Answer and Explanation:
- A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).
- The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?
1st cross:
Parental) grogro ro+ro+ x gro+gro+ roro
F1) gro+gro ro+ro
2nd cross:
Parental) gro+gro ro+ro x grogro roro
Gametes) gro+ro+ gro ro
gro+ro gro ro
gro ro+ gro ro
gro ro gro ro
Punnet square)
gro+ro+ gro+ro gro ro+ gro ro
gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro
gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro
gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro
gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro
F2)
0.518 grogro ro+ro (518 individuals)
0.471 gro+gro roro (471 individuals)
0.006 grogro roro (6 individuals)
0.005 gro+gro ro+ro (5 individuals)
Total number of individuals 1000
<u><em>Note</em></u>: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.
To know if two genes are linked, we must observe the progeny distribution. <em>If individuals, whos </em><em>genes assort independently,</em><em> are test crossed, they produce a progeny with equal </em><em>phenotypic frequencies 1:1:1:1</em>. <em>If</em> we observe a <em>different distribution</em>, that is that <em>phenotypes appear in different proportions</em>, we can assume that<em> genes are linked in the double heterozygote parent</em>.
In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.
By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are <em>gro+ro+</em> and <em>gro ro</em>, while the parental gametes are <em>gro+ro</em> and <em>gro ro+.</em>
So, the genotype, in linked gene format, of the double heterozygote individual in the <u>F1</u> is gro+ro/gro ro+.
To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (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.
The recombination frequency is:
P = Recombinant number / Total of individuals
P = 6 + 5 / 1000
P = 11 / 1000
P = 0.011
The <u>genetic distance between genes,</u> is 0.011 x 100= 1.1 MU.
<u>Genetic Linkage Map:</u>
Parental Phenotypes)
-----gro+------ro---- -----gro------ro+----
----- gro ------ro---- ---- gro------ ro ----
Recombinant phenotypes)
-----gro+------ro+---- -----gro------ro----
----- gro ------ ro---- -----gro------ro----
<u>If the genes were unlinked</u> and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been
4/16 = 1/4 gro+gro ro+ro
4/16 = 1/4 gro+gro roro
4/16 = 1/4 grogro ro+ro
4/16 = 1/4 grogro roro
Their phenotypic frequencies would be 1:1:1:1 related.
