Question

and long). thus, each gamete will contain one of the coat - color alleles, c or c and one of the hair - length alleles, b or b. in replicated crosses of a specific dark, short - haired dog with an albino, long - haired dog, all the offspring were dark with short hair, as shown in cross i. however, in subsequent crosses of another dark, short - haired dog with a dark, long - haired dog, the ratios shown in cross ii below were obtained.
cross | parents | offspring
1 | dark, short hair x albino, long hair | all dark, short hair
ii | dark, short hair x dark, long hair | 3 dark, short hair
| | 3 dark, long hair
| | 1 albino, short hair
| | 1 albino, long hair
in cross ii, the genotype of the dark, short - haired parent is
ccbb
ccbb
ccbb
ccbb
ccbb

Explanation:

Step1: Analyze Coat Color Inheritance

From Cross I (Dark, short hair \( \times \) Albino, long hair), all offspring are dark, short hair. This implies dark (C) is dominant over albino (c), and short hair (B) is dominant over long hair (b). So the dark, short - haired parent in Cross I likely has genotype \( CCBB \), and the albino, long - haired parent has \( ccbb \), producing \( CcBb \) offspring.

Step2: Analyze Cross II Offspring Ratios

In Cross II, parents are Dark, short hair \( \times \) Dark, long hair. The offspring ratios for coat color: Dark : Albino = \( (3 + 3+1 + 1):(1 + 1)=8:2 = 4:1\)? Wait, no, let's count the number of dark and albino offspring. Dark offspring: \( 3\) (dark, short) \(+ 3\) (dark, long) \(= 6\); Albino offspring: \(1\) (albino, short) \(+ 1\) (albino, long) \(= 2\). So Dark : Albino = \( 6:2=3:1\). For hair length: Short hair offspring: \( 3\) (dark, short) \(+ 1\) (albino, short) \(= 4\); Long hair offspring: \( 3\) (dark, long) \(+ 1\) (albino, long) \(= 4\). So Short : Long = \( 1:1\).

Step3: Determine Genotype of Dark, Short - Haired Parent in Cross II

For coat color (C - c), a \( 3:1\) ratio of dark to albino suggests that both parents are heterozygous (\( Cc \)) for coat color (since \( Cc\times Cc
ightarrow 3C\_:1cc \)). For hair length (B - b), a \( 1:1\) ratio of short to long hair suggests that one parent is heterozygous (\( Bb \)) and the other is homozygous recessive (\( bb \)) (since \( Bb\times bb
ightarrow 1Bb:1bb \)). The dark, short - haired parent has dark coat (so at least one C) and short hair (so at least one B). From the above, for coat color, it's \( Cc \), and for hair length, it's \( Bb \)? Wait, no. Wait the dark, long - haired parent has genotype \( Ccbb \) (because it is dark, so \( C\_ \), and long hair, so \( bb \), and to get a \( 3:1\) ratio for coat color, it must be \( Cc \)). Then the dark, short - haired parent: for coat color, to get \( 3:1\) ratio (dark:albino), it must be \( Cc \) (since the other parent is \( Cc \)). For hair length, the dark, long - haired parent is \( bb \), and the offspring have short:long = \( 1:1\), so the dark, short - haired parent must be \( Bb \) (because \( Bb\times bb
ightarrow Bb \) (short) and \( bb \) (long) in 1:1 ratio). Wait, but let's check the options. The options are CcBb, ccbb, CCBB, CCbb, ccBB. Wait, maybe I made a mistake. Wait in Cross I, all offspring are \( CcBb \). In Cross II, the dark, long - haired parent: let's re - evaluate. Wait the dark, long - haired parent: long hair is \( bb \), and dark coat. If we look at the offspring of Cross II, the albino offspring have genotype \( cc \). So both parents must contribute a \( c \) allele. So the dark, short - haired parent must be \( Cc \) (because to get \( cc \) offspring, both parents need to have a \( c \) allele). For hair length: the dark, long - haired parent is \( bb \), and the offspring have short and long hair in 1:1 ratio. So the dark, short - haired parent must be \( Bb \) (because \( Bb\times bb
ightarrow Bb \) (short) and \( bb \) (long) in 1:1). So the genotype of the dark, short - haired parent in Cross II is \( CcBb \).

Answer:

CcBb