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21. rudolph, the red - nosed reindeer has two black - nosed parents. dr…

Question

  1. rudolph, the red - nosed reindeer has two black - nosed parents. draw a punnett square that describes how two black - nosed reindeer can have a red - nosed baby write the probabilities of each phenotype in the space provided.

red - nose
black - nose

  1. mendel knows how important it is for santa’s reindeer to fly. prancer can fly, but her mate is a heterozygous non - flyer named larry. draw a punnett square to show the probability that their offspring will be able to fly?

flyer
non - flyer

  1. santa recently learned of two islands where the first magical reindeer came from. all of the reindeer on one island were purebred for tall antlers and all of the reindeer on the other island had pure short antlers. he brought all of these reindeer together in the north pole and they produced over 1,000 baby reindeer. when santa saw that all of the babies had tall antlers, he was scared that short - antlered reindeer would go extinct. use two punnett squares to show santa that just because all had tall antlers, it didn’t mean that the short - antlered reindeer were extinct. explain your answer in the space provided.

generation 2 generation 3

  1. in the space to the right, draw a reindeer with the following genotype:

aa nn ff

Explanation:

Question 21 Solution:

Step1: Determine Genotypes

Black nose is dominant (let \( B \) = black, \( b \) = red). Parents are black - nosed but have a red - nosed offspring, so parents are heterozygous (\( Bb \)).

Step2: Set Up Punnett Square

\( B \)\( b \)
\( b \)\( Bb \)\( bb \)

Step3: Calculate Probabilities

Red - nose (\( bb \)): \( \frac{1}{4}=25\% \), Black - nose (\( BB, Bb \)): \( \frac{3}{4} = 75\% \)

Step1: Determine Genotypes

Let \( F \) = fly (dominant), \( f \) = non - fly. Prancer (fly) is homozygous dominant (\( FF \)) (since she can fly and to have a heterozygous non - flyer mate, her genotype for flying should be dominant homozygous to be a flyer), Larry is heterozygous non - flyer? Wait, no: If Larry is a heterozygous non - flyer, that doesn't make sense. Wait, correct: Let \( F \) = fly (dominant), \( f \) = non - fly. Prancer can fly, so her genotype is \( FF \) (assuming flying is dominant). Larry is a heterozygous non - flyer? No, non - flying should be recessive. So Prancer: \( FF \), Larry: \( Ff \) (heterozygous, but he is a non - flyer? Wait, no, maybe flying is recessive. Let's re - define: Let \( f \) = fly (recessive), \( F \) = non - fly. Then Prancer (fly) is \( ff \), Larry is heterozygous non - flyer (\( Ff \)).

Step2: Set Up Punnett Square

\( F \)\( f \)
\( f \)\( Ff \)\( ff \)

Step3: Calculate Probability of Flyer (\( ff \))

Flyer (\( ff \)): \( \frac{2}{4}=50\% \)

Step1: Generation 2 (Parental Cross)

Let \( T \) = tall (dominant), \( t \) = short. Purebred tall (\( TT \)) × purebred short (\( tt \)).
Punnett Square for Generation 2:

\( T \)\( T \)
\( t \)\( Tt \)\( Tt \)

All offspring (\( Tt \)) have tall antlers.

Step2: Generation 3 (Cross of Generation 2 Offspring)

Cross \( Tt \) × \( Tt \)
Punnett Square for Generation 3:

\( T \)\( t \)
\( t \)\( Tt \)\( tt \)

Short - antlered (\( tt \)) offspring: \( \frac{1}{4}=25\% \), so short - antlered trait is still present (recessive allele \( t \) is in heterozygous \( Tt \) of Generation 2 and can be expressed in Generation 3).

Answer:

Red - nose: \( 25\% \) (or \( \frac{1}{4} \))
Black - nose: \( 75\% \) (or \( \frac{3}{4} \))

Question 22 Solution: