How does a Punnet square help us to understand an offspring's phenotype & genotype?
483751
a punnets square allows you to graphicially understand all possible offspring created by two parents. the phenotype is the outward expression of the gene you are studying like a rabbit being brown or white. the gene is usually coded by the first letter of the dominant phenotype. suppose brown was thought to be dominant and white was recessive. brown would be called BB if it was homozygous or Bb if if was heterozygous. white would be called bb. the heterozygous organism is always the dominant phenotype. (except in codominance etc.) punnet squares can be used to answer questions like is it possible for two blondes to have a brunette baby? or is it possible for two purple plants to produce seeds that would make white flowers? your book or your notes will show you how to set up a punnet square properly.
976375
AA = p(squared)\nAa = 2pq\naa = q(squared)\np + q = 1\n\nFirst you need to interpret "9% have the disease". This means that 9% of the population have both recessive alleles, which means the frequency of aa is 0.09. You are asked to calculate the percentage (or frequency) of the resistant (or dominant allele), which is denoted A. So by using the above equations:\n\naa = 0.09, which can also be expressed as a(squared) = 0.09\n\na = square root of a(squared) = square root of 0.09 = 0.3\n\nThe recessive allele (a in this case) is always donated by q, and the dominant allele (A in this case) is always denoted by p. So therefore; a = q.\n\np + q = 1\np + 0.3 = 1\np = 1 - 0.3\np = 0.7\n\nSince p = A, the frequency of the dominant allele (A) equals 0.7
881626
Construct a punnet square:\n\nMom=  I^A & i\nDad= I^B & i\nOffspring will have a 1/4 chance of being: I^AI^B, I^Bi, I^Ai, ii\n\nIf the mother is I^Ai and the father is I^Bi, the possible blood phenotypes in the offspring are AB, B, A, and O.  There is a 25% chance of each blood type.
1017859
Well, the equation is : (p^2) + (2pq)+(q^2)= 1....where (p^2) and (q^2) represent the frequency of homozygous alleles and (pq+qp=2pq) represent the frequency of heterozygous alleles.  \n\nThe way to use this formula is for example : " if you have a plant population that consists of 84% plants with Red flowers (RR, Rr -alleles) and 16% White flowers ( rr-alleles)," you use the formula above as follows:\n\n(q^2) = .16= White flowered plants (rr trait)\n(p^2)+ (2pq)= .84= Red flowered plants (RR, or Rr traits)\n\n\n\nNow there is another equation : p+q=1\n\nSo, take the square root of (q^2)=.16, q=.4 and then plug that into p+q=1. Now solve for p,  p should be .6.  Now  note that q=.4 and p=.6 and with that said plug it into the original equation in order to find the frequency. \n\n(p^2) +(2pq)+(q^2)=1\n\n(p^2) = (.6) (.6) = .36  or 36%  for the homozygous dominant trait (RR)\n\n(q^2)= (.4)(.4)= .16 or 16% for the homozygous recessive trait- (rr)\n\n(2pq) = (2)(.6)(.4)= .48 or 48% for the heterozygous trait - (Rr).\n\nAll of this should equall 100%.  So 36%+16%+48%=100%.\n\nSo, basically the Hardy-Weinberg equation is to see the frequency of the alleles (RR,Rr, and rr) in a population.
807540
First thing you want to check is to see if the trait skips a generation or not.\n\nIf it does then chances are that it's a recessive trait.  If it doesn't then chances are that it's a dominant trait.  \n\nI want to stress the word "chances are" because it's possible that a trait NOT skip a generation yet still be a recessive trait.  Blood O type, for example, is recessive whereas A and B are dominant.  If both parents are blood type O, the child will be blood type O as well.  \n\nNext thing you want to check out is count the number of traits in the males compared to females.  \nIf it's about a 50:50 ratio, then it's an autosomal trait.  \nIf it's mostly males and virtually no females, then it's an X-linked recessive.  \nIf it's about 1/3 male and about 2/3 females, then it's X-linked dominant (doesn't happen very much but still possible)\n\nGood luck!  Hope this helps!
830750
1. heterozygous: Punnett square is Pp x pp\n2. dominant b/c if it was recessive that would be the only type of allele the parents would pass on and all their children would have to be intensely freckled\n3. the Punnett square would be a 4x4 box with the B's and C's independant of each other. so on one side you have BC, Bc, bC, and bc with all of those showing up on the other side as well b/c the female was the same genotype as the male.  Obviously the black color is dominant b'c the male is heterozygous and black so all heterozygous offspring and all homozygous dominant offspring will be black as well.  the question gets a little fuzzy here though, i think you may have left some out b/c we don't know which of the two genes controls color or if they both do how the interaction between them works.  I'd be glad to help further if you have any more information to supply.
