Question

question 4
a male dog that is heterozygous for long ears (ll) and heterozygous for a pointed nose (nn) is mated with a female dog that is also heterozygous for both traits.
what is the phenotypic ratio for this cross?

Explanation:

Step1: Determine gametes

For a heterozygous \( LlNn \) individual, using the law of independent assortment, the possible gametes are \( LN \), \( Ln \), \( lN \), \( ln \).

Step2: Set up Punnett square

	\( LN \)	\( Ln \)	\( lN \)	\( ln \)
\( Ln \)	\( LLNn \)	\( LLnn \)	\( LlNn \)	\( Ll nn \)
\( lN \)	\( LlNN \)	\( LlNn \)	\( llNN \)	\( llNn \)
\( ln \)	\( LlNn \)	\( Ll nn \)	\( llNn \)	\( ll nn \)

Step3: Determine phenotypes

• \( L - N - \) (long ears, pointed nose): \( LLNN(1)+LLNn(2)+LlNN(2)+LlNn(4)=9 \)
• \( L - nn \) (long ears, non - pointed nose): \( LLnn(1)+Ll nn(2)=3 \)
• \( llN - \) (short ears, pointed nose): \( llNN(1)+llNn(2)=3 \)
• \( ll nn \) (short ears, non - pointed nose): \( 1 \)

Answer:

\( 9:3:3:1 \)