Pedigree Analysis

The pedigree shows X-linked recessive inheritance. In $F_2$, the mother is a carrier female ($X^C X$) and the father is unaffected ($XY$). The offspring genotypes are: $X^C X$ (carrier female, 1/4), $XX$ (normal female, 1/4), $X^C Y$ (affected male, 1/4), $XY$ (normal male, 1/4). The probability of an unaffected child being a carrier is 1/4.

Pedigree symbols: affected individuals; mating; mating between relatives (consanguineous mating); parents with male child affected with disease. [add image]

In the chart the father is affected and the mother is unaffected; among the four children, two daughters and one son are affected and one son is unaffected.
Take the trait as recessive: the affected father is homozygous recessive (aa).
Affected children (aa) must receive one recessive allele from each parent, so the unaffected mother must carry the recessive allele: she is heterozygous (Aa).
The cross $aa \times Aa$ gives affected (aa) and unaffected (Aa) children in a 1 : 1 ratio, as seen. So a normal daughter (Aa) was also possible.
The same pattern fits an X-linked recessive trait such as colour-blindness (father $X^cY$, mother $X^CX^c$), so colour-blindness cannot be ruled out; and the father, being affected, cannot be homozygous dominant.
Hence the correct conclusion is that the female parent is heterozygous.

Standard pedigree symbols: a square represents a male and a circle represents a female; an open (unshaded) symbol is an unaffected individual and a filled (shaded) symbol is an affected individual.
A diamond represents an individual whose sex is unspecified, so a filled diamond is not 'male affected'.
An open circle is an unaffected female, not an unaffected male.
An open square is an unaffected male, not an unaffected female.
A single horizontal line joining a square and a circle shows mating; a double line shows mating between relatives (consanguineous mating).
Hence the correct representation is the square and circle joined by a double line = mating between relatives.

In the chart, unaffected parents have affected children; this is possible only if the trait is recessive and both parents are heterozygous carriers.
The affected individuals include a female born to an unaffected father. For an X-linked recessive trait an affected daughter ($X^aX^a$) must have an affected father ($X^aY$), so sex-linked inheritance like haemophilia is ruled out.
The trait is therefore autosomal recessive: carrier parents $Aa \times Aa$ give affected children (aa) of either sex.
Phenylketonuria is an inborn error of metabolism inherited as an autosomal recessive trait (it is not sex-linked).
Hence the chart shows inheritance of a condition like phenylketonuria as an autosomal recessive trait.

The parents are both unaffected, but some children are affected, so the disease is recessive; a dominant disease would appear in at least one parent.
Only sons are affected and the daughters are normal, which points to a sex-linked (X-linked) recessive trait.
The mother must be a carrier ($X^aX$) and the father normal ($XY$).
Each son receives his only X chromosome from the mother, so half the sons are expected to be affected ($X^aY$); here 3 out of 5 sons are affected.
The daughters receive the father's normal X, so none of them is affected (half of them are carriers).
So the suggested mode of inheritance is sex-linked recessive (like haemophilia).