September 1, 2007 at 6:43 pm #8160
This is my first genetics assignment, and I know have very little understanding of the information so far. Any help would be greatly appreciated. Here’s one of my questions.
Two homozygous strains of corn are hybridized. They are distinguished by 6 different pairs of genes, all of which assort independently and produce and independent phenotypic effect. The F1 hybrid is selfed to give and F2 generation.
What is the # of possible genotypes in the F2 plants?
How many of these genotypes will be homozygous at all six gene loci?
If all gene pairs act in a dominant-recessive fashion, what proportion of the F2 plants will be homozygous for all dominants? What proportion of the F2 plants will be homozygous for all dominants?
What proportion of the F2 will show all dominant phenotypes?
How should I begin this problem? If there are 6 different pairs of genes, do I need to do a trihybrid cross?
September 2, 2007 at 7:53 pm #75739dan167Participant
Hey, I am only in yr 13 so this is probably wrong 😳 , but I am bored so gonna give it a go.
The corn is Homozygous but dosent say if its dominant or recessive, so it could be:
AABBCCDDEEFF X AABBCCDDEEFF or AABBCCDDEEFF X aabbccddeeff or aabbccddeeff X aabbccddeeff
F1 would be:
AABBCCDDEEFF or AaBbCcDdEeFf or aabbccddeeff
F2 would be: (I am not sure what selfed means but im guessing breed with each other)
AABBCCDDEEFF or aabbccddeeff or
Parents: AaBbCcDdEeFf X AaBbCcDdEeFf
Gametes: So many to work out 2 genes is 4, 3 genes is 9 so I guess you square it so 36 different gametes? X abcdef
Dunno if there is a fast way to work out the gametes…
September 3, 2007 at 1:28 am #75744mithParticipant
*hint* you’re going to find a lot of different combos, maybe pascal’s triangle would help.
September 3, 2007 at 9:44 am #75752BDeisParticipant
This is where knowing your genotype and phenotype ratios comes in handy.
A punette’s square will show that the only genotype in F1 is Aa for all six genes
1) The hetrozygous geneotype formula is 3 to the nth power where n = the number of segregating gene pairs. So the answer would be 3 to the 6th power.
2) The genotype ratio for each gene is 1/4 AA: 1/2 Aa and 1/4 aa. So half of each gene will be homozygous (AA or aa) and half will be heterozygous.
If you multiply the ratio for each of the six pairs you get your answer. In this case it is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 or 1/2 to the 6th power
3) This is done the same way as #2 except you use the Aa ratio
4) This is done like 2 & 3 but you use the phenotype ratios. AA and Aa will express the dominate trait so 3/4of F2 will express the dominate phenotype (1/4 + 1/2).
Hope this helped.
September 3, 2007 at 11:19 am #75753
Thank you so much BDeis. I hope that you won’t mind me asking more questions in the future. It is so hard to get these problems started on my own, but what you explained made perfect sense. I am very appreciative.
September 3, 2007 at 11:26 am #75755BDeisParticipant
No problem. In fact, I am studying the same thing right now.
September 3, 2007 at 11:46 am #75756
Wonderful….I’m about to go to work, but I’ll post one of my other problems later. I just want to get a grip on this early stuff, so I wont be so clueless later. I don’t mind putting in the many hours of studying, but it sucks when I still don’t get it. Ahh genetics! Glad one of us is "getting it". 😉
September 3, 2007 at 4:44 pm #75763
Here’s another one that I’ve been working on figuring out.
A corn geneticist has three pure lines of genotypes a/a; B/B; C/C, A/A; b/b; C/C, and A/A; B/B; c/c. All the phenotypes determined by a, b, and c will increase the market value of the corn; so naturally he wants to combine them all in one pure line of genotype a/a; b/b; c/c.
a. Outline an effective crossing program that can be used to obtain the a/a; b/b; c/c pure line.
b. At each stage, state exactly which phenotypes will be selected and give their expected frequencies.
c. Is there more than one way to obtain the desired genotype? Which is the best way?
Assume independent assortment of the three gene pairs. (Note: Corn will self or cross-pollinate easily.)
Do I mate 2 of the pure lines of genotypes, self the F1 and then cross the remaining line of genotypes with the F2?
What do you look at when attempting to answer these types of questions? I want to know how to start breaking down the question so that I can figure it out. Thanks again!
September 3, 2007 at 5:36 pm #75766Final_KingParticipant
Two homozygous strains of corn are hybridized. They are distinguished by 6 different pairs of genes, all of which assort independently and produce and independent phenotypic effect. The F1 hybrid is ❗ selfed ❗ to give and F2 generation.
are u refering to the plant as a sexual or b sexual reproducer. A i think is going to reproduce anther and b sexual reproduces by itslef so if it the corn plant reproduces with its self the corn plant all i know is its dna should be the same unless there is a mutation wihten the gentetic code but if it needs to be pollenated with anther corn plant or farmer JOe decides to cross polinate it with anther species fo plant like tomatoe[pplant then the corn and corn plant would have slightly diffrent dna i think hey i just got into this subject and the corn breeded with the tomatoe plant iare breeded together there should be a radical change in dna rather than small change from a mutation or breeding with anther corn plant. i would apreciati it if a mderator could clerafy my statment seeing as how i stink at speeking/typing/ or writing im not that conversational.
September 3, 2007 at 5:38 pm #75767Final_KingParticipantquote Final_King:
September 3, 2007 at 11:41 pm #75774
Still not very clear on how to set up the corn question…if anyone else could help that would be wonderful.
September 4, 2007 at 4:29 am #75779mithParticipant
Try working backwards, to get a pure line of all recessives, you need either two pure lines of recessives or a heterozygous line mating with a heterozygous line.
September 5, 2007 at 5:35 pm #75818
For anyone who this helps, here is how I now know how to do it. Mate the first two pure lines, self the outcome, and use a branch diagram to find the phenotypic ratios. Use the oucome that has the ratio of 1/16 and mate that with the 3rd pure line. Self the outcome, and determine the phenotypic ratios. aa bb cc was formed in a phenotypic ratio of 1/64. Thanks to the forum folks for all of your help!
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