Consider a cross between a plant with two flower colors, yellow and red. Yellow (Y) exhibits complete dominance over red (y). It also has either single blossoms or double blossoms, single blossoms (S) exhibit complete dominance over double blossoms (s). A dihybrid yellow, single blossomed flower is crossed with a red flowered and double blossomed plant. You observe the following phenotype frequencies among the 1545 offspring plants: 500 yellow flowers and single blossom plants
515 red flowers and double blossom plants
260 yellow flowers and double blossom plants 270 red flower and single blossom plants.
What is the recombination frequency between these two genes?
a. (260+270) 1545
b. 270/1545
c. 1545/ (260+270)
d. 260/1545
e. (500+515)/1545
f. 1545/260
g. 1545/270
From these results one can conclude that:.
a. the two genes are linked
b. the two genes are not linked
c. if the genes are linked, they are separated far enough apart to appear to assort independently

1) option a is correct. (260+270)/1545

(Note about this option: In the statement, it is written as (260+270) 1545. Lacks the division symbol, / )

2) From these results one can conclude that the two genes are linked (option a)

Explanation:

To know if two genes are linked, we must observe the progeny distribution. If heterozygous individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, phenotypes appearing in different proportions, we can assume linked genes in the double heterozygote parent.  

We might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny. Phenotypes with the highest frequencies represent the parental gametes.

To calculate the recombination frequency, we will use the following formula: P = Recombinant number / Total of individuals.  

In the present example:

Parental)        YySs                        x                 yyss

Gametes)  YS parental type                        ys, ys, ys, ys

                 ys parental type

                 Ys recombinant type

                 yS recombinante type

The total number of individuals in the offspring: 1545

Phenotypic class Number of offspring  

According to this information, the phenotypic frequencies of the progeny differ from the phenotypic ratio 1:1:1:1. We can assume then, that these genes are linked.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 260 + 270 / 1545

P = 530/1545

P = 0.343

The genetic distance between genes is 0.343 x 100= 34.4 MU.

did you fail the test? xd

what is shown in the image?

eukaryote

chloroplast

mitochondrion

submitted