Genetics in-lab Assignment Biol 199 Week 9 /40 Marks To help those that are filling this in on word -- -here is a Punnett square that you can copy and paste if you need to draw one for you answer....

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Answered 1 days AfterJul 23, 2021

Answer To: Genetics in-lab Assignment Biol 199 Week 9 /40 Marks To help those that are filling this in on word...

Divya answered on Jul 25 2021
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Genetics in-lab Assignment
Biol 199 Week 9 /40 Marks

To help those that are filling this in on word -- -here is a Punnett square that you can copy and paste if you
need to draw one for you answer.

Solving Monohybrid Punnett Squares (DRAW PUNNETT SQUARES FOR THIS SECTION SO WE CAN SEE
YOUR WORK):

1. An autosomal gene that codes for black fu
r color in guinea pigs has two alleles. Two recessive copies
of this allele will result in a guinea pig that cannot produce black fur; it will therefore have white fur.
Construct a Punnett square to show all possible combinations of gametes that could result from a
cross of a black guinea pig that is heterozygous for black fur and a guinea pig that is homozygous
recessive for white fur (which results in white fur coloration).
Answer:
The Punnett square is depicted below.
a. What is a probability that the offspring will have black fur?
Answer:
The probability that the offspring will have black fur is 50%. This progeny will be
heterozygous for black fur.
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b. What percentage of the offspring will be homozygous recessive? [2 marks]
Answer:
The percentage of the homozygous recessive offspring is 50%.

2. The seed shape of a pea can be round (R) or wrinkled (r). A homozygous plant for the round (dominant)
genotype is crossed with a homozygous plant for the wrinkled genotype to create the F1 generation.
The F1 generation were crossed with each other to produce an F2 generation.
Answer:
The Punnett square is depicted below.
a. What is the probability of the F1 generation plants having a wrinkled phenotype?
Answer:
The probability of the F1 generation plants having a wrinkled phenotype is nil.
b. What percentage of the F2 generation will be heterozygous?
Answer:
50% of the F2 generation will be heterozygous.
c. What is the phenotypic ratio for the F2 generation? [3.5 marks]
Answer:
The phenotypic ratio of the F2 generation is 3:1.
3. Aaron is the youngest child in a family that includes 4 children. All of the children have the recessive
phenotype – attached earlobes. Yet, both parents have unattached earlobes, the dominant trait. What
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are the possible genotypes for the parents? Use Punnett squares to prove your answer (include
Punnett Squares of options that do NOT work). This is an autosomal trait. [2.5 marks]
Answer:
The possible genotypes for the parents are Ee and Ee i.e., both parents are heterozygous for
unattached earlobes, the dominant trait. The Punnett square explaining the same is outlined below.













Solving Problems with Other Types of Dominance:
4. Feather colour is co-dominant in chickens, with heterozygous chickens exhibiting both black and white
feather phenotypes. Create a legend for the different feather colours and draw a Punnett square of
the cross between a homozygous chicken with black wings and a homozygous chicken with white
wings. What are the genotypic and phenotypic ratios of the F1 generation? [2 marks]
Answer:
The Punnett square is depicted below.
The different feather colors would be:
4
1. BB – Black
2. WW – White
3. BW – Both black and white
The genotypic and phenotypic ratios of the F1 are the same. 100% of the progeny will have both black
and white feathers with the genetic composition BW.
5. The color of fruit for a mystery plant is determined by two alleles. When two plants with orange fruits
are crossed the following phenotypic ratios are present in the offspring: 25% red fruit, 50% orange
fruit, 25% yellow fruit.
a. What are the genotypes of the parent orange-fruited plants? Show justification of genotype by
demonstrating the...
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