probability of having a colour blind son (the grandson of the original couple) is $1/4$ or $0.25$.

(C) The correct answer is $(C)$ Autosomal dominant.
Analysis of the pedigree:
$1$. The trait appears in every generation (vertical transmission),which is a characteristic of dominant traits.
$2$. Affected individuals have at least one affected parent.
$3$. Both males and females are affected,and affected fathers can pass the trait to their sons,which rules out $X$-linked recessive inheritance.
$4$. Since the trait does not skip generations and appears in both sexes with equal frequency,it is identified as an autosomal dominant trait.

(C) Colour blindness is an $X$-linked recessive disorder.
$1$. The man's father was colour blind,but the man himself is normal because he inherits his $Y$ chromosome from his father and his $X$ chromosome from his mother. Thus,the man's genotype is $XY$.
$2$. The woman had a colour blind mother $(X^cX^c)$ and a normal father $(XY)$. She must have inherited an $X^c$ chromosome from her mother and an $X$ chromosome from her father. Thus,the woman's genotype is $X^cX$.
$3$. Cross: $XY \times X^cX$
- Offspring genotypes: $XX, X^cX, XY, X^cY$
- The male children are $XY$ (normal) and $X^cY$ (colour blind).
$4$. Out of the male children,$50\%$ are colour blind.

(A) : Haemophilia is a sex-linked recessive disease,also known as bleeder's disease.
In this condition,a single protein involved in the cascade of blood clotting is affected (e.g.,antihaemophilic globulin or factor $VIII$ in haemophilia-$A$,or plasma thromboplastin factor $IX$ in haemophilia-$B$).
Because the patient lacks these essential clotting factors,even a minor cut leads to continuous bleeding,which can be fatal.
It is caused by a recessive gene $h$ located on the $X$-chromosome.
$A$ female becomes haemophiliac only if both her $X$-chromosomes carry the recessive gene $(X^hX^h)$.
$A$ female with one recessive allele $(XX^h)$ is a carrier and appears normal because the dominant allele on the other $X$-chromosome ensures normal clotting.
Males are more frequently affected because they have only one $X$-chromosome $(X^hY)$,meaning a single recessive gene is sufficient to express the disease.

(A) Thalassaemia is an autosomal recessive blood disorder.
Let the normal allele be $T$ and the recessive allele for thalassaemia be $t$.
Both parents are carriers,meaning their genotype is $Tt$.
When crossing $Tt \times Tt$,the possible genotypes of the offspring are $TT$ (normal),$Tt$ (carrier),$Tt$ (carrier),and $tt$ (affected).
The probability of an affected child $(tt)$ is $1/4$ or $0.25$.

(A) : Pedigree analysis is a method of studying the inheritance of specific traits over several generations in a family.
It does not confirm that $DNA$ is the carrier of genetic information,as this is a tool for observing phenotypic patterns of inheritance.
Experiments to prove that $DNA$ is the genetic material require molecular biology techniques,such as the Hershey-Chase experiment.
Therefore,the statement that it confirms $DNA$ as the carrier of genetic information is incorrect.

(B) Colour blindness is an $X$-linked recessive trait. Let $X^C$ be the allele for colour blindness and $X$ be the normal allele.
$1$. The man is normal-visioned,so his genotype is $XY$. His father was colour-blind,but this does not affect the man's genotype as he inherits his $Y$ chromosome from his father.
$2$. The woman's father was colour-blind,meaning she must be a carrier of the trait. Her genotype is $XX^C$.
$3$. The cross between the man $(XY)$ and the carrier woman $(XX^C)$ is: $XY \times XX^C$.
$4$. The possible genotypes of the offspring are: $XX$ (normal daughter),$XX^C$ (carrier daughter),$XY$ (normal son),and $X^CY$ (colour-blind son).
$5$. The probability of having a colour-blind daughter is $0$ because a daughter needs to inherit the $X^C$ allele from both parents to be colour-blind,but the father is normal $(XY)$.

