Purebred lines of fruit flies with wild type (tan) body color and stubby bristles are mated to flies with ebony bodies and normal bristles. the resulting fl offspring all have a normal wild-type body color and stubby bristles. the fl flies are crossed with flies recessive for both traits (i. e. ebony bodies and normal bristles). what phenotypes and ratios of phenotypes would you expect to find among the offspring produced by crossing the fl with flies recessive for both traits? assume you found the following results among the offspring.
wild type body, stubby bristles 430
ebony body, stubby bristles 55
wild type body, normal bristles 65
ebony body, normal bristles 450
how could you explain these results?

Given that ebony bodies and normal bristles are recessive traits.

Parent 1: Pure breeding tan body and stubby bristles (AABB)

Parent 2: Pure breeding ebony body and normal bristles (aabb)

AABB X aabb = AaBb

Hence in F1 progeny entire offspring has tan body and stubby bristles

When AaBb X aabb, expected progeny would be :

AaBb

aaBb

Aabb

aabb

Since the different type of gametes are expected to form in same number, phenotypic ratio of this progeny would be 1:1:1:1

But as given, the offspring were not produced in this ratio. Flies with parental traits ( AABB and aabb ) were far more in number than flies with the recombinant traits ( aaBb and Aabb ). This phenomena is called as linkage. The genes for both the traits were linked so when the gametes were formed, they did not show independent assortment. Parental gametes were produced more in number resulting in more offspring with parental phenotype.

i can't see the full page, therefore, can't answer it

explanation:

you are certainly talking about this picture:

a: is the label of the white square, it is showing the  replication fork, formed by helicase (d) ==> so it is the site of dna separation and replication 

b: is a new dna brand that is following directly the replication fork, so it is continuous ==> built continuously from 5' to 3

c: dna polymerase cannot replicate dna on its own, it needs the of their proteins, and one of them plays the role of a clamp which fixes  the dna with the dna template and increases  the processivity of the enzyme. ==>   stops polymerase from falling off the  template.

d: is the enzyme responsible for forming the replication fork (helicase) ==> unwinds and separates dna into two strands

e: is the delayed strand, it is built  in the opposite sense of the helicase, so dna will be built in fragments ==> built  fragments from 5' to 3 joins

and finally f: is the enzyme that builds  the new dna strand (dna polymerase) ==>   adds nucleotides to make a new dna helix.

credit: gradman555p98x9p

read more on -