Question

You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait.

• allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A
• allele B is dominant to b; phenotype b = mutant for trait b; phenotype B = wild type for trait B
• allele C is dominant to c; phenotype c = mutant for trait c; phenotype C = wild type for trait C

Note: phenotypes can be represented by single letters. For example phenotype A = genotypes Aa or AA; phenotype a = genotype aa. Assume that phenotype ab = mutant phenotype for traits a and b, and wild type phenotype for trait C.

You cross an individual heterozygote for all three genes, with an individual who is homozygote recessive for all three. Out of 10,000 offspring you get the following phenotypes and amounts:

• phenotype a - 98
• phenotype b - 150
• phenotype c - 4751
• phenotype ab - 4749
• phenotype ac - 147
• phenotype bc - 99
• phenotype abc - 3
• wild type - 3

Use this information to answer the following questions.

Question 9

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You cross an individual heterozygote for genes A, B and C, with an individual who is homozygote recessive for all three. Assuming independent assortment for all three genes what do you expect to see out of 10,000 offspring?

Remember:

• for all three traits wild type phenotype is dominant to mutant phenotype
• allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A
• allele B is dominant to b; phenotype b = mutant for trait b; phenotype B = wild type for trait B
• allele C is dominant to c; phenotype c = mutant for trait c; phenotype C = wild type for trait C

Select one:

a. Three different phenotypes among the offspring; approximately 3333 offspring of each phenotype.

b. Eight different phenotypes among the offspring; approximately 1250 offspring of each phenotype.

c. Approximately 156 offspring will have the recessive phenotype for all three traits.

d. Approximately 2963 offspring will be wild type (dominant phenotype for all three traits).

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Question 10

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For each phenotype determine if the number of offspring you are observing is greater or less than you'd expect if all three genes are sorting independently (expectation = independent assortment).

Use the information at the top of this page to choose the correct answer.

For phenotype a there are  lessmore offspring observed than expected.

For phenotype b there are  lessmore  offspring observed than expected.

For phenotype c there are  lessmore  offspring observed than expected.

For phenotype ab there are  lessmore  offspring observed than expected.

For phenotype ac there are  lessmore  offspring observed than expected.

For phenotype bc there are  lessmore  offspring observed than expected.

For phenotype abc there are  lessmore  offspring observed than expected.

For wild type there are  lessmore  offspring observed than expected.

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Based on the information given and your answer to the previous questions, what was the genotype of the heterozygous parent, in the correct order?

Select one:

a. BAc/baC

b. ACB/acb

c. ABc/abC

d. AcB/aCb

e. ABC/abc

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Question 12

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Given the information at the top of the page, determine the recombination frequency for A-C. Show your work here!

Note: You will choose the correct answer in the question below (multiple choice). This is for you to show your work.

You can show your work by typing out the math (fractions, % or decimals; can use any format you like) or explaining how you found the answer in words, or a combination of both. Basically any description/showing of your work even if your answer is wrong = 2 marks.

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Given the information at the top of the page, what is the recombination frequency for A-C? Frequency is given as a number from 0 to 1.

Select one:

a. 0.03

b. 0.1

c. 0.02

d. 0.05

e. 0.015

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Answer 1

Answer 9:)

The parental genotypes:

Heterozygous = AaBbCc

Homozygous = aabbcc

The parental cross:

AaBbCc           x          aabbcc

F1 generation:

Gametes	abc
ABC	AaBbCc
ABc	AaBbcc
AbC	AabbCc
Abc	Aabbcc
aBC	aaBbCc
aBc	aaBbcc
abC	aabbCc
abc	Aabbcc

Therefore, we would get 8 genotypes. There is an independent assortment in the cross. Therefore, each genotype will get an equal ratio in the total population of the F1 generation.

Number of offspring in each genotype = 10000/8

Number of offspring in each genotype = 1250

Therefore, option b is the correct answer.

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