(C) The pedigree shows an affected father and an unaffected mother producing both affected and unaffected offspring. This pattern is consistent with an autosomal dominant trait.
Let the dominant allele be $A$ and the recessive allele be $a$.
The affected father must be heterozygous $(Aa)$ because he has an unaffected son $(aa)$.
The unaffected mother must be homozygous recessive $(aa)$.
The cross is $Aa \times aa$.
The offspring genotypes are $Aa$ (affected) and $aa$ (unaffected).
Since the mother is $aa$,she contributes an $a$ allele to all children. The father contributes either $A$ or $a$.
Therefore,the female parent is homozygous recessive $(aa)$,and the male parent is heterozygous $(Aa)$.
Option $(A)$ is incorrect because the female parent is homozygous recessive.
Option $(B)$ is incorrect because they can have a normal daughter $(aa)$.
Option $(C)$ is correct because if the trait were colour blindness ($X$-linked recessive),an affected father would pass the trait to all his daughters,but here he has an unaffected daughter (if we assume the circle represents a daughter,though the pedigree shows two affected daughters and one unaffected son and one affected son). Actually,looking at the pedigree,the father is affected and the mother is unaffected. They have two affected daughters,one unaffected son,and one affected son. This is consistent with an autosomal dominant trait. If it were $X$-linked recessive,an affected father $(X^aY)$ and normal mother $(X^AX^A)$ would produce normal daughters $(X^AX^a)$ and normal sons $(X^AY)$. Thus,it cannot be $X$-linked recessive (colour blindness).

(B) $Sickle-cell$ anaemia is an autosomal recessive genetic disorder in which erythrocytes become sickle-shaped.
It is caused by the formation of an abnormal haemoglobin called $Haemoglobin-S$ $(HbS)$.
$Haemoglobin-S$ is formed when the $6^{th}$ amino acid of the $\beta$-globin chain, $i.e.$, glutamic acid, is replaced by valine due to a point mutation.
This occurs due to a single nucleotide change ($GAG \rightarrow GTG$ at the $DNA$ level, corresponding to $A \rightarrow T$ substitution) in the $\beta$-globin gene.
In the normal $\beta$-globin gene, the $DNA$ sequence is $CCTGAGGAG$, while in $sickle-cell$ anaemia, the sequence is $CCTGTGGAG$.

(A) : This chart shows the inheritance of an autosomal recessive trait like phenylketonuria. An autosomal recessive trait may skip a generation. It appears in the case of marriage between two heterozygous individuals $(Aa \times Aa = 3\ Aa + 1\ aa)$,a recessive individual with a hybrid $(Aa \times aa = 2\ Aa + 2\ aa)$,and two recessive individuals $(aa \times aa = \text{all } aa)$. Phenylketonuria is an inborn,autosomal,recessive metabolic disorder in which the homozygous recessive individual lacks the enzyme phenylalanine hydroxylase. The heterozygous individuals are normal but carriers.

(C) : $A$ congenital disorder is a medical condition that is present at birth. Congenital disorders can be a result of genetic abnormalities,the intrauterine environment,or unknown factors.
Sickle cell anaemia is a group of genetic disorders caused by sickle haemoglobin $(HbS)$. $HbS$ molecules tend to clump together,making red blood cells sticky,stiff,and more fragile,causing them to form into a curved,sickle shape.
Huntington's chorea is an inherited disorder characterized by degenerative changes in the basal ganglia structures,which ultimately result in a severely shrunken brain and enlarged ventricles,abnormal body movements called chorea,and loss of memory.
Both are genetic in origin and present from birth,hence they are classified as congenital disorders.

(A) Colour blindness is a recessive sex-linked trait caused by a gene on the $X$ chromosome.
Let $X^c$ represent the allele for colour blindness and $X$ represent the normal allele.
$A$ colour blind woman has the genotype $X^c X^c$.
$A$ normal visioned man has the genotype $XY$.
When they reproduce,the cross is $X^c X^c \times XY$.
The possible genotypes for the offspring are $X^c X$ (carrier daughters) and $X^c Y$ (colour blind sons).
Since all sons receive their $X$ chromosome from their mother,and the mother is colour blind $(X^c X^c)$,all sons will inherit the $X^c$ allele.
Therefore,all sons will be colour blind.

(C) $1$. Analysis of the pedigree chart shows that the trait appears in the offspring of unaffected parents (the parents in the first generation are both unaffected,but they have an affected son and an affected daughter). This indicates that the trait is recessive.
$2$. Since the trait appears in both males and females (an affected son and an affected daughter in the first generation),and it skips generations,it is autosomal recessive. Specifically,the presence of an affected daughter from unaffected parents confirms it is autosomal,as a sex-linked recessive trait would require the father to be affected to pass it to his daughter.

(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,leading to mental retardation.
Option $(D)$ is incorrect because the failure of cytokinesis leads to polyploidy (e.g.,Down syndrome),not $PKU$.

