ex if a pig with white dominant coat mates with a pig with black recessive coat what is outcome and please go into extreme step-by-step detail!!!
123283
First of all you have to find out if your dominant trait (white coat) is homozygous (WW) or heterozygous (Ww). Your recessive trait has to be (ww)(b/c both alleles must be recessive in order for the black coat to be expressed.) After you find this out you can use a Punnett Square to plug in your information and find the outcome. This site will step you through drawing a punnett square step by step. http://www.borg.com/~lubehawk/psquare.htm
381195
B\nwhen the white and purple were bred, they were all purple so it must be dominant. those plants would then be Pp, and when bred with pp half the plants were purple (Pp) and half were white (pp)
390816
animal farm was a good book about Stalin's regime and commuism. as u probably already read, not all animals were created equally. the differences showed the ability of the animals to think and their intelligence. the pigs were the most articulate and therefore the obvious leader animals of the barn animals. they could get thier piont across  and could. none of the other animals could write or read as well as the pigs. the speech made by major showed that the animals were not al equal because the pigs sat in the first row, you could see from the begining that pigs thought they were better than the other aniamls.
350011
the genes of the chickens are complementary to each other. the black is AA and the gray is Aa. when a punet square is completed, the result is half black and half grey. The white chicken is aa. when a cros between the grey Aa and white aa, the resukt will be half white and half grey. if the cross was white crossed to black the result is all grey (test cross).
657112
firstly ... 4 all those who say pigs dont hav wings \nhttp://perdurabo10.tripod.com/sitebuildercontent/sitebuilderpictures/wingedpigs.jpg\nhihihi ;)\n..... n now BABy ... yes its very true ... a recent news says a few pigs hav been endowed with wings by tissue culture ... the experiment was done on ten piggies n round 8 hav grown wings... 4 of them hav got wings to a length of about 3 feet ... n the rest 4 hav got wings varying from 2 cms to 13 cms. ... n scientists claim that these wings can regenerate just as starfishes do ... the tissues implanted hav been cultured with the genes of starfish as well ... this is the cause y scientists claim they can regenerate .... hope this piece helped :)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nomg!! ... im  too good ... i just made it up .. n it sounds sooo true n factual  to me!! .... hihihi ;)
32902
kurdiatcha's mostly got it - I've just got a few things to add:\n\n1) The original transmission of HIV from monkies to humans actually occurred in a hunting accident during which the blood of a monkey mingled with the blood of a wounded hunter. Consuming well-cooked monkey meat probably would not pose a significant threat of HIV transmission.\n\n2) Creating a monkey/human hybrid is not possible, and it's not likely that it will ever be possible. Yes, humans and monkies share much of the same genetic material, but it's important to note that we also share much of that same genetic material with plants, fishes, etc. (think about it from an evolutionary perspective, and it makes sense). If you're taking chromosome numbers into account, humans (46 chromosomes) are actually more closely related to bats (44) or tobacco plants (48), although you can imagine what a disaster it would be trying to cross a human with one of these two organisms. For comparison, rhesus monkies have 42 chromosomes.\n\nIf you've had general biology at the college level, you're aware that chromosome numbers are crucial in reproduction. Considering the cases in which breeding among different species is successful (think male donkey [62 chromosomes] + female horse [64] = mule [63]), the progeny will not be capable of reproduction due to its odd number of chromosomes. (Each diploid individual contributes only half of his/her genetic material, remember, forming a new diploid individual from the two haploid sets of chromosomes.) In short, human and monkey chromosomes combined could not create progeny (viable or sterile).\n\n3) Tons of people still cling to the idea that molecular biology (genetics) can accomplish anything. This is not true - yet. While the field is advancing at an incredibly rapid rate, we're nowhere near successfully using gene therapy to cure diseases, for example. (The France gene therapy experiment resulted in fatalities.) Sure - we have the human genome sequenced, but we're still figuring out exactly how everything works. It's unlikely that science will be interested in attempting to cross humans and monkies when so much more important genetic research has to be done.\n\nI'm done. :D
