Alright Mbaliey — let's do this. 12.1 a) Step 1: Identify individual 4 in the pedigree. Individual 4 is an unshaded circle. Step 2: Refer to the key. An unshaded circle represents a normal female. a) The phenotype of individual 4 is: $$\boxed{\text{Normal female}}$$ b) Step 1: Identify individual 2 in the pedigree. Individual 2 is an unshaded circle, so she is phenotypically normal. Step 2: Observe her children. Individuals 6 and 7 are shaded squares, meaning they are haemophiliac males. Step 3: Determine the genotype of a haemophiliac male. Since haemophilia is X-linked recessive ($X^h$), a haemophiliac male has the genotype $X^h Y$. Step 4: Males inherit their X chromosome from their mother. Therefore, individual 2 must have passed on the $X^h$ allele to her sons (individuals 6 and 7). Step 5: Since individual 2 is phenotypically normal, she must also possess the dominant normal allele $X^H$. b) The genotype of individual 2 is: $$\boxed{X^H X^h}$$ 12.2 Haemophilia is an X-linked recessive disorder. Females have two X chromosomes ($XX$), while males have one X and one Y chromosome ($XY$). For a female to suffer from haemophilia, she must inherit two copies of the recessive allele ($X^h X^h$), one from each parent. However, if she inherits one normal allele ($X^H$) and one haemophilia allele ($X^h$), she will be a carrier but phenotypically normal because the dominant normal allele masks the recessive one. Males, having only one X chromosome, will express the disorder if they inherit the single recessive allele ($X^h Y$). Therefore, it is less likely for a female to inherit two recessive alleles than for a male to inherit one. 12.3 Step 1: Determine the genotypes of individuals 13 and 14. • Individual 13 is an unshaded square, meaning he is a Normal male. His genotype is $X^H Y$. • Individual 14 is a shaded circle, meaning she is a Haemophiliac female. Her genotype is $X^h X^h$. Step 2: Set up a genetic cross (Punnett square) between individual 13 ($X^H Y$) and individual 14 ($X^h X^h$). Parents: $X^H Y \times X^h X^h$ Gametes from individual 13: $X^H$, $Y$ Gametes from individual 14: $X^h$, $X^h$ Punnett Square: $$ \begin{array}{|c|c|c|} \hline \textbf{Gametes} & X^h & X^h \\ \hline X^H & X^H X^h & X^H X^h \\ \hline Y & X^h Y & X^h Y \\ \hline \end{array} $$ Step 3: Analyze the offspring genotypes and phenotypes. • $X^H X^h$: Normal female (carrier) • $X^h Y$: Haemophiliac male Step 4: Determine the percentage chance of having a haemophiliac son. From the Punnett square, there are 2 out of 4 possible offspring that are haemophiliac sons ($X^h Y$). Percentage chance = $\frac{2}{4} \times 100\% = 50\%$. The percentage chance of individuals 13 and 14 having a haemophiliac son is: $$\boxed{50\%}$$ Send me the next one 📸