969016
I believe that these questions  relate to the "Hardy-Weinberg equalibrium" for population genetics.\nP(purple/dominant)=700 or .7\nq(White/recess.)=300 or .3\nP+q= 100% or 1.00\nPlug these numbers into the quadratic equation:\nP squared + 2Pq + q squared  = 1.00\nnote: the 2Pq represents the hybrib condition\n\n#1 P=.7 , q =.3      Ans.  B\n#2  2(.7)(.3) = .42        Ans. ????\n#3      Ans. A ??\n#4  q = .0001 Ans. A\n#5 2(.9999)(.0001) = .00019   Ans.A
51603
A punnett square helps to organize all of the possible results and from that you can see the odds of different genes and their characteristics appearing in offspring.
426265
Sure, I do.\n\nStep one: identify your question.\n\nStep two: figure out as much as you can about the phenotypes of the parents and the offspring\n\nStep three: some of the genotypes may be obvious from the phenotypes. \n\nStep four: figure out what kind of gametes an individual can make. \n\nStep five: Are you in control of any crosses? Or are the crosses already done? \n\nStep six: If there is any need, do Punnett squares of those crosses you can't just figure out by looking.
566012
EEEEH. I'm not really good with this mumbojumbo stuff but here is a site on genotypes and stuff. -> http://janis7hepc.com/Genotypes.htm <-
197178
Erase, cancel out all the introns, promoter and tata box, and you are left with those sequence you want. Then use the dna translation table to figure out the amino acid sequence. Then use those amino acid sequence to figure out what phenotype they correspond to via doll codes.
903177
Take each of the three traits seperately and compare to expected mendalian ratios. (Testcross means the male was completely recessive. If the female is a heterozygote then the expected ratio is 1:1, if homozygous dominant 1:0, if homozygous recessive 0:1). \n\nFor example\n243+241+15+10=509 black (1000/509=2)\n235+226+12+18=491 not black (2)\n==>approximately 1:1\nThe female had to be Bb because both types are visable.\n\nscarlet =491\nnot scarlet=509\n==>1:1, STst\n\nhooked=514\nnot hooked=486\n==>1:1, HKhk\n\nFemale is B b, ST st, HK hk\n\nFor linkage/mapping distance, take the total progeny that do not look like either parent and had to come from a crossover event and divide by the total number of progeny. Assume that the combinations you see most often are parental  (243+241+235+226). Work with two traits at a time. Significant deviation from mendalian ratios indicates linkage.\n\nb and st\nBST(parental)=235+18=253\nbst(parental)=243+10=253\nBst=226+12=238\nbST=241+15=256\nparental=506, recombination=494...approximately1:1 no linkage \n\nb and hk\nBHK=12+18\nbhk=15+10\nBhk(parental)=235+226\nbHK(parental)=243+241\nparental=945, recombination=55   \n55/1000=.055(100)=5.5cM between b and hk\n\nYour map distance is cM.
1024331
I apologize but all that i can come up with is one.\n\nYou could raise fruit flies (Drosphila) and relate them to Gentics. i.e. punnet sqaures, Gregory Mendel, common characteristics, principle of dominance and recessiveness, etc.\n\nif you would like me to go further into detail about this project, please e-mail me at \n\nthe_lonley_kender@yahoo.com or send me an IM on Yahoo! at the_lonley_kender or on AIM at  thelonleykender
266686
i took biology way back ..lets see...8 years back, :)\nbut this is what i found on a website...\n\n\nProblem 5: Dash Technique\n\nRecall that a phenotype is the outward expression of an organism's genes or genotype. For example, type A blood (the phenotype) is the expression of the genes, AA or AO (the genotype).\n\nNote that we do not know for certain whether the correct genotype is actually AA or AO since both genotypes will produce the type A phenotype. For this reason, when solving many blood type genetics problems we let a "dash" represent the uncertain gene In the example of the type A blood, we could represent the genotype as A-.\n\nTheres a sample problem that the website has solved!\n\nHappy genetics..long live Mendel..but hes dead..isnt he? that was dumb!
348504
Genotype of the man-Pp\nGenotype of the woman-Pp\nPossible genotypes of children-PP, 2Pp, pp\nPossible phenotypes of children-2 carriers, one that has it, one that is completly free of it\nProbability of PKU disease in their children-25%
427517
Something to do w/ genetics............\n\n-Mr. Happy-\n:o)
191231
i'm not sure if it's the "dash" technique, but sometimes, when u're writing genotypes and the knowledge of only one allele is req'd... u can just write A/- ... meaning A/A or A/a.\n\nhope that's useful.