You are interested in mapping the location of two loci in crested geckos that control different aspects of the crest phenotype. Gene D controls the extension of the crest along the back. 'd codes for spikes that continue along the back past the neck and is recessive to 'D' which codes for spikes that stop at the neck. Gene e controls the length of spikes; short spikes (ee) are recessive to long spikes (E_). (For you crestie fans, yes, this is totally made up).
1. You testcross double heterozygote geckos and observe the phenotypes listed below in the offspring
Phenotypic class Number of offspring Space for notes
Spikes along back and long 79
Spikes along back and short 12
Spikes only on neck and long 10
Spikes only on neck and short 76
a. What was the genotype, in linked gene format, of the double heterozygote parent? Explain how you know.
b. What is the recombination frequency between these two genes? Show your work.

a. Ed/eD

b. RF=0.12

Explanation:

The alleles for genes D/d and E/e are:

After testcrossing a double heterozygote (DdEe x ddee) there are 4 types of offspring, two of them much more abundant than the other two. The homozygous recessive parent can only produce ed gametes, so the phenotypes of the offspring depend on the gametes that the double heterozygous parent produced.

The offspring was:

Total: 177

a) This result suggests that the genes are linked. Since recombination is a rare event, the most abundant phenotypes always come from the parental gametes, and the least abundant come from the recombinant gametes.

Therefore, the genotype of the doube heterozygote parent was Ed/eD.

b) Recombination frequency (RF) = Recombinants / Total

  RF = (12+10)/177

  RF = 0.12

a. The genotype, in linked gene format, of the double heterozygote parent is De/dE

b. The recombination frequency between these two genes is 0.124

Explanation:

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

In the present example, the genotype of the double heterozygote parent is De/dE.

In this way, we might recognize which are the recombinant gametes produced by the di-hybrid, by looking at the phenotypes with lower frequencies in the progeny.

To calculate the recombination frequency we will make use of the next formula: P = Number of Recombinant individuals/ Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

In the present example:

Parental) De/dE   x   de/de

Gametes) De parental type

                 dE parental type

                 DE recombinant type

                 de recombinante type

Phenotypic class Number of offspring

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 12 + 10 / 79 + 12 + 10 + 76

P = 22 / 177

P = 0.124

The genetic distance between genes is 0.124 x 100= 12.4 MU.                                            

i believe it would be the lungs.