Question

5. You started working in a Drospohila lab. You are studying two different autosomal traits, body color and antennae length. Gray body (B) is completely dominant to black body (b) and long antennae length (A) is completely dominant to short antennae length (a). You cross true-breeding gray bodied, long antennae files to true-breeding black bodied, short antennae. The F1 progeny all have gray bodies and long antennae. a. You perform a testcross to the F1 progeny produced from the parent cross above and obtain 1,900 progeny. 743 gray, long; 777 black, short; 202 gray, short; 178 black, long i. What is the recombination frequency between these two genes? Are the two genes linked to each other? If so, what type of linkage do they have complete or incomplete) and how do you know? ii. Write the genotype of the heterozygous F1 fly as the alleles appear on chromosomes. Are the alleles organized in cis (Coupling) or trans (repulsion)? b. Several other genes in Drosophila are: straight wing shape (W) or curly wings (w); red eyes (E) or purple eyes (e) and straight legs (L) or bent legs (1) and recombination frequencies were calculated. Use these recombination frequencies, as well as the recombination frequency you calculated between the B and A genes to construct a single map. On the map mark the distance between each gene pair in map units. A-E = 41% L-B = 28% B-W = 16% W - A = 35% W-E = 7% L-W = 43% B-A = _ % E-B = 22% L-A = 10% L-E = _% c. What is the map distance between the L and E genes? What is the recombination frequency between the L and E genes? Are they the same? Why or why not?

Answer 1

Answer:

5). a).

i).

AB/AB x ab/ab ----- parents

AB/ab -------------------------F1

AB/ab x ab/ab ----- test cross

According to the Mendel experiments, phenotypic ratio of dihybrid testcross ratio is 1:1:1:1 when the two genes are assorted independently.

Based on this, each phenotype has an equal proportion, 1900/4= 475.

As the observed number of each phenotype is not fit in the ratio. This indicates the two genes are linked.

Hint: Always parental combinations are more numbered than recombinant combinations in linkage.

Recombination frequency= (no. of recombinants/total progeny)100

RF= (202+178/1900)100= 20%

ii).

The linkage is incomplete linkage because four types of phenotypes are produced.

b). The genes are in cis (coupling) linkage as dominant allele of both genes are on same Chromosome.