(C) is incorrect because sickle-cell anaemia is an autosomal recessive trait,not sex-linked.
$R$ is incorrect because sickle-cell anaemia is controlled by a single pair of alleles ($Hb^A$ and $Hb^S$),not by more than two pairs (which would be polygenic inheritance).
Therefore,both $A$ and $R$ are incorrect.

(B) In the given pedigree chart,both parents are unaffected (unshaded),but they have affected (shaded) offspring. This indicates that the trait is recessive.
If the trait were sex-linked recessive,an affected daughter must have an affected father,which is not the case here (the father is unaffected).
Therefore,the trait must be autosomal recessive,where both parents are heterozygous carriers $(Aa)$.
This allows for the possibility of having affected $(aa)$ children,as seen in the chart.

(B) Haemophilia is a genetic disorder caused by a defect in the genes responsible for blood clotting.
It is a $X$-linked recessive trait,meaning the gene responsible for the disease is located on the $X$ chromosome.
Since it is recessive,a female must have the defective gene on both $X$ chromosomes to express the disease,while a male only needs one defective $X$ chromosome to express it because they have only one $X$ chromosome.
Therefore,it is classified as a sex-linked recessive disorder.

(D) Colour blindness is a sex-linked recessive disorder due to a defect in either red or green cone of the eye,resulting in failure to discriminate between red and green colour.
This defect is due to a mutation in certain genes present in the $X$ chromosome.
It occurs in about $8\%$ of males and only about $0.4\%$ of females because the genes that lead to red-green colour blindness are on the $X$ chromosome.
Since males have only one $X$ chromosome and females have two,the incidence is much higher in males.

(D) Mendelian disorders are primarily determined by alteration or mutation in the single gene. Examples include Haemophilia,Colour blindness,Sickle cell anaemia,Phenylketonuria,Thalassemia,and Cystic fibrosis.
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.

(B) Colour blindness is an $X$-linked recessive disorder.
Let $X^c$ represent the allele for colour blindness and $X$ represent the allele for normal vision.
The genotype of the colour blind man is $X^cY$.
The genotype of the homozygous normal vision woman is $XX$.
When they cross: $X^cY \times XX$.
The possible offspring genotypes are $XX^c$ (carrier daughter) and $XY$ (normal son).
Since the son receives the $Y$ chromosome from the father and the $X$ chromosome from the mother (who is homozygous normal),all sons will have normal vision.
Therefore,the probability of their son being colour blind is $0\%$.

(D) Haemophilia is a sex-linked recessive disorder,which means the gene for the disease is located on the $X$ chromosome.
Because it is recessive,a female must have the mutation on both $X$ chromosomes to express the disease,while a male only needs one affected $X$ chromosome.
Queen Victoria was a known carrier of haemophilia,and her pedigree shows the transmission of the disease to several royal families in Europe.
The disease is caused by a defect in a single protein that is part of the cascade of proteins involved in blood clotting.
Therefore,the statement that it is a sex-linked dominant disease is incorrect.

(D) An autosome-linked recessive trait is controlled by a gene located on one of the autosomes (non-sex chromosomes).
Since autosomes are present in equal numbers in both males and females,the inheritance pattern does not show sex-linkage.
For a recessive trait to be phenotypically expressed,the individual must possess two copies of the recessive allele (homozygous recessive,e.g.,$aa$).
If an individual is heterozygous (e.g.,$Aa$),they are a carrier but do not express the disease.
Therefore,the disease is expressed only in the homozygous condition.

(D) The characteristics are analyzed as follows:
$(i)$ Sickle-shaped $RBC$ is a hallmark of sickle-cell anaemia (autosomal recessive).
$(ii)$ Non-stop bleeding due to lack of clotting factors is characteristic of haemophilia ($X$-linked recessive).
$(iii)$ Haemophilia is $X$-linked recessive; thus,a heterozygous female (carrier) transmits the disease to her sons.
$(iv)$ Sickle-cell anaemia is autosomal recessive; thus,both heterozygous parents (carriers) can transmit the disease to their offspring.
Therefore,$(i)$ and $(iv)$ correspond to sickle-cell anaemia,while $(ii)$ and $(iii)$ correspond to haemophilia.

(A) Colorblindness is an $X$-linked recessive disorder.
Let $X^C$ be the allele for colorblindness and $X$ be the allele for normal vision.
The husband has normal vision,so his genotype is $XY$.
The wife has normal vision,but since her father was colorblind,she must be a carrier,so her genotype is $XX^C$.
When we cross $XY \times XX^C$,the possible genotypes for their children are:
$1. XX$ (Normal daughter)
$2. XX^C$ (Carrier daughter)
$3. XY$ (Normal son)
$4. X^CY$ (Colorblind son)
Since colorblindness is $X$-linked recessive,a daughter can only be colorblind if she inherits the $X^C$ allele from both parents. In this case,the father is normal $(XY)$,so he cannot pass the $X^C$ allele to his daughter.
Therefore,the probability of their daughter being colorblind is $0\%$.

