00:01
To answer this question we talk about inheritance.
00:03
It says, for the following problem, calculate the probability of the off -remid ring the cross using the assume and product rule.
00:09
The parent generation has a genotype heterocygios a, heterozygos -a, heterocygous -a -heteros -c, homo -sacos -ecesive -d, and heterocyg theroseph -eus -e, with a heterocyg therocytocis -a, hithelis -sic -c, heteros -dd6 -5, and it says, what is the probability that the f -1 generation would be like this, homocygogynoseph a, heterocygios b, heterocygotechus b, hiterozygose d, and hitrocygos e.
00:39
Okay, so in this case, let's make five monohybarypanene squares.
00:43
You have to cross each the a gene here with the aging here, the b gene here, and the same for all the genes.
00:49
So you have heterozygosis a with heterotigos a.
00:56
You get this, this, this, and this, that heterocygoyles b with the homocygogynophagus a, hitro -sigos, gytrocygos, homo -sidesive, and homo -scygisive.
01:09
And the hitro -sai -s -c, with the hetero -sai -o -sig -si -gos -i -gisif, and the homo -sai -gisive -d.
01:19
And homo -sai -sidesive -d with the homo -sai -sidesive -d with a homo -saigo -visive -visive -visive -visive.
01:25
You have the hittler -sigital -6 -5 -5 -6 -5 -6 -6 -6 -5 -6 -6 -6 -6 -6 -6 -6 -6 -6 -s.
01:40
So this is your panezware.
01:42
Now, it says here that you want this genotype, right? so in this case, you have to find the probabilities to get homozygo recessive a.
01:53
You have that out of four possibilities, only one is homozygous decisive a.
01:57
So you have one out of four, that is one quarter...