Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Breeding show chickens, you identify a novel feather pattern gene, Pg. Segregati

ID: 215853 • Letter: B

Question

Breeding show chickens, you identify a novel feather pattern gene, Pg. Segregation ratios between it and another feather color gene, Melanotic (Ml) seem a little off and you believe there might be genetic linkage between these 2 loci. To test this hypothesis, you perform some mapping crosses as follows:

P:       Pg/Pg ; ml/ml                                 X            pg/pg ; Ml/Ml

           (patterned, pale pigment)                (unpatterned, dark pigment)

F1:     Pg/pg ; Ml/ml                       X           pg/pg ; ml/ml

           (patterned, dark pigment)               (unpatterned, pale pigment)

Progeny:        

Phenotype

Number observed

patterned, pale pigment

29

unpatterned, dark pigment

30

patterned, dark pigment

22

unpatterned, pale pigment

19

Does this data support the hypothesis of genetic linkage? Run a Chi-square test. Show how you determined this.

Phenotype

Number observed

patterned, pale pigment

29

unpatterned, dark pigment

30

patterned, dark pigment

22

unpatterned, pale pigment

19

Explanation / Answer

null hypothesis = genes are independently assorted and show no linkage.

This is a test cross (F1 crossed wih homozygous recessive)

so the ratio should be 1:1:1:1

total individuals is 100 so expected number of offsprings for each genotype will be 25

Chi square= (Observed- expected)2/expected

(29-25)2/25+ (30-25)2/25+ (22-25)2/25+ (19-25)2/25

16+25+9+36/25= 3.44

degree of freedom= 3

P value at 0.05 for df =3 is 7.84

P value is greater than 3.44 so we accept the null hypothesis and reject the alternate hypothesis that genes show linkage.