In: Biology
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.
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:
Use this information to answer the following questions.
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. ABC/abc
c. AcB/aCb
d. ACB/acb
e. ABc/abC
Answer:
c. 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 )100
= (494/10000)100 = 4.94%
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 )100
= (203/10000)100 = 2.03%
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 )100
= (303/10000)100 = 3.03%
Recombination frequency (%)= distance between the genes (map units)
The order of genes = a-----3.03--------c-----2.03m.u.------- b
Therefore, the genotype of heterozygous parent = AcB / aCb