00:01
Hello everyone in this question we have been given that in fruit flies there is brown body that is denoted by the letter t here and this is a dominant character over the early that is small t that represents tan body we have the wings that is represented by w l and it is dominant that has wings and it is dominant over small w small l and this is recessive and has wingless character and the both of the genes that is t and that is the gene for the color and the wings are 15 map units apart 15 enchant the t and wlg can produce the recombination at chromosome 11.
01:17
So a true breeding tan fly that is small t small t with wings that is capital w capital w.
01:33
Let's represent this wl as capital w and this small w small l as small w.
01:41
Okay.
01:42
So here we have capital w capital w as it has the wings and it is true breeding.
01:49
It is crossed with a with a fly having the brown body that means capital t and capital t and it will be small t.
02:02
Small w and small w for the wing list and all the f1 generation have the brown wings that is capital t and it will be in the heterozygous condition because in f1 only the dominant allele is only the dominant phenotype is shown and the offsprings are always in the heterozygous condition in the f1 generation so this is the genotype of the f1 generation that is the genotype of the f1 generation that is the has brown color with wings.
02:39
Now the f1 female, that is capital t small t, capital w, small w is crossed with the testmate, that is small t, small t, and when we do the test cross, we have the recessive alleys.
03:04
So, these are crossed then we have to tell that how many progeny of each phenotype would be produced from this cross so we have been told that in total we have one thousand six hundred and forty eight of springs okay so now we need to tell that how many of them told that these two genes are 15 map units apart that is 15 percent of recombinant frequency is possible so two types of the recombinant frequency two type of the recombinant offspring will have the 7 percent of the chance from this offspring that they are recombinant so set divided by hundred multiplied by 1648 will be equals to 115 .36 so two of the recombinants will have 115 .3 six so two of the recombinants will have one one five 0 .36 progenies and the rest 85 % will be the parental type so that is 42 .5 % of them of one type of genotype will be parental and the 42 .5 % of the other type of genotype will be parental...