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Is normal color vision dominant or recessive Is colorblindness dominant or recessive?
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Is colorblindness a disability?
Not necessarily. The allele for colorblindness is recessive. For a female, in order to be colorblind she must have to recessive alleles for colorblindness. Example: XcXc would be colorblind. XCXc would be a carrier for colorblindness, but not colorblind. For a male, because colorblindness is a sex-linked gene, he only needs one allele to be colorblind. Example: XcY is colorblind. XCY is not colorblind.
What would be the color vision of children if their mom is colorblind and their dad is normal?
Colorblindness is a recessive, sex-linked trait, and the gene that causes it occurs on the X chromosome. For the mother to be colorblind, she must have two copies of the gene and be homozygous recessive. The father, on the other hand, can not have the gene, because he (as a male) only has one X chromosome. So, if we let Xc represent the recessive gene for colorblindness and Xn represent the normal gene, the the mother is XcXc and the father is XnY. All of their children will receive the recessive gene from their mother. In the males, this means that they will be colorblind, because the chromosome they get from their father will by the Y chromosome. The daughters, however, will get the Xn gene, which is dominant and will override the gene for colorblindness. Thus, all of the couple's sons will be colorblind, and none of their daughters will be.
What is the probability that a woman who is a carrier of the colorblind gene and a color blind man will have a first son who will be color blind?
The probability is 0 (but the daughter will be a carrier of the color blind gene). This is because the gene dictating whether someone is color blind or not is linked to the X chromosome (and not the Y). The color blind gene is a recessive gene whilst the normal color vision gene is a dominant gene. Hence if a girl (XX) has one normal vision gene (from one parent) and one color blind gene (from the other parent), her normal vision gene will be dominant to the recessive color blind gene and hence she will have normal vision (but she will be a carrier of the color blind gene). If both her parents contribute the recessive color blind gene to her, then she will be color blind. For a woman (XX) to be color blind, she needs to be have both genes to be recessive (ie where there is no dominant normal color vision gene to dominate). For a man (XY), as long as the X gene contributed by his mother is a color blind gene, he will be color blind because he has no other X chromosome where a dominant normal color gene could reside. Hence, to answer the question, a man with normal color vision (XY, with a dominant normal color vision X gene since the gene can't be the recessive color blind gene otherwise he will be colorblind) and a colorblind woman (XX, both recessive color blind genes), will each contribute an X each the child. The man will contribute his only X chromosome which carries the normal color vision X gene and the woman can only contribute a recessive color blind gene. The man's normal color vision X gene will be dominant, and hence the daughter will definitely have normal vision (despite being a carrier).
What is an example of recessive human trait?
1. freckles: dominant no freckles: recessive 2. dark eyes: dominant light eyes: recessive 3. free earlobe:dominant attached earlbe: recessive 4. polydactilism (6 fingers or toes): dominant 5 fingers or toes: recessive 5. normal chin: recessivecleft chin: dominant 6. can roll tongue: dominant cannot roll tongue: recessive 7. cannot fold tongue: dominant can roll tongue: recessive 8. straight pinkie: dominant crooked pinkie: recessive 9. widow's peak: dominant straight hairline: recessive 10. separate eyebrows: dominant uni-brow: recessive
A male who has normal vision marries a female who is a carrier for colorblindness can they have a colorblind daughter?
Assuming that the man who has normal vision is homozygous for normal vision, the couple's daughter will either be homozygous for normal vision or heterozygous (normal vision but carrier for color blindness) for normal vision. In light of this, the couple's daughter will not be color blind.
Which parental pair could produce females with colorblindness?
Yes. Two normal-visioned parents can produce a color blind child only if both the parents have are heterozygous. To determine the phenotypes of the parents, you will have to look at their parents.
What is the probability that a colorblind woman and a man with normal vision will have a colorblind daughter?
It all will depend on what kind of (recessive or dominant) alleles are responsible for the colorblind characteristic and what kind of alleles do the parental genes have.
What are the male genotype for colorblindness?
For example, if a mother is a carrier for colorblindness (X+Xc), and a father has normal vision X+Y, then their sons have a 50% chance of colorblindness because they inherit their X chromosome from their mother and their Y chromosome from their father.
What is the proper name for colorblindness?
The proper name for colorblindness is color vision deficiency.
How do you express sex linked genes?
Let's take red-green color vision as an example. You could use the letter n to represent the dominant and recessive alleles. You could use a capital letter N to represent the normal allele, which is dominant, and a lowercase letter n to represent the defective allele, which is recessive. The genotypes for the sex-linked trait would be as follows: female, homozygous dominant: XNXN; female, heterozygous: XNXn; male dominant, XNY; male recessive, XnY.
How do you show a punnett square of a woman who is color blind and a man who has normal vision?
Colorblindness is a sex-linked trait carried on the X chromosome. Males have an X and a Y chromosome (XY) and females have two X chromosomes (XX). Normal vision is dominant over colorblindness. This means that the man must have an allele for normal vision on his only X chromosome. This also means that the woman must have two colorblind alleles on both of her X chromosomes. If we use a capital 'B' to represent normal vision and a lowercase 'b' to represent colorblindness, the genotype of the male would be XBY, and the genotype of the female would be XbXb. Now that we know the genotypes, setting up the Punnett square should be easy. Here is a direct image link to the Punnett square: http://i.imgur.com/gDfUv.png This Punnett square tells us that there is a 50% chance of having a female that has normal vision and a 50% chance of having a colorblind male.
A women wo isn't colorblind but has an allele for color blindness reproduces with a man who has normal vision what is the chance that they'll have a colorblind daughter?
There is no chance for a colorblind daughter because the x-chromosome that comes from the male is normal and then it does not matter which x-chromosome comes from the female because colorblindness is a recessive gene, therefore there is no chance.