(D) 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 frequently affected than males.

(D) 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,which leads to the accumulation of phenylalanine in the body.
Cataract,leprosy,and night blindness are generally not classified as genetic disorders in the context of Mendelian inheritance; leprosy is an infectious disease,while cataract and night blindness can be caused by environmental factors,aging,or nutritional deficiencies.

(A) Color blindness is an $X$-linked recessive trait. For a girl to be color-blind,she must inherit the recessive allele $(X^c)$ from both parents. This means her father must be color-blind $(X^cY)$ and her mother must be either a carrier $(X^CX^c)$ or color-blind $(X^cX^c)$. Since the question implies the rarity of the condition,the most specific scenario is that the father is color-blind and the mother is at least a carrier. However,looking at the provided options,the condition where a girl is color-blind requires the father to be color-blind and the mother to carry the gene. Option $A$ is the most accurate description of the genetic requirement.

(A) Color blindness is an $X$-linked recessive disorder.
Let the genotype of the color-blind mother be $X^cX^c$ and the normal father be $XY$.
When these parents cross:
- The mother produces gametes $X^c$ and $X^c$.
- The father produces gametes $X$ and $Y$.
- The offspring genotypes will be:
- $X^cX$ (Carrier daughter)
- $X^cY$ (Color-blind son)
Therefore,all sons will be color-blind and all daughters will be normal carriers.

(C) Hemophilia is an $X$-linked recessive disorder.
If the mother is a carrier $(X^HX^h)$ and the father is normal $(X^HY)$,the cross results in:
$1$. $X^HX^H$ (Normal daughter)
$2$. $X^HX^h$ (Carrier daughter)
$3$. $X^HY$ (Normal son)
$4$. $X^hY$ (Hemophilic son)
In this scenario,$50\%$ of sons are hemophilic and $50\%$ of daughters are carriers. This confirms that the mother carries the recessive gene on one of her $X$ chromosomes.

(D) Color blindness is an $X$-linked recessive disorder.
Let the normal vision allele be $X^C$ and the color-blind allele be $X^c$.
$A$ normal male is $X^CY$ and a normal female is $X^CX^C$ or $X^CX^c$ (carrier).
If the mother is a carrier $(X^CX^c)$ and the father is normal $(X^CY)$,the cross is $X^CX^c \times X^CY$.
The offspring genotypes are $X^CX^C$ (normal female),$X^CX^c$ (carrier female),$X^CY$ (normal male),and $X^cY$ (color-blind male).
Thus,a color-blind male child is possible if the mother is a carrier. $A$ woman is a carrier if her father was color-blind (since she receives one $X$ chromosome from her father).

(D) Albinism is an autosomal recessive trait. Let $A$ be the dominant allele for normal skin pigmentation and $a$ be the recessive allele for albinism.
$1$. The woman's father was an albino $(aa)$,so she must have inherited one recessive allele $(a)$ from him. Since she is phenotypically normal,her genotype must be $Aa$.
$2$. The man is an albino,so his genotype must be $aa$.
$3$. The cross between the woman $(Aa)$ and the man $(aa)$ is: $Aa \times aa$.
$4$. The resulting offspring genotypes are $Aa$ (Normal) and $aa$ (Albino) in a $1:1$ ratio.
Therefore,the ratio of their offspring will be $1$ Normal : $1$ Albino.

(D) Color blindness is an $X$-linked recessive disorder. Let the allele for normal vision be $X^C$ and the allele for color blindness be $X^c$.
The genotype of a color-blind woman is $X^cX^c$.
The genotype of a man with normal vision is $X^CY$.
When these two individuals cross:
Parents: $X^cX^c$ (female) $\times$ $X^CY$ (male)
Gametes: $X^c$ and $X^c$ (from mother); $X^C$ and $Y$ (from father)
Offspring genotypes:
$X^CX^c$ (Carrier daughter, normal vision)
$X^CX^c$ (Carrier daughter, normal vision)
$X^cY$ (Color-blind son)
$X^cY$ (Color-blind son)
Therefore, all sons will be color-blind, and all daughters will be carriers with normal vision.