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.

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 ( less/more) offspring observed than expected.

For phenotype b there are ( less/more) offspring observed than expected.

For phenotype c there are ( less/more) offspring observed than expected.

For phenotype ab there are ( less/more) offspring observed than expected.

For phenotype ac there are (less/more) offspring observed than expected.

For phenotype bc there are ( less/more) offspring observed than expected.

For phenotype abc there are ( less/more) offspring observed than expected.

For wild type there are ( less/more) offspring observed than expected.

Question 11

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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

Clear my choice

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

Answer 1

10).

Trihybrid testcross ratio would be 1:1:1:1:1:1:1:1. It means that each phenotype would be an equal number, i.e., 10000/8 = 1250.

phenotype a – 98= observed 1250

phenotype b – 150= observed 1250

phenotype c – 4751= observed 1250

phenotype ab – 4749= observed 1250

phenotype ac – 147= observed 1250

phenotype bc – 99= observed 1250

phenotype abc – 3= observed 1250

wild type – 3= observed 1250

For phenotype a there are (less) offspring observed than expected.

For phenotype b there are (less) offspring observed than expected.

For phenotype c there are (more) offspring observed than expected.

For phenotype ab there are (more) offspring observed than expected.

For phenotype ac there are (less) offspring observed than expected.

For phenotype bc there are (less) offspring observed than expected.

For phenotype abc there are (less) offspring observed than expected.

For wild type there are (less) offspring observed than expected.

11).  d. AcB/aCb

Explanation:

phenotype a – 98= aBC

phenotype b – 150=AbC

phenotype c – 4751=ABc

phenotype ab – 4749=abC

phenotype ac – 147=aBc

phenotype bc – 99=Abc

phenotype abc – 3=abc

wild type – 3=ABC

Hint: Always non- recombinant genotypes are large numbered than the recombinant genotypes.

Hence, the parental (non-recombinant) genotypes is ABc/abC

1).

If single cross over occurs between a&b

Normal order = AB/ab

After crossing over = Ab/aB

Ab Progeny= 99+150 = 249

aB progeny = 98+147 = 245

Total progeny = 10000

Total progeny = 494

Recombination frequency between a&b = (Number of recombinants / Total progeny )

= (494/10000) = 0.05%

2).

If single cross over occurs between b&c

Normal order = Bc/bC

After crossing over = BC/bc

BC Progeny= 99+3 = 102

bc progeny = 98+3 = 101

Total progeny = 203

Recombination frequency between b&c = (Number of recombinants / Total progeny )

= (203/10000) = .03%

3).

If single cross over occurs between a&c

Normal order = Ac/aC

After crossing over = AC/ac

AC Progeny= 3+150 = 153

ac progeny = 3+147 = 150

Total progeny = 303

Recombination frequency between a&c = (Number of recombinants / Total progeny )

= (303/10000)= 0.03%

Recombination frequency (%)= distance between the genes (map units)

The order of genes = a-----0.03--------c----0.03m.u.------- b

Therefore, the genotype of heterozygous parent = AcB / aCb

12). a). 0.03