Biology: genotypes...anyone understand this?
921440
You need to cross it with another individual that displays the recessive phenotype (which you would know to be homozygous recessive).\nIf your plant is heterozygous, then about half the progeny will display the recessive phenotype.\nIf your plant is homozygous dominant, then none of the progeny will display the recessive phenotype.
784007
I don't think they can be on the same chromosome. Epistatic alleles are alleles that assort independently, but can affect the function of another allele. It could be genotypic (one allele preventing transcription of another) or phenotypic (both alleles are transcribed, but one hides the effect of the other). If they are on the same chromosome, they wouldn't always sort independently; there'd be a high degree of linkage (like blond hair and blue eyes).
117215
The closest answer is D\n\nI suggest you to give it a try yourself first.\n\nFirst try to visualize how many different types of gamates will be there. You will see two types A and a\n\nNow, draw a punnet square & cross them. You will get only one possible offspring i.e. Aa in the first filial generation.(F1 generation)\n\nNow, on performing self cross on two Aa individuals in the similar fashion, you will get a 3:1 phenotypic ratio and a 1:2:1 genotypic ratio in the F2 generation.
292522
The term comes from oldish genetics vocabulary and it refers to the most common allele (in terms of proportion) for a given gene in a population. \nBefore modern population genetics research, it was believed that most genes (and associated characteristics) usually had one "normal" form (that would be the wild-type allele, present in most individuals) and another "mutant" form (the least common one). \nNow we know that natural populations are polymorphic (have usually several alleles) for any given gene, and there may be one or more than one more frequent alleles.
126548
it's at least 99.99%... the 0.01% difference is what pharmacogenetics/genomics is trying to exploit... their are hap-maps and genetic diversity studies being generated by companies such as perlegen and affymetrix... check out their sites.\n\nthe single nucleotide polymorphisms (SNPs) make genomic profiles unique, even among the unique profiles, there are clusters or groupings with similar patterns...
435531
The 1:4 ratio others are giving you are wrong.  If there are two separate traits you want to find the recombination ratio for you need to do a punnet square (google that I'm not explaining).  Long story short, if the parents are heterzygous for the dominant gene (they carry one dominant and one recessive gene), there will be a 1:16 chance that the child will inherit *both* homozygous recessive traits.\n\nHere, the recessive traits are blue eyes and blong hair; dominant traits are brown eyes and black hair.
85122
Genotype is the trait inherited\nPhenotype is the physical attribute\n\nE.g.:\na. Genotype- height\n   Phenotype- short or tall\n\nb. Genotype- fur colour\n   Phenotype- black or brown (taking into account that it is a mouse)\n\nc. Genotype- hair\n   Phenotype- straight or curly
835767
The word dominant should give you a clue ;)\n\nNote for the first person who answered this question- if the plant has 2 recessive alleles, then it won't have the dominant allele. Haha :p\n\nIn any case, the plant will be tall. :)
708035
I can only answer number one.\nThe possible traits can be determined.  But there's no way of know the exact traits of the offspring unless the parents have either two dominate or two recessive genes with every specif trait.  For instance, using punnett square:\n \n----d----d\nD  Dd Dd\nD  Dd Dd\n\nAll the possible genotypes are the same.\nSorry I couldn't answer your other questions!
477782
Genetics and evolution.
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.
39991
Alleles are different 'forms' of the same gene.\n\nFor example, there are three alleles of the gene for blood groups; alleles A, B and O.\n\nIn diploid organisms like ourselves we have two complete sets of genes, so there are various combinations of alleles. One allele is inherited from the mother and one from the father.\nAA - blood group A\nAO - blood group A (because A is dominant to O)\nBB - group B\nBO - group B (B dominant to O)\nAB - group AB (alleles are codominant - equal dominance)\nOO - group O\n\nGeneologically, we can follow the inheritance of various characteristics (determined by alleles) and similarly blood groups can be used to eliminate or confirm parentage.\n\nThe 'classic' geneological example is the family tree of Queen Victoria of Great Britain. This follows the inheritance of the sex-linked gene for haemophilia.
273434
There are multiple start codons, depending on the species.  But I'm pretty sure that there are always a few oddities in how different species might encode information is slightly non-standard ways.
47775
A genotype is the actual genetic code of an organism, the sequence of the base pairs on the DNA. The phenotype is what those genes actually express, what you would see if you looked at the organism.\n\nSay there's a gene that controls the color of a mouse's fur. We'll call the dominant form of the gene which gives brown fur, B. The recessive form, for white fur, will be b.\n\nA mouse gets one copy of the gene (called an allele) from each parent, for a total of two. That means there are three combinations of alleles: BB, bb, and Bb. A mouse with two recessive alleles will be white, and a mouse with two dominant alleles will be brown. A heterozygous mouse (with one dominant and one recessive copy) will also be brown.\n\nGenotype          Phenotype\nbb                White\nBB                Brown\nBb                Brown\n\nSo even though the BB and Bb mice have different genotypes, they have the same phenotype.\n\nIn the example above, the phenotypic ratio would reflect the amount of mice of a different color in a population. For example, if you crossed two heterozygous mice, the brown to white ratio of the offspring would be 3:1. The genotypic ratio, reflecting the genetic code of the offspring (BB to Bb to bb), would be 1:2:1.
527680
genotype... ;p\n\nthe genotype, or the identity of the alleles that an individual carries at one or more positions on the chromosomes, is the specific genetic make-up or genome of an individual in the form of dna. together with the environmental variation that influences the individual, it codes for the phenotype of that individual.
901637
Let B mean the genotype for dominant black fur and b mean the genotype for brown fur. \n\nRecall that when a gene is dominant, it is always expressed when  it is present. When a gene is recessive then it is ONLY expressed  when there is no alternative.\n\na) Pure black mice would have B1B2 and brown mice must be b1b2 because it is a recessive gene. (I've put in the numbers to act as subscripts so it makes more sense.) Crossing these, you would get offspring that inherit one gene from each parent.\nB1b1 ==> black\nB1b2 ==> black\nB2b1 ==> black\nB2b2 ==> black\n\nSo false because all the offspring from pure black and brown rats would be black. \n\nb) True. Pure black mice have BB. As both parents have BB then there's not a chance for bb to occur!\n\nc) False. Brown mice are always bb (recessive) so bb x bb can never result in Bb or BB. So offspring of 2 brown mice are always brown.\n\nd) Heterozygous black mice are Bb. So crossed with brown mouse bb (b1b2):\nBb1 ==> black\nbb1 ==> brown\nBb2 ==> black\nbb2 ==> brown\nSo you get equal numbers of black and brown offspring. True.
121961
bud_leaf_... answered it for me, but generally, if you put all of the generation on a pedigree (a family tree that lists similar traits), the carrier always passes the genes on to the offsspring. because the carrier gences are both recessive, if the person mates with another carrier, or a victim the offspring WILL get the disease, unless if it is genetically altered. if the carrier mates it a non carrier with  dominant gene (example, Tt) the offspring will be a carrier. If the mate does not have the disease with both dominant(TT) the offspring will either not be a carrier, or be a carrier. but with the case of hemophelia,it is a sex-linked disease, so the males will be more likely to get it because they have ONLY ONE X CHROMOSOME, WHILE FEMALES HAVE TWO, WHERE THE SECOND  X ACTS AS A BACKUP. THe females will be carriers, while, sadly, the males will be victims. That is a example of a SEX-LINKED DISEASE. It really depends on what type the disease is, though (sex-linked, or just genetic)