All of the offspring resulting frin a cross between a red snapdragon and a white snapdragon are pink.  What is a possible explanation for this?
213783
I'm not totally sure on this, just guessing what my mum has told me, and seeing how she gardens a lot, I'll assume it's true: The red and white seeds can grow together, to form a hybrid, adding a percentage of both colors to form pink. Hope it helps!
120047
Let's assume that "tongue rolling" is a dominant trait (I believe it is), and that it is not sex-linked. Let's call the dominant allele "R" and the recessive allele "r". The female must be rr, because the heterozygous condition, Rr, will yield a person who is able to roll her tongue.\n\nAll offspring will be able to roll their tongues.\n\nHere is the cross:\n\n-----R----R----\nr | Rr -- Rr\n\nr | Rr -- Rr\n\nAs you can see, all F1 offspring will be heterozygous, and will therefore all be able to roll their tongues.\n\nBy the way, Grumpykiwi, it doesn't mean that it isn't dominant. It merely means that your mother was HETEROZYGOUS, rather than homozygous dominant, as was the man in the problem.
182103
Are the genes linked? I don't have my book with me here.\n\nIf not, mate a male with one of them. Take the male progeny from that mating and cross with the other mutant. Select hermaphrodites from the period of egg laying when the sex ratio is 1:1 (say the second 8-hour period after they're together). Allow these hermaphrodites to self and select double mutants from the progeny. I think this should do it, but you'd better draw it out.\n\nIf they're linked I think it will just be harder to find double mutant progeny in the final stage, again draw it out to check.
360986
Guinea pigs are not Mendel's peas.  As a child, I tried to breed a race of gold-coated, straight hair Guinea pigs.  It turned out that only males could be gold-straight.  The females were never gold, and only sometimes straight, no matter what I did in back-crossing.  Perhaps one of these traits is sex linked?  I think hair color is too complicated to be dealt with on a simple dominant/recessive cross grid.
250820
well, i hope it wont happen that, but id like it for one reason : my friend has a problem :she will go at the prom, and she doesn't know what to chose between orange & green (serious problem!!! :)) so, if "orange and green got married and were always together ", she has 2 choices : to give up and chose other colour, or to wear them both! :)
36636
well if it's the classic example..it's pink..the intermediate color between red and white.
493131
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.
256028
http://www.pepto-bismol.com/history.shtml\n\nIt says a Dr. mixed up the first form of Pepto and added colorant to make it pink, it doesn't say why he wanted it to be pink.  Probably because once he mixed up the ingredients it looked nasty and he added color to make it look better and pink is just the color that happened.
544196
"A" is the answer.  It happens in prophase of meiosis 1 and it does exactly what choice "A" states.  The rest of the choices happen during meiosis as well, but do not explain the importance of crossing over.
546115
I asked that question, and there is no known mechanism. There is however a mechanism that keeps different species from being able to interbreed.
280779
The answer is "b"\n\nAssuming the dwarf characteristic is recessive, since about half of the offspring were dwarf, that means that half the offspring have to by homozygous for the dwarf genotype (tt).  The only way to get that is to cross the Tt and tt.  Thus you get half the offspring being Tt (tall) and half being tt (dwarf).  If you cross a homozygous dominant (TT) with anything else, all the offspring will be tall, because all will carry at least 1 "T" gene.  If you cross "Tt" and Tt" then you get 1/4 TT (tall), 1/2 Tt (tall), and 1/4 tt (dwarf).
521562
During prophase 1 of meiosis, the chromosomes have duplicated and have joined with their homologous pair to form a 4 stranded tetrad.\n\nBecause you get one chromosome in each homologous pair from your mother and one from your father, although they code for the same concepts, they may code slightly differently (ie blue eyes vs brown eyes).\n\nTo exchange genetic material, during prophase one the two chromatids closest together in the tetrad (one from each mom and dad- the outer two are too far away to do this, so they remain untouched) "cross over".  Literally, the ends of the chromatids cross each other and bits of the chromosomes break off of one and reattach on the other, giving genetic variability.\n\nOk, so for example: the chromosome your mother gave you coded for blue eyes, blonde hair, curly hair, and long fingers.\nThe chromosome from your father codes for the same principles, just differently- say brown eyes, red hair, straight hair and short fingers.\n\nSo crossing over between these two occur in prophase 1, and some codes are swapped.\n\nNow we have a tetrad (4 chromatids) that code for the following.\n(We'll say the chromosome from mother is on the left and from father is on the right)\nStarting with the leftmost and going to the rightmost:\nChromatid 1 codes for: \n-blue eyes, blonde hair, curly hair, and long fingers\nChromatid 2 codes for:\n-blue eyes, brown hair, curly hair, and short fingers.\nChromatid 3 codes for:\n-brown eyes, blonde hair, straight hair, and long fingers\nChromatid 4 codes for:\n-brown eyes, brown hair, straight hair and short fingers.\n\nSo now we have the genetic variability, and 4 completely unique chromatids, that are still homologous pairs, just giving us four different codings for the same traits.\n\nHope this helps.