643081
Most if not all species face extinction. the question is thus what is the fate of the Panda sibling species.\nalas with no clear ability to switch habitat and food, and with the destruction of their existig habitat and isolated inbreeding 9they are not inbreeding tolerant) the fate is likely extinction of that particyular gene pool line\n\nBut cheer up, the end might be in sight as humans are only a transient blip on this planet. Their extinction is essentially guaranteed (daft non-recombining X / Y system for example)  the only questions are when (UNKNOWN) and will it be just the species with survival of postcestors (no noticed difference), or the exctinction of all sibling species and thus end of the line
555026
T-dominant- means no tay-sachs\nt-recessive- tay sachs\n\nto have Tay Sachs you must have 2 recessive alleles so tt (this is homozygous)\nThe only way the parents could have a child with tay sachs is for both to be heterozygous for tay sachs which means Tt.\n\nYour punnent square would look like this:\n\nMom goes across and dad goes down\n\nMom-T    t\n''''''''''''_________\nD  T  I  TT I  Tt  I\nA---------------------              \nD  t  I  Tt    I  tt    I\n------------------------\n\nTo do a punnent square you write out one parent's alleles across the top and the other along the side.  Then in each box you write the alleles that are mix in each box.  You can see that the parents would produce in the first box TT which means that person is homozygous (both the same) (also since both capitols would never pass on tay sachs).   The second box across Tt which is heterozygous.  We call this person a carrier.  A carrier is someone who does NOT have tay sachs, but can give it to someone else.  The first box in the second row is Tt too.  A carrier as well.  The second box in the second row is tt.  This is homozygous again, but since a lower case t represents tay sachs this person has it.  \nSo there are 4 boxes.  1/4 of the time TT- no tay sachs\n                                     2/4 of the time Tt-carrier but no tay sachs\n                             and  1/4 of the time tt-tay sachs\n\n1 divided by 4 is .25 which is 25%.  \nThese parents have a 25% chance of having another child with Tay sachs.\n\nSorry about the punnent sq. I couldn't get it to draw right.
413964
Chickens have 1 head and 2 feet, pigs have 1 head and 4 feet.\n\nLet c = # of chickens, p = # of pigs.\n\nSo we have the following:\n\nc + p = 24\n2c + 4p = 80\n\nYou can solve 2 equations in 2 unknowns.  Multiply the first equation by 2, then subtract it from the second to solve for p:\n\n2c + 4p = 80\n2c + 2p = 48\n--------------------\n0c + 2p = 32\nSo .... p = 16\n\nNow substitute p=16 into your first equation to solve for c:\n\nc+16 = 24\nSo ... c=8\n\nANSWER:\nChickens = 8\nPigs = 16\n\n~ ♥ ~
484891
this question depends on the brown hamster. if its BB crossed with the white (must be bb) then you will get all brown offspring.  However if the brown parent is Bb, the the offspring will be 1:1.  Two white offspring and two brown offspring.
513540
let x be chicken\ny be the pigs\n\nx + y = 36\n2x + 4y = 100\n\n2x + 2y = 72\n2x + 4y = 100\nsubtract\n2y = 28\ny = 14 pigs\n36 -14 = x\nx = 22 chicken\n\n^_^
230159
1. The barber shaves himself. Because you have written "He shaves ALL and...". All includes himself as well.\n\n2. No, it dosen't support the logic. Ravens could be "Non-black", but still be Ravens.\n\n3. He will hand the baby back. If he was so intent of eating the baby, he would probably have not proposed the question to Melissa in the first place.\n\n4. Lets take two cases...\n\nCase 1...Suppose A is dog.\n\nThen, according to the conditions given,\nB is dog, C is wolf, D is dog, E is wolf, and A is wolf...which contradicts our assumption.\n\nCase 2...Suppose A is wolf.\n\nThen, according to the conditions given,\nB is wolf, C is dog, D is wolf, E is dog, and A is dog...which contradicts our assumption again.\n\nThus, there are no fixed species of animails here, or the data given is insufficient.
