all about punnett squares
January 28, 2006 at 10:21 am #3381
i am a bit confused about punnett squares. one thing i can’t understand is how those letters get there and how they got to be combined like that in the squares. also, i can’t understand how to make a punnett square. all i know is make a square, cut it into four and place letters. but how am i supposed to place the letters if i don’t know where (in exact place) should i put them? please help me. my teacher does not discuss it briefly and neither one of us in class understands it at all. could you please help me? 😥
January 28, 2006 at 1:35 pm #38735
January 29, 2006 at 5:22 am #38802Dr.SteinParticipant
How many Punnett Square thread in this forum? I count there is something like four or five… hmmm 😕
January 30, 2006 at 4:58 am #38865
February 2, 2006 at 10:37 am #39310
Thanks!!!! but i also want to ask. which is more easy to use when solving genetic problems: algebraic solutions or the punnett square?
February 2, 2006 at 8:00 pm #39348
It depends on the letter (I mean the characteristic number) number you are given. If you are given like 20 letters, I don’t advise you to use punnet square. 🙂
February 3, 2006 at 12:03 pm #39455
could there be a difference between the two? i mean the algebraic solutions and punnett square? what if the letters are only four? which is the easiest and more advisable way?
February 3, 2006 at 12:52 pm #39469MrMisteryParticipant
well the punnet sqaure will probably take you more time. If you have 4 alleles it will take you some time. But punnet squares guarantee accuracy, unlike algebra where you can easily make a mistake(like me)
February 3, 2006 at 8:03 pm #39503
The results will be the same, if you don’t make mistake. Well, I don’t like maths. So if the letter number is 4 I advice you to use punnet square. Maybe it may take some time, but, as Andrew said the probability of making mistake when using punnet square is less than the other.
February 4, 2006 at 4:08 am #39536
another question. if i am performing a dihybrid cross, which shoul i prefer to use: punnet square or the algebra? i am not that fond in math but i’ve got good grades mind you. 😉 8)
February 4, 2006 at 8:02 am #39559M.Shanti PriyaParticipant
For a dihybrid cross you can use punnett square. From trihybrid better to go for dichotomous / branching method.
February 4, 2006 at 9:27 am #39563
what’s a dichotomous / branching method? i’ve never heard of it, sorry. could you explain it please?!
February 4, 2006 at 9:35 am #39564
another question. how do i solve this problem:
a cross of F1 plants heterozygous for height should produce a ratio of tall to short plants of 3:1. Three seeds from such a cross have been grown and they ahave produced three tall plants. what are the chances that a fourth seed will produce a short plant? And what do you mean by probability?
February 4, 2006 at 5:10 pm #39633
Have you made apunnet square? Try it. Then think that tall is dominant over short ( you can understand this (even if you don’t know) from the ratio given. Make the punnet square you will findand understand.)
For probability: Lets say you made a punnet square and crossed. You think the whole as 100. then you divide the whole into percentage (%).
And a hint for you: pay attention to the ratio and the words "heterozygote F1". 🙂
February 6, 2006 at 9:41 am #39851
oh. now i understand. though i haven’t actually done the punnet square yet because i’m quite confused with the question, i understand your point. but now, my question is what’s a dichotomous / branching method? is there really a trihybrid cross because all i know are mono and dihybrid cross. i’ve never heard of it, sorry. could you explain it please?! 😕
March 2, 2006 at 11:15 am #42156
how do you solve this kind of problem? I never seem to have any idea on how to answer this. i even have no idea on how to make a punnet square on this problem. help me. onegai!
A normal visioned and bald woman whose father is bald and a mother who is red-green color blind marries a man who is nonbald but is red-green color blind.
a. What could be the probability that their children will be:
1. bald girls?
2. red-green color blind boys?
3. bald boys?
4. normal visioned girls?
5. nonbald and normal visioned girls and boys?
6. red-green color blind and bald boys and girls?
7. nonbald and red-green color blind boys and girls?
8. normal visioned and bald boys and girls?
b. Give the phenotypic ratio and genotypic ratio.
c. What could be the probability that they’re children will be both normal to girls (in percent)?
please help me! i really have no idea on how to solve this!
- You must be logged in to reply to this topic.