▶️ Answer/Explanation
Answer: (C) 50% TT, 50% Tt
Parent 1 is homozygous dominant (TT), and parent 2 is heterozygous (Tt). The cross can be set up as:
Parent 1 gametes: T, T
Parent 2 gametes: T, t
Using a Punnett square: \[ \begin{array}{c|c c} & T & t \\ \hline T & TT & Tt \\ T & TT & Tt \\ \end{array} \] The offspring genotypes are: 2 TT and 2 Tt. This gives a 50% chance of TT and 50% chance of Tt. No tt genotype appears because one parent provides only the dominant allele (T).
▶️ Answer/Explanation
The heterozygous parent has genotype \(Hh\) (short-haired). The long-haired parent must be homozygous recessive \(hh\) (since long hair is recessive). A cross between \(Hh \times hh\) can be analyzed using a Punnett square: gametes from the heterozygous parent are \(H\) and \(h\); from the long-haired parent, all are \(h\). The offspring genotypes are \(Hh\) and \(hh\) in equal proportions. Phenotypically, \(Hh\) is short-haired and \(hh\) is long-haired. Therefore, the expected phenotypic ratio is 50% short-haired and 50% long-haired.
✅ Answer: (C)
✅ Answer: (C)
Question
The diagram shows part of a genetic diagram for the inheritance of sex in humans.
Answer/Explanation
Ans: B