328093
I don't see how it is false either. I mapped it out, and here's what I got. The eldest lady (grandma) could be XcXc (homozygous for colorblind). Grandpa must be XhY (hemizygous for hemophilia). This makes normal woman XcXh (the Xc would be normal for hemophilia and the Xh normal for color, masking the expression, resulting in normal). Husband must be XcY(hemizygous for colorblind). Son could inherit Xc from mom, thus being Xc (colorblind). Daughter could inherit Xh from mom and must inherit Xc from dad, thus being XcXh (the Xc would be normal for hemophilia and the Xh normal for color, masking the expression, resulting in normal). Please let me know if there is an error.\n\nSon cannot be normal; he'd be either Xc or Xh which is either colorblind or hemophilia.
494052
First, note the word "true-breeding".  That indicates each plant is homozygous for all those traits, which means that you don't have to worry about that plant being het for any of those genes.  And it means that they only make a single kind of gamete with regard to thsoe 4 genes.\n\nTechnically, these loci might all be linked to each other, you would need F2 data to know for sure.  But you are supposed to assume that four traits are controlled by 4 genes.\n\nNow, since the parents were homozygous for all their genotypes, the offspring will all be identical, they will all be het for those genotypes.  And which ever phenotype they are, that's the dominant one for that trait.  So purple, green, terminal, and yellow are the dominant phenotypes.\n\nAnd there are 16 different gametes to be made, because there are 4 genes (presumably) and each is het in the F1.\n\nThe only way for an F2 to show all 4 recessive phenotypes would be for it to get a gamete from each parent taht also had all four recessive alleles.  So 1 in16 gametes will fit the bill, so the odds of getting that gamete from on parent is 1 in 16, the odds of getting it from the other parent is 1/16, so the odds of getting it from both is 1/256.
398609
A)\n\nAs far as pod size, at least one parent must be homozygous for H, since all the offspring have one pod.  The other parent coudl have any genotype for that gene\n\nThe leaf shape is showing the 3:1 ratio characteristic of a dihybrind cross.  So both parents must be het at the leaf locus.\n\nB) is similar, since all the offspring have 3 pods, the parents must be homozygous for that recessive allele.  The other parent coudl have any genotype for that gene.\n\nThe 3:1 ratio of leaf shape shows that the parents must again both be het for the leaf shape locus.\n\nC)  If all the offspring show the dominant pheontype, one parent must be homozygous for the pod number, and one must be homozygous for the leaf shape.  Either the same parent is homozygous for both, or each is homozygous for one trait.  Again, if one parent is homozygous for a gene, the other parent coudl have any genotype at that locus.\n\nD)  The ratio of one-pod/3-pod is again, 3:1, so both parents must be het for that gene.  Since half the offspring have the dominant leaf shape, and half have the recessive leaf shape, that means that one parent must be homozygous for the recessive allel, and the other must be het.\n\ne) The ratio of one-pod/3-pod is closest to 3:1, so both parents must be het for that gene.  again, the ration is leaf shapes is closest to 1:1, so one parent must be homozygous recessive, the other het.\n\nBut the ratios aren't quite right, there's something non-Mendelian affecting the offspring here.
483836
from parents XXh and XY the offspring would be:\n\nXX (a noncarrying female), XY (a normal male), XXh (a carrier female), and XhY (a hemophiliac male)
201110
It has to do with when the nipples develop in utero (when we are a fetus)...\nOne interesting consequence of the developmental set-up just described is that during the very early stages of fetal life, before the sex hormones have had a chance to do their stuff, all humans are basically bisexual. Among other things, you have two sets of primitive plumbing--one male, one female. Only one set develops into a mature urogenital system, but you retain traces of the other for the rest of your life.\nHuman nipples appear in the third or fourth week of development, well before the sex characteristics. (The sex hormones start to assert themselves at seven weeks.) As many as seven pairs of nipples are arranged along either side of a "milk line," a ridge of skin that runs from the upper chest to the navel. \nhttp://www.straightdope.com/classics/a1_...\n\nMale nipples aren't exactly a genetic glitch: they are evidence of our developmental clock. In the early stages of life from conception until about 14 weeks, all human fetuses look the same, regardless of gender. At the tender age of 14 weeks post-fertilization , genetically-male fetuses begin to produce male hormones including testosterone. These hormones turn the androgynous fetus into a bouncing baby boy. \nHere's where the developmental clock comes in. By 14 weeks, when the hormones turn on, the nipples have already formed. So, while our male fetus goes on to become a baby boy, he keeps his nipples, reminding all of us that people, male and female, started off the same way. \nMale nipples aren't exactly a genetic glitch: they are evidence of our developmental clock. In the early stages of life from conception until about 14 weeks, all human fetuses look the same, regardless of gender. At the tender age of 14 weeks post-fertilization , genetically-male fetuses begin to produce male hormones including testosterone. These hormones turn the androgynous fetus into a bouncing baby boy. \nHere's where the developmental clock comes in. By 14 weeks, when the hormones turn on, the nipples have already formed. So, while our male fetus goes on to become a baby boy, he keeps his nipples, reminding all of us that people, male and female, started off the same way. http://amos.indiana.edu/library/scripts/nipples.html