The recessive alleles al and dp of two genes on chromosome 2 in Drosophila determine the presence of antenna a
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As I recall, the map unit is a 1% chance that two genes will sort independently. So, 13% will be al/+ phenotype, 13% dp/+, 37% are al/dp, and the other 37% are wildtype. I think... definitely check this before you accept it as true.
392240
they have a gene that causes them not to be hairy
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The "I" stands for the isoagglutinogen gene.  Everybody has two copies of this gene, but know that this gene can come in different "flavors", what scientists call "alleles".  These flavors are Ia, Ib and Io.  The "I" indicates a single gene, and the a,b and o indicate the different versions, or "alleles", of this gene.  The "o" is termed recessive, because it "invisible" to the immune system, while "a" and "b" are dominant, meaning the immune system can identify and react with them.
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That's sick....\n\nAnyway, I did my assignment on albinism, specifically OCA1-ts albanism.  Do an OMIM search on the NCBI website, that should get you started.  The one mutation I picked was R402Q.
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Do refer to websites, such as Wikipedia, Vivisimo or Google; you will find countless articles, links and so on. Start with any scientific term, such as 'chromosome' and it will get you somewhere. Kimballs Biology Pages are a wonderful bet too.
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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.
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Each chromosome in your body has an "identical pair"- there are two chromosomes that code for the same functions- one from your mother and one from your father.\n\nBecause they come from two sources, they are coding for the same functions, but may code differently for them (ie blue eyes as opposed to brown eyes).\n\nDuring prophase one, each chromosome duplicates into two sister chromatids joined by a centromere, and then the double stranded chromosome finds it's partner which is coding for the same thing (just came from the opposite parent), and is also double stranded.  Now you have a 4 stranded tetrad.\n\nGenetic variability comes from crossing over.  This is where the two chromatids closest together in the tetrad exchange some genetic information.\n\nFor example, two of the chromatids code for blue eyes, brown hair and shortsightedness.\nThe other two chromatids code for brown eyes, red hair and normal vision.\n\nDuring crossing over one chromatid from each pair forming the tetrad literally have ends "crossing over" and bits of information are exchanged.\n\nAfter crossing over, you have 4 chromatids with unique codings.  Using the example above, we may now have:\n-blue eyes, brown hair, shortsightedness\n-blue eyes, red hair, normal vision\n-brown eyes, brown hair, shortsightedness\n-brown eyes, red hair, normal vision\n\nWe now have genetic variability in each gamete produced from this meiotic division.
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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.
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http://www.google.co.in/search?hl=en&q=multiple+genes&btnG=Search&meta=\n\nhttp://www.newton.dep.anl.gov/askasci/bio99/bio99889.htm\n\nhttp://www.jimmunol.org/cgi/content/full/173/11/6501
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Color blindness is a sex-linked thing (giggles from the 5th grader)\nThe gene is on the X chromosome.  Men get one X and one Y chromosome.  If their X has the gene for color blindness, there is no matching gene on the Y chromosome to "cover" for the defective gene.  The color blindness is expressed.\n\nWomen, however, are blessed with 2 X chromosomes.  If they get a gene for color blindness, the other X usually has a normal gene that can "do the work" that the abnormal gene cannot.  Color blindness is not expressed, and she is a carrier of the gene.\n\nWomen can be color blind if their color blind Dad marries a carrier Mom, and she gets 2 defective genes.\n\nHow's that?
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chromosomes from different parents determine the genetic structure of the child - with recessive genes only becoming apaprent if they are present from both parents
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Yes.  The woman is heterozygous for colorblindness (mutant from mom, normal from dad).  Since colorblindness is sex-linked recessive (i.e. carried on the x-chromosome, so males only have one copy and females have two), the son inherited the mutant allele from his mother, and non from his father.  The daughter inherited the normal allele for color vision from her mom and the mutant from her dad.  Since colorblindness is recessive, she has normal color vision.\n\nThe woman is also a carrier for hemophilia (mutant from dad, normal from mom).  The daughter would be normal no matter what unless her dad has hemophilia because she gets a normal allele from him.  She could be a carrier or could be homozygous dominant.  The son inherited the normal allele from his mom, and nothing from his dad.\n\nBy the way, contrary to the previous answer, both colorblindness and hemophilia are sex-linked _recessive_ traits.  And expression of a given allele does not "skip a generation," no matter what; it is always the result of the combination of alleles, not whether the gene was expressed in the previous generation.\n\n__________\nThat's interesting, because my textbook from AP bio states, "color-blindness is a mild disorder inherited as a sex-linked trait," but maybe you know something that the professors who wrote my textbook do not?  There's also the minor detail that both my AP bio teacher last year and my genetics professor this year used colorblindess as an example of a sex-linked trait.  And for what it's worth, Wikipedia states, "the genes for the red and green color receptors are located on the X chromosome."\n\nAutosomal traits are, by definition, not sex-influenced, as autosomal means carried on any chromosome other than the sex choromosomes.
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dominant e.g. Aa, AA - Recessive -aa
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Very simply put, dominance and recessiveness are characteristics of genes.\n\nA dominant gene is expressed (shows up visibly in the offspring) when only one copy of the gene is present, on one chromosome from one parent. Let's say Tallness is a Dominant gene. If you have ONE tall parent, and get ONE copy of the Tall gene, you will probably be tall. If your tall parent was the son or daughter of a TALL and a SHORT person, they may only have ONE copy of that gene. If you get both SHORT chromosomes from Mom and DAD, you will NOT be tall. So you have 50 to 100 chance of being tall with a Dominant gene.\n\nNow, let's take the gene for BLUE eyes. It is recessive. You need TWO copies of this gene to get blue eyes. If Mom has blue eyes and Dad has blue eyes, and you get copies of both blue eye genes, you will have blue eyes. But if Mom has blue eyes, and Dad has brown eyes, you might end up with Brown eyes. Don't that make your blue eyes brown, so to speak.\n\nBUT..if Dad's mom had blue eyes and Dad's dad brown (your grandparents) and if Dad has one blue eye gene and one brown eye gene and has brown eyes AND if your Mom still has those blue peepers, you have a 1/4 chance of ending up with one blue eyed gene from AND a blue eyed gene from Dad and you will also have blue eyes like your Mum. Don't that make your brown eyes...blue?
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I don't like your teacher.  I think he/she is looking for B. genes, but then he/she should not have put chromosomes as an alternative choice.  Mendel's theory of independent assortment was only true because the genes he studied were on separate chromosomes.  It could be argued that what he was actually studying was chromosomal inheritance.
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Your friend is correct. I just studied this in school. Dark hair always dominates light hair. And the allele for green eyes overpowers the allele for blue eyes.
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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.