533126
Ok, I'll help you out, but I'm not going to give you an answer because I'm not going to do the work.  Besides, you were pretty sloppy here, calling dogs kittens and #3 makes no sense if you're talking about whole oz.  So reword that ridiculous statment to say that dog B weighs EXACTLY the same as dog F.\n\nFirst thought, always:  "How many unknowns do I have"\n\nYou have seven:  A-G\n\nSecond thought, always:  "I MUST have 7 different equations or I cannot solve the problem."\n\nNote:  In this problem if you have less than 7 different independent equations, it is IMPOSSIBLE to solve the question (that is, without some other constraint.  For example if you had some problem asking you to find the total number of puppies in the kennel, and the numbers worked out you'd have 2 puppies or 0.5 puppies, you have to go with 2 because you can't have half a puppy unless you're a sicko).\n\nOk, you need 7 equations.  Start writing.\n\nA+B+C = D+E+F+G\n\nC=B+2=A+6\nF=G+12=E+3=D+13\nB=F\n\nThere are your seven equations.  I leave it to you to understand fully where those seven come from.  From here, you start substituting, until you get the first equation in terms of one single variable.  Without giving it too much thought, I would suggest "F".  Solve for that one variable, then solve for the other variables.  Since you have 7 equations and 7 unknowns, there is no "minimum" weight:  you will have the exact weight of the kittens.
354994
There is no crossing over during prophase 2! It's a fact. That answers your question pretty nicely.
197609
You will get a rabbit that is temporarily immune to the bacterial antigen that was injected into the first rabbit. This would be like giving a type O mother Rhogam.
532895
To start with, You need more information as to the Parental genotype for each question, \nFor example in question (a) since you have only mentioned full dominance, we can either assume both parents will be A/A x A/A. However this cross is useless because you know there will not be variation in the F1 progeny because both are dominant. thus the ratio is 100% all dominant A/A.\nThe second assumption is to cross a dominant parent A/A with a recessive parent a/a. If that's the case then you choose one gamete from each parent to form the F1 as follow\n\nParental Genotype        A/A    x    a/a\nGametes                          A     x    a\n\nF1                                         A/a\n\nfrom here, if you need to get genotype and phenotype ratio for your F2, you need information as to what kind of cross you want do. You can self the F1 or you can testcross with a recessive parent. In either case the ratio is different\n\nSelf-cross                  A/a    x      A/a\nGametes              1/2 A , 1/2 a    x   1/2 A , 1/2 a\n\nF2  punnet square\n..............1/2  A..........1/2 a\n\n1/2 A......1/4 A/A........1/4 A/a\n\n1/2 a.......1/4 A/a........1/4 a/a\n\nPhenotypic ratio of F2:  1/4 A/A; 2/4 A/a; 1/4 a/a = 1:2:1\nGenotypic ratio of F2:  (1/4 A/A + 2/4 A/a); 1/4 a/a = 3:1\n\n\n\nTest-cross                  A/a    x      a/a\nGametes              1/2 A , 1/2 a    x   1/2 a , 1/2 a\n\nF2  punnet square\n..............1/2  A..........1/2 a\n\n1/2 a......1/4 A/a........1/4 a/a\n\n1/2 a.......1/4 A/a.......1/4 a/a\n\nPhenotypic ratio of F2:  1/2 A/a; 1/2 a/a = 1:1\nGenotypic ratio of F2: same as genotypic ratio\n\nquestion (a) is not formulated in a clear way. for example. question (a) should have mentioned something like this:\n\na) The F2 of monohybrid cross with both parents full dominance, or 1 parent full dominance, 1 parent recessive.\n\nIs hard to work in genetics problems when the problem lacks crucial information. let me know if there is more information that you did not write down on your question.
389934
No, it will perish.  Species take thousands of years to evolve.  If you put the common pig into the wild you have to consider that for it to adapt to the new environment it will take a process where thousands of pigs would have to be involved, most will die, and a few will inbreed traits that will help them survive in their new environment.  If you release just one pig, even if it were to survive, it will die in time being the same pig you released.