a) Step 1: Identify individual 4 in the pedigree. Individual 4 is an unshaded circle.

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: 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: 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 × X^h X^h Gametes from individual 13: X^H, Y Gametes from individual 14: X^h, X^h Punnett Square: |c|c|c| Gametes & X^h & X^h \\ X^H & X^H X^h & X^H X^h \\ Y & X^h Y & X^h Y \\ 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 = (2)/(4) × 100\% = 50\%. The percentage chance of individuals 13 and 14 having a haemophiliac son is: 50\% Send me the next one 📸