What is the expected genotypic ratio for a one-factor cross of two hybrid organisms?
149630
1:2:1\n1 homozygous dominant\n2 heterozygous (the hybrid)\n1 homozygous recessive
706626
When it states the frequency of the dominant allele in a population, think of p as one allele that 36% of the population is carrying. so, we can set p=36%, or .36\n\nThe next part of the problem asks for the frequency not of the recessive allele, but of the recessive trait. In order to carry the recessive trait, the person must have both alleles. The probability that both parents passed down the allele for the person to have the trait, using the multiplicative rule of probability would be q*q. We are solving for q^2 since we are looking for the trait not the allele. \n\n1-p=q\n1-.36=.64\nq*q=.64*.64=0.4096\n0.4096*100%=40.96%
813775
Monohybrid inheritance is the inheritance of a single characteristic. The different forms of the characteristic are usually controlled by different alleles of the same gene. For example, a monohybrid cross between two pure-breeding plants (homozygous for their respective traits), one with yellow seeds and one with green seeds, would be expected to produce an F1 (first) generation with only yellow seeds because the allele for yellow seeds is dominant to that of green.\n\nSource:\nhttp://en.wikipedia.org/wiki/Monohybrid
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.
653912
Realistically, none, because the chromosomes will cross over during meiosis, so none of the chromsomes will be exactly as they descended from the parents.\n\nBut if the haploid number is 6, and you pretend there's no crossing over, then there are 2^6 ways for the chromosomes to assort, and 2 of them will be 6 chromosomes from a single parent.  So that's 1 in 32, or about 3%.
439187
Genes that are close together will tend to stay together after a cross-over. The distance between the genes will be roughly proportional to the frequency at which they are separated by cross-overs.
251834
mendel tested genes in peas. he cross bred smooth and wrinkled peas and found that all the offspring grew smooth. then he bred the offspring and they grew some smooth and some wrikled at a ratio of 3:1.\nthis maade him think about dominance, in this case, the smooth allele was dominant to the wrinkled allele. he also did work on smooth gen peas and wrinkled yellow peas. When he bred these, he found a ratio of 9(sreen smooth):3(green wrinlked):3(yellow smooth):1(yellow winkled). this showeed him that the green and smooth alleles were dominant to yellow and wrinkled.\nhe deided that the rules of breeding ratios would either be 9:3:3:1,3:1,2:1, or 1:1 when dealing with dominance
716113
There is no crossing over during prophase 2! It's a fact. That answers your question pretty nicely.
190009
yeah like the first person says. it all depends because there are dominant and recesive genes or chromosomes or whatever, and i think green is dominant, but if you get get 2 blues they dominate over 1 green, i think that is how it works.
425063
2 recessive alleles - Homozygotes
341693
The mother and child will have *nearly* identical DNA.  There will be a few mutations because mutations are ubiquitous, but 99.99+% of their DNA will be the same.
1169109
Crossing over increases genetic diversity.
217273
This is a fairly long answer... first you need to get the initial allele frequencies. equation is p2 + 2pq + q2 (p2 and q2 from here on in is p squared and q squared)\nSince I am left to assume that counter clockwise is recessive then .36 is the initial q2. Taking the square root of that we get .6 as the initial allele frequency for q. 1-q=p so p=1-.6=.4. In other words, the island initally has 80p (clockwise)alleles and 120q (counterclockwise) alleles since humans are diploid and have two copies of each gene.\nNow, the 25 sailors have have added 50 copies of the q allele thereby changing the allele frequencies. The new allele frequencies are (120+50)/250 for q and 80/250 for p. when you do the math you get .32 for p and .68 for q. This will be the allele frequencies for the next generation too. \nNow plugging those numbers into the equation that I sited earlier you get .1024 for p2, .4352 for 2pq and .4624 for q2. You can give those a dominant/recessive notation such as WW for p2, Ww for 2pq, and ww for q2 to show genotype frequencies. \n\nThe evolutionary force at work is migration. \nhope this is what you were looking for.
374218
not an exact science. has to do with dominant and recessive genes. which ever the genes fit together will determine the output. Like Punnet Squares. Could be shorter, in the middle, or taller than both of you.
909716
Mother's genotype is X*X and dad's genotype is XY.\nX*X -------> 1/2 X* & 1/2 X\nXY --------> 1/2 X & 1/2 Y\n\nKids genotype will be: 1/4X*X, 1/4XX, 1/4X*Y, 1/4XY\nTherefore, there is a 25% chance of having a daughter who is a carrier, 25% chance of having a daughter who is not a carrier, 25% chance of having a hemophilic son and 25% chance of having a healthy son.
373156
according to genetics,if both parents are homozygous black(both of each parent's alleles or genotypes are the same),the odds are ALWAYS zero.eg:both parents' genotype is BB(B-black colour is the dominant while b-white colour is recessive)the children are ALL BB,so they are All black. \n\nif one parent is heterozygous(genotypes are different),same as above.eg:if the mother is BB & the father is Bb, 2 children are BB while the other 2 are Bb,which means All are black\n\n  if both parents are heterozygous the odds are i in every 4 children. eg:both parents are Bb, 1 child is BB-black , 2 children are Bb-black & 1 child is bb-white.\n\ni hope u have studied genetics in bio-if not,then ur lost!good luck!!!
905475
A. You have co-dominance in a heterozygous match, so the off spring ratio would not be the normal heterozygous, but would come out as a homozygous cross.\n\nB. Call your recessive allele, a. If you did the cross with a non-recessive you would have 1/4 homozygous,aa, progeny. Lethals, when homozygous, will die. You can see that messes with the Mendelian ratio.