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1. Consider the problem of comb shape inheritance in chickens, where walnut, rose, pea, and single...

1. Consider the problem of comb shape inheritance in chickens, where walnut, rose, pea, and single ae observed as distinct phenotypes. How is comb shape inherited, and what are the genotypes of the P1 generation of each cross. Use the following data to answer these questions. Test the results in cross 4 to your hypothesis. (30 pts.) Cross 1: single X single all single Cross 2: walnut X walnut all walnut Cross 3: rose X pea all walnut Cross 4: F1 X F1 of Cross 3 walnut X walnut  98 walnut 28 rose 34 pea 9 single

F. Write out the genotypes of the four crosses. Do they make sense in terms of the offspring phenotypes and the hypothesis that you made in E? G. How do you test your hypothesis from the information in Cross 4? H. Execute the test. How do you get the expected numbers for each class? I. How many degrees of freedom are there? What is the cutoff value for significance? J. How do you interpret the result?

Solutions

Expert Solution

Null hypothesis -

Comb shape in poultry is inherited as a non mendelian character. It is an epistatic phenomena. Epistasis is a non mendelian inheritance in which alleles at one gene locus mask the effect of alleles at other gene locus.

Phenotype Genotype
A_bb rose
aaB_ pea
aabb single
A_B_ walnut

1. aabb × aabb = aabb (single)

2. AaBB × AABb = A_B_ (walnut)

3. AAbb × aaBB = AaBb (walnut)

4. AaBb × AaBb =

A_B_ = walnut = 98

A_bb = rose = 28

aaB_ = pea = 34

aabb = single = 9

# progenies = 98 + 28 + 34 + 9 = 169

Test for cross 4 -

Phenotype #observed #expected (O-E)^2/E
Walnut 98 9/16 × 169 = 95.06 0.0909
Pea 34 3/16 × 169 = 31.68 0.128
Rose 28 3/16 × 169 = 31.68 0.427
Single 9 1/16 × 169 = 10.56 0.23

Calculated chi square value = 0.0909 + 0.128 + 0.427 + 0.23 = 0.8759

Degrees of freedom = 4 - 1 = 3

P value = 0.9 to 0.95

Interpretation : null hypothesis accepted

Comb shape in poultry is inherited as an epistatic character.

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gladiator answered 2 years ago
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