who is good at genetics

[BD_Paintball]

so i have one problem left and i cant figure it out. it goes like this

A Particular variety of pig is highly prized in some parts of the world because of a feature in its nose that enables the pig to efficiently find truffles. This variety of pig is called a truffle pig. A mating of two truffle pigs resulted in the following offspring: 10 Truffle pigs and 5 normal pigs.
Suggest the genotype of the parents in the mating and describe the nature of genes involved in efficient truffle locating in these pigs.

now i think i know that the gene interaction is Suppression where it restores wild type which is normal pigs(non truffle finders) and the ratio of that is ~10:6. i just can not for the life of me find a genotype that will result in a 10:6 ratio.


HELP.

[calebh]

a little complicated for my genetic knowledge... i took biology 1 year, and am utterly clueless. all i can figure is that multiple genes are at work. at least, i think so. lol

[personman]

AA = dead
Aa = trufflez
aa = no trufflezzzzz

Both parents are heterozygous for the truffle sniffing gene. When they mate, the punnet square looks like this:

   A     a

A  AA   Aa

a  Aa   aa

And if the pig is homozygous dominant for the truffle gene, he dies. They mate 20 times, make 15 little piggys. 10 of them are heterozygous and 5 of them are homozygous recessive.
That's the only way I can think of to do it... maybe I should have paid more attention in AP bio.

[calebh]

AA = dead
Aa = trufflez
aa = no trufflezzzzz

"dead?" they never told me i could use that lol. do they really die that way? that would make the truffles gene dominant, which i dont think it is. unless it really works out that way.

[BD_Paintball]

AA = dead
Aa = trufflez
aa = no trufflezzzzz

Both parents are heterozygous for the truffle sniffing gene. When they mate, the punnet square looks like this:

   A     a

A  AA   Aa

a  Aa   aa

And if the pig is homozygous dominant for the truffle gene, he dies. They mate 20 times, make 15 little piggys. 10 of them are heterozygous and 5 of them are homozygous recessive.
That's the only way I can think of to do it... maybe I should have paid more attention in AP bio.

that was my first guess but the double mutant (aa) will die not the AA. the aa is a synthetic lethal and the AA is a wind type. what i have right now is the parents are both a+/a ; s+/s. then the double mutants (a+/a+ ; s+s+ and a/a ; s/s) both are "normal" pigs. this will result in a 10:6 ratio and the double mutants will have the suppressive gene and the ones with mutants on each gene (a and s) will have the mutant gene. i dont know if this is right but im positive it has to be a suppressive gene. hmmmmm. but then again i might be making this to complicated and the AA will be a normal pig and the aa will die. or the pig has to be hetero for the trait and if its homozygous it will be a wild type. ahhhhhhh.

[personman]

i dont know what the hell all those terms mean, ive been sleeping through ap bio. but I'm pretty sure the way I said would work. there are genes that will kill you when they are homozygous dominant but when they are heterozygous you live.. or something.. i cant remember.

yea im going to fail the ap exam. oh well.

[jenarelJAM]

and <i>this</i> little piggy went...

[ScatterPlot]

(Gotta preface this by saying I've had barely any Bio)

If there are 2 genes responsible, say A and B, here's the square you get-

        AB         Ab        aB         ab
AB      AABB       AABb      AaBB       AaBb
Ab      AABb       AAbb      AaBb       Aabb
aB      AaBB       AaBb      AaBB       aaBb
ab      AaBb       Aabb      aaBb       aabb

Looks to me like if you have either an AA or a BB or both in there, then you get a normal non-truffle pig. There are 7 of those in there, leaving 9 that are not- 7:9 is close to 6:10

[BD_Paintball]

(Gotta preface this by saying I've had barely any Bio)

If there are 2 genes responsible, say A and B, here's the square you get-

        AB         Ab        aB         ab
AB      AABB       AABb      AaBB       AaBb
Ab      AABb       AAbb      AaBb       Aabb
aB      AaBB       AaBb      AaBB       aaBb
ab      AaBb       Aabb      aaBb       aabb

Looks to me like if you have either an AA or a BB or both in there, then you get a normal non-truffle pig. There are 7 of those in there, leaving 9 that are not- 7:9 is close to 6:10

well i turned in the problem today and the 9:7 ratio means that the genes are in the same pathway and these genes aren't. i already did a problem where the genes were in the same pathway so it cant be that. i put something down that i think is right, i will let you know if i was when i get it back in a few days.