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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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 fur 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). a. What is a probability that the offspring will have black fur? i. 50% B b b Bb bb b Bb bb b. What percentage of the offspring will be homozygous recessive? [2 marks] i. 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. a. What is the probability of the F1 generation plants having a wrinkled phenotype? R R r Rr Rr r Rr Rr i. 100% b. What percentage of the F2 generation will be heterozygous? R r R RR Rr r Rr rr i. 25% c. What is the phenotypic ratio for the F2 generation? [3.5 marks] R R r RR Rr r Rr rr i. 1:2:1 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 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] E e E EE Ee e Ee ee a. 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] 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 predicted genotypes produced by the known phenotypic ratio of the F1 generation. [2 marks] b. What type of dominance is this? [1 mark] Blood Type Genotype A IAIA or IAi B IBIB or IBi AB IAIB O ii 6. In blood typing, the gene for type A and the gene for type B are codominant. The gene for type O is recessive. Use Punnett squares to show your work. Use the table provided for the symbols you use. a. One parent is homozygous for type A, and the other is AB. What are the possible genotypes and phenotypes of their offspring? [2 mark] b. Is it possible for an individual with blood type A, and an individual with blood type B, to have a child with blood type O? What is the percent chance? [2 mark] Solving Sex-linked Trait Problems: DRAW PUNNETT SQUARES FOR THIS SECTION SO WE CAN SEE YOUR WORK · Traits can be inherited by a gene located on a sex chromosome; this is called a sex-linked gene Male = XYFemale = XX · Remember: females receive an X chromosome from both their father and mother (one copy from each), whereas the male receives an X chromosome from only his mother (his dad gives him a Y). i.e. If a father had a certain trait on his X chromosome, the daughter must receive that same trait on her X chromosome · Ex. Colourblindness. The gene for normal vision and colourblindness are on the X chromosome. The colourblindness is recessive. Let normal vision be B; colour blindness be b Unaffected female XB XB Carrier female XB Xb Colorblind female Xb Xb Unaffected male XBY Colorblind male XbY 7. A woman carrying the colour blindness trait (i.e. she has normal vision but has one mutant gene) marries a colourblind man. What are the chances that the offspring will be colorblind if they are female? If they are male? [2 marks] 8. A colorblind female marries a normal male. How many of the female offspring will be carriers of the colorblind allele? [2 mark] 9. A woman with normal vision, whose father was colorblind, married a man who had normal vision. The colorblindness characteristic is X-linked recessive. What are the expected F1 genotype and phenotype ratio from this mating? [3 marks] PART 2 Analyzing Fruit Fly Phenotypes and Genotypes The students from BIOL199 2019 scoring the F1 generation formed by crossing females that are homozygous for the white gene and heterozygous for the curly gene to pure-breeding wild-type males (i.e. red eyes and straight winged). Use this chart to record the data from the class data for F1 flies. Be sure to enter the complete phenotype (i.e. wing structure and eye colour) in the "phenotype" column. Phenotype # males # females White eyes, curly wings 102 0 White eyes, straight wings 99 0 Red eyes, curly wings 0 101 Red eyes, straight wings 0 103 1. What eye colour phenotype(s) is (are) present in the F1? Is there a difference in phenotype between males and females? [1 mark] 2. What wing phenotype(s) is(are) present in the F1? Is there a difference in phenotype between males and females? [1 mark] 3. Based on your answer to question 1, is the white locus on the autosome or on the X chromosome? Justify your answer by drawing out the Punnett square(s) for the cross for both an autosomal and a sex linked cross scenario. [3 marks] 4. Based on your answer to question 3, is the mutant white allele recessive or dominant to its corresponding wild-type allele? Justify your answer (refer to the Punnett square(s) above if necessary). [1 mark] 5. Based on your answer to question 2, is the mutant curly allele recessive or dominant to its corresponding wild-type allele? Justify your answer by drawing out the Punnett square(s) for the cross [3 marks] 6. Based on your answer to question 5, is the curly locus on an autosome or on the X chromosome? Justify your answer (refer to the Punnett squares above if necessary). [1 mark] BIOL 199 Spring 2021 4 Solving Pedigree Problems: · Pedigrees are used to determine the inheritance patterns of particular human traits · There are 2 questions to consider while solving pedigree problems: 1. Is the condition dominant or recessive? 2. Is it autosomal or sex-linked? · Also keep in mind that an individual can be a carrier of a condition, without presenting the phenotype. Pedigree Symbolism: Example: Below is a pedigree for a neurological disease. The son is affected (solid square). a. If the disease is caused by a mutation in the gene on the X chromosome, is the mutation recessive or dominant? Recessive b. From which parent did the son inherit the disease gene? Mother 7. Answer the questions below using this pedigree. a. Using the previous pedigree, what is the most likely mode of inheritance for this pedigree? Why? [2 marks] b. State the genotypes of individuals # 6-8 in the following table”. Use the uppercase letter to represent the dominant allele and lowercase letter to represent the recessive allele. [1 mark] Individual Genotype #6 #7 #8 Legend: c. If individuals #6 and #7 have another daughter what are the chances that she will be affected? Show the work using a Punnett square. [2 mark] 8. Answer the questions below using this pedigree. The mode of inheritance is sex-linked recessive. Using Punnett squares show that it is possible for this pedigree to be sex-linked recessive. [1 marks] 9
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.
2
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
3
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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