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Man who is color blind marries a woman not colorblind and doesn't have recessive allele will their children be carriers of the color blind allele?
If a colour blind man married a carrier woman they could produce a carrier daughter, a colour blind daughter, a normal son or a colour blind son. The probability of each phenotype occurring is 25%.
If XC represents the normal allele for seeing colour and Xc represents the colour blind allele the genotypes of the possible offspring would be as follows:
Carrier daughter = XCXc
Colour blind daughter = XcXc
Normal son = XCY
Colour blind son = XcY
This information is incorrect. In fact a woman can be color blind. My mother is color blind as are my brothers. My sister and I are not though we carry the gene. I have two daughters and one is color blind and the other is not color blind.
The information I gave is not incorrect - I have included the possibility of that 'mating' producing a colour blind female child.
What else can I help you with?
If Tall is dominant over short a homozygous dominant man marries a homozygous recessive woman what of their children are expected to be short?
All of their children are expected to be heterozygous for the trait (Tt), which means they would be tall (dominant phenotype). The recessive trait for being short would only manifest if both copies of the gene were recessive.
A man homozygous for the perfect pitch trait marries a women without perfect pitch What is the probability that any of their three children will NOT have perfect pitch?
If perfect pitch is dominant you could have this. PP X pp Then all of the children will have perfect pitch. If perfect pitch is recessive, then; pp X Pp then there is only a 50% chance of any one child having perfect pitch. In the future be more precise.
Red-green color is inherited as a sex linked recessive trait If a color-blind woman marries a man with normal vision what would be the expectes phenotypes of their daughter?
Zero chance for any daughter (she needs an X from mom and an X from dad and if dad is normal, then the daughter will only receive the normal X from him. So even if she gets a colorblind X from mom, she will not be colorblind). 50% chance for sons. Dad only contributes an X to the son. But mom can give either a normal X or a colorblind X. So there's a 50% chance of him being colorblind.
What is the phenotypic ratio of the offspring when a tall man marries a short woman?
The phenotypic ratio of the offspring when a tall man marries a short woman is 3:1 (tall/short). The phenotypic ratio is figured by using the punnet square with the dominant allele for tall and the recessive allele for the short gene.
Curly hair is recessive and straight hair is dominant A woman with curly hair marries a man who is homozygous dominant for straight hair. What are the outcomes for their children?
All of their children will have straight hair, as the father can only pass on the dominant straight hair allele. The children will inherit one straight hair allele from their father and one curly hair allele from their mother, but the dominant straight hair allele will mask the recessive curly hair allele.
In humans normal color vision is dominant over colorblindness. A colorblind male marries a female who is a carrier for color blindness. What is the probability that a colorblind child would be born to?
The probability of a colorblind child being born is 50%. This is because the male passes his Y chromosome to all his sons, and since he is colorblind, his sons will inherit the colorblind gene from him. The daughters will inherit their X chromosome from the mother and have a 50% chance of being carriers like her.
If a normal woman whose father is color-blind marries a colorblind man what are their chances of having a colorblind son?
50%
If a color blind woman marries a man with normal vision the children will be?
there is a 50% chance that the child will be colorblind. If it is a boy, it will be colorblind, but if it is a girl, it will only be a carrier. Mother's chromosome is XrXr and Father's Chromosome is XRY, which means the children's genotypes will be XRXr if girl and XrY if a boy.
A womens father is colorblind She marries a colorblind man Will there son or daughter be colorblind?
Colorblindness is an X-linked recessive disorder. This means girls (who have the sex chromosomes XX) must have a colorblind X from dad and a colorblind X from mom. Boys only need to have one colorblind X to be colorblind because they have sex chromosomes XY (and have only 1 X). If the dad has it, he has the colorblind X. If the daughter has it, she must have gotten her mom's colorblind X. If the mom is colorblind, then every child they have will be colorblind. If the mom is not colorblind, then she must be a carrier - she must have 1 normal X and 1 colorblind X. Mom is either colorblind (with 2 colorblind Xs) or she is a carrier. Dad is definitely colorblind.
If Tall is dominant over short a homozygous dominant man marries a homozygous recessive woman what of their children are expected to be short?
All of their children are expected to be heterozygous for the trait (Tt), which means they would be tall (dominant phenotype). The recessive trait for being short would only manifest if both copies of the gene were recessive.
If someone in a family is a purebred and he marries someone with a recessive trait will his children still be considered purebreds?
