Question

genetic crosses that involve 2 traits - floppy eared bunnies
name: jarrett. sanchez date: feb 05
in rabbits, black hair is dominant to brown hair. also in rabbits, long straight ears are dominant to floppy ears.
these letters represent the genotypes and phenotypes of the rabbits:
bb = black nose \t ee = long ears
bb = black nose \t ee = long ears
bb = pink nose \t ee = floppy ears

1. a male rabbit with the genotype bbee is crossed with a female rabbit with the genotype bbee the square is set up below. fill it out and determine the phenotypes and proportions in the offspring.

(there is a punnett square here with rows labeled be, be, be, be and columns labeled be, be, be, be, and some handwritten genotypes inside the squares. also, there are questions on the right: how many out of 16 have:
black noses and long ears?
black noses and floppy ears?
pink noses and long ears?
pink noses and floppy ears?
then another part: show the cross: bbee x bbee
(there is an empty punnett square here)
how many out of 16 have:
black noses and long ears?
black noses and floppy ears?
pink noses and long ears?
pink noses and floppy ears?

Answer:

To solve the genetic cross \( BbEe \times bbEe \), we use a Punnett square. Here are the steps:

Step 1: Determine the gametes for each parent

• For the male parent \( BbEe \), the possible gametes are formed by independent assortment of the two traits. The alleles for the first trait (nose color: \( B \) and \( b \)) and the second trait (ear type: \( E \) and \( e \)) combine. So the gametes are: \( BE \), \( Be \), \( bE \), \( be \).
• For the female parent \( bbEe \), the alleles for the first trait are both \( b \), and for the second trait are \( E \) and \( e \). So the gametes are: \( bE \), \( be \), \( bE \), \( be \) (or simplified as \( bE \) and \( be \) each with a frequency of 2 out of 4, but for the Punnett square, we can list them as \( bE, bE, be, be \)).

Step 2: Set up the Punnett square

We create a 4x4 Punnett square (since each parent has 4 gametes, although the female's gametes are repeated). The rows are the female gametes (\( bE, bE, be, be \)) and the columns are the male gametes (\( BE, Be, bE, be \)).

	\( BE \)	\( Be \)	\( bE \)	\( be \)
\( bE \)	\( BbEE \)	\( BbEe \)	\( bbEE \)	\( bbEe \)
\( be \)	\( BbEe \)	\( Bbee \)	\( bbEe \)	\( bbee \)
\( be \)	\( BbEe \)	\( Bbee \)	\( bbEe \)	\( bbee \)

Step 3: Analyze the phenotypes

• Black nose ( \( B\_ \)) and long ears ( \( E\_ \)):
• Genotypes: \( BbEE, BbEe, BbEE, BbEe, BbEe, BbEe, bbEE, bbEe, bbEE, bbEe \)
• Let's count:
• \( BbEE \): 2
• \( BbEe \): 4
• \( bbEE \): 2
• \( bbEe \): 2
• Total: \( 2 + 4 + 2 + 2 = 10 \)? Wait, no, let's count again from the Punnett square:
• First row (\( bE \)): \( BbEE \) (1), \( BbEe \) (1), \( bbEE \) (1), \( bbEe \) (1)
• Second row (\( bE \)): \( BbEE \) (1), \( BbEe \) (1), \( bbEE \) (1), \( bbEe \) (1)
• Third row (\( be \)): \( BbEe \) (1), \( Bbee \) (1), \( bbEe \) (1), \( bbee \) (1)
• Fourth row (\( be \)): \( BbEe \) (1), \( Bbee \) (1), \( bbEe \) (1), \( bbee \) (1)

Now, let's categorize by phenotype:

• Black nose (\( B\_ \)): genotypes with \( B \) ( \( BbEE, BbEe, BbEe, BbEe, BbEe, BbEE, BbEE, Bbee, Bbee \)) Wait, no, better to use the phenotype definitions:
• Black nose: \( Bb \) (since \( B \) is dominant, \( Bb \) or \( BB \), but here male is \( Bb \), female is \( bb \), so offspring with \( B \) are \( Bb \) (black nose), offspring with \( bb \) are pink nose.
• Long ears: \( EE \) or \( Ee \) (since \( E \) is dominant), floppy ears: \( ee \).

Let's re - categorize each cell:

• \( BbEE \): black nose, long ears
• \( BbEe \): black nose, long ears
• \( bbEE \): pink nose, long ears
• \( bbEe \): pink nose, long ears
• \( BbEE \): black nose, long ears
• \( BbEe \): black nose, long ears
• \( bbEE \): pink nose, long ears
• \( bbEe \): pink nose, long ears
• \( BbEe \): black nose, long ears
• \( Bbee \): black nose, floppy ears
• \( bbEe \): pink nose, long ears
• \( bbee \): pink nose, floppy ears
• \( BbEe \): black nose, long ears
• \( Bbee \): black nose, floppy ears
• \( bbEe \): pink nose, long ears
• \( bbee \): pink nose, floppy ears

Step 3: Count the phenotypes

• Black noses and long ears (\( B\_ \) and \( E\_ \)):
• Let's identify the cells with \( B \) (black nose) and \( E \) (long ears, i.e., \( EE \) or \( Ee \)):
• \( BbEE \): 2 cells (first and second row, first column)
• \( BbEe \): 4 cells (first row second column, second row second column, third row first column, fourth row first column)
• Wait, actually, from the Punnett square:
• \( BbEE \): 2
• \( BbEe \): 4
• Total: \( 2 + 4=6 \) (wait, let's count again. The cells with \( B \) and \( E \) (long ears) are:
• \( BbEE \): 2
• \( BbEe \): 4
• Total of \( 2 + 4 = 6 \) out of 16.
• Black noses and floppy ears (\( B\_ \) and \( ee \)):
• Cells with \( B \) and \( ee \): \( Bbee \)
• From the Punnett square, \( Bbee \) appears 2 times (third row second column, fourth row second column)
• So 2 out of 16.
• Pink noses and long ears (\( bb \) and \( E\_ \)):
• Cells with \( bb \) and \( E \) (long ears, \( EE \) or \( Ee \)):
• \( bbEE \): 2
• \( bbEe \): 4
• Total: \( 2+4 = 6 \) out of 16.
• Pink noses and floppy ears (\( bb \) and \( ee \)):
• Cells with \( bb \) and \( ee \): \( bbee \)
• From the Punnett square, \( bbee \) appears 2 times (third row fourth column, fourth row fourth column)
• So 2 out of 16.

Final Counts (out of 16)

• Black noses and long ears: \( \boldsymbol{6} \)
• Black noses and floppy ears: \( \boldsymbol{2} \)
• Pink noses and long ears: \( \boldsymbol{6} \)
• Pink noses and floppy ears: \( \boldsymbol{2} \)