Describe Charly's phenotype, Portia's genotype, Bill's genotype and explain why Patrick is normal but a carrier of Tay-Sachs disease.

Step 1: (a) Charly’s phenotype
Tay-Sachs disease is an autosomal recessive disorder. Affected individuals have genotype $tt$, while $TT$ or $Tt$ are normal.
In the pedigree, Charly's symbol is shaded, indicating affected status.
Charly’s phenotype: affected (has Tay-Sachs disease).
affected (has Tay-Sachs disease)

Step 2: (b) Portia’s genotype
Portia’s symbol is unshaded: normal phenotype ($TT$ or $Tt$).
Portia and Patrick (normal) have children Elizabeth and Sam, both shaded (affected, genotype $tt$).
To produce $tt$ offspring, each parent must contribute one $t$ allele.
Thus, Portia’s genotype must be $Tt$.
Tt

Step 3: (c) Bill’s genotype
Bill’s symbol is unshaded: normal phenotype ($TT$ or $Tt$).
Bill and Pam (normal) have child Charly (affected, $tt$), plus normal children Helen and Portia.
To produce a $tt$ child, both parents must contribute $t$, so both are $Tt$.
Bill’s genotype: $Tt$.
Tt

Step 4: 2.5.2 Explain why Patrick is normal, but a carrier of Tay-Sachs.
Patrick’s symbol is unshaded: normal phenotype (genotype $TT$ or $Tt$).
Patrick and Portia ($Tt$) produced affected children (Elizabeth and Sam, both $tt$).
Each $tt$ child requires one $t$ from Patrick.
If Patrick were $TT$, all children would be $T-$ (normal), which is impossible.
Thus, Patrick must be $Tt$: normal phenotype, but carrier.
Patrick is a carrier ($Tt$) because he transmits the recessive $t$ allele to his affected children.