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.

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

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

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

Answer 1

The information given in question says that there are 3 genes and each have and allele that show dominant/recessive relationship. As A is Dominant to a ; B is Dominant to b and C is Dominant to c.

But the question also talks about wild and mutant type.

Wild type = it is naturally occurring type.

Here for the three genes you will get the wild type as when -- wild type for A needs to be expressed it must be AA or Aa as per dominant/ recessive relation.

This will be followed in case of B and C also. Wild type when BB, Bb and CC, Cc.

But the information also says that when any of 3 genes are present in recessive form in an individual then it will show the mutant type or say the recessive allele is expressed. E.g. if genotype is AA Bb cc --- phenotye c is expressed ; when AaBbcc --- phenotye c ; aaBbcc -- phenotye ac is expressed.

This is the basic idea that the question is talking about and the example given in question also demonstrate the above nature.

so in question we have asked that when an individual heterozygote for A, B and C is crossed with individual Homozygous.

AaBbCc ----- heterozygote individual

aabbcc​ ------ homozygote individual

Gametes by AaBbCc ---- total 8 Gametes ---- ABC,ABc, AbC, Abc, aBC, abC, aBc and abc.

Gametes by aabbcc --- one kind of Gamete ---- abc

On crossing these we get,

	abc	phenotye
ABC	AaBbCc	Wild-- ABC
ABc	AaBbcc	phenotye c
AbC	AabbCc	phenotye b
Abc	Aabbcc	phenotye bc
aBC	aaBbCc	phenotye a
abC	aabbCc	phenotye ab
aBc	aaBbcc	phenotye ac
abc	aabbcc	phenotye abc
		8 phenotyes

Each phenotye have a ratio of 1/8 to be expressed and hence number of progenies for each phenotye will be from 10,000 offsprings.

1/8 * 10,000

= 1250.

So answer is option d. Eight phenotyes each with 1250 offsprings.

Thank you....

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