Question

Biol 210: Genetics. Assign. #1 #5) You are studying a gene for body colour in house flies. The wild-type allele (B) has a dark blue phenotype. In population screens you find 6 recessive mutants B' through B$ and establish homozygous strains, all of which have a pale blue phenotype. You cross each mutant strain to itself and to all the other mutant and wild type lines and observe the following results (+ = wild- type phenotype in F1, -= mutant (light blue) phenotype in Ft). B + B + B2 + B3 + B4 + B5 B Which mutations are allelic (i.e., how many loci are found and what are the alleles of each locus)? #6) The Black Langshan breed of chickens has feathered shanks. When Langshan are crossed to the Buff Rock breed with unfeathered shanks, all the Fhave feathered shanks. Out of 360 F2 progeny, 24 were found to have non-feathered shanks and 336 had feathered shanks. a. What type of gene interaction would explain the observed F2 phenotypic ratios? b. What proportion of the feathered F2 would be expected to be heterozygous at one locus and homozygous at the other?

Answer 1

5) After crossing each mutant strain to other mutant strains if we find wild type (Dark blue) phenotype in F₁, then complementation can occur between the two mutant strains & mutation of the two mutants are in different genes. But if we observe mutant phenotype (Light blue) in F₁, then no complementation occur & mutation of the two mutants are in the same gene.

From the table we find that, progeny of cross between mutants B¹ & B⁵ produced mutant phenotype. So, no complementation occur between them & mutation of the two mutants are in the same gene. Also, progeny of cross between mutants B², B⁴ & B⁶ produced mutant phenotype. So, no complementation occur between them & mutation of these mutants are in the same gene. Finally, progeny of cross between B³ & all other mutants showed wild type phenotype. So, complementation can occur between B³ & all other mutant strains. So, mutation of B³ is in different gene from all other mutants. So, finally we find that 3 genes (loci) are involved (Alleles of each locus: Mutant B¹ & B⁵; Mutant B², B⁴ & B⁶; Mutant B³).