No, the children would not be considered purebred. Their genetic makeup would be a mix of the purebred parent's dominant traits and the recessive traits from the other parent, making them a hybrid of the two.
If a color blind male who has normal clotting blood marries a female who is a carrier of hemophilia and has normal color vision could they have a color blind child?
In short, hemophilia has nothing to do with colorblindness, but YES, they could have a colorblind child if she is a carrier for the colorblindness gene. Color blindness is an X-linked trait. That means it is carried in the X chromosome, which differentiates whether a baby will be a girl or a boy. Women have two X chromosomes (XX), and men have an XY combination. If a woman is a carrier for color blindness, only one of her chromosomes will be affected (we'll call it a little "x"), and for that reason she will not be colorblind. Men, on the other hand, only have one X chromosome, so any time they carry the colorblindness gene, they will be colorblind. A woman will carry the colorblindness gene if: a. Her father is colorblind b. Any of her offpsring are colorblind She may carry the colorblindness gene if: a. Male family members (brothers, uncles, etc.) are colorblind A child inherits one chromosome from each parent. He/She will get an X chromosome from his/her mother, and an X from her father (if a girl) or a Y from his father (if a boy). So, If a woman has normal vision (assuming she does not have a family history of colorblindness), XX, and a man is colorblind, xY, they have several different chances for different offspring: Xx (a normal girl who carries the colorblindness gene) XY (a normal boy) Xx (a normal girl who carries the colorblindness gene) XY (a normal boy) The short answer is that ALL CHILDREN WILL HAVE NORMAL VISION. However, all daughters will be CARRIERS, meaning they can pass colorblindness on to their children.
Why are males momre likely than females to have sex-linked traits controlled by a recessive allele?
Males are more likely than females to have sex-linked traits controlled by a recessive allele because they only need one recessive allele to have the sex-linked trait. In contrast, females need two recessive alleles to have the sex-linked trait, so they have a lower probability of having it.This is best viewed with a Punnet square. Say the recessive allele that controls the sex-linked trait is Xa. XA is the dominant allele and Y is the male chromosome.Scenario #1If the mom is XA XA and marries a man with the sex-linked trait Xa Y, then none of the sons will have the sex-linked trait. All the daughters will have the mutant allele, but they will all be carriers with normal phenotypes since they only have one mutant allele.XaYXAXA XaXa YXAXA XaXa YScenario #2If the mom is XA Xa and marries a man with the sex-linked trait Xa Y, then there is a 50% chance that each child will have the sex-linked trait, regardless of sex.XaYXAXA XaXA YXaXa XaXa YScenario #3If the mom is a carrier XA Xa and marries a normal man XA Y, then there is a 50% chance each son will have the sex-linked trait. The daughters may be carriers, but none of them will have the sex-linked trait.XAYXAXA XAXA YXaXA XaXa YScenario #4If the mom has the sex-linked trait Xa Xa and marries a normal man XA Y, then all of the sons will have the sex-linked trait. The daughters will be carriers, but none of them will have the sex-linked trait.XAYXaXA XaXa YXaXA XaXa YScenario #5If the mom has the sex-linked trait Xa Xa and marries a man who also has the sex-linked trait Xa Y, then all of their children will have the sex-linked trait.XaYXaXa XaXa YXaXa XaXa YAs you can see, there are many more strikethrough outcomes (8) where the male has the sex-linked trait controlled by a recessive allele than bold outcomes (3) where the female has the sex-linked trait. Thus, males are more likely than females to have sex-linked traits controlled by a recessive allele.
Male pattern baldness is a recessive sex-linked trait on the X chromosome Xb A woman whose father had male pattern baldness marries a man with this trait What is the probability that any son boor?
k so. i think you meant marries a woman? lol k so its recessive carried on the X chromosome. Her father genetic make up is X^b Y. marries a woman who has the trait is X^H X^h (since its recessive and dominant overthrows recessive and she has the trait h). so use a punnet square. you should get.. two girls. one is afflicted, the other one carries the trait. two boys: a normal son, and an afflicted son. Hope this helps!
A man with hemophilia marries a woman without the disease What is the possible offspring?
All Girls will be carriers of Hemophilia. All Boys will be unaffected (they won't have Hemophilia).
A man homozygous for the perfect pitch trait marries a women without perfect pitch What is the probability that any of their three children will NOT have perfect pitch?
If perfect pitch is dominant you could have this. PP X pp Then all of the children will have perfect pitch. If perfect pitch is recessive, then; pp X Pp then there is only a 50% chance of any one child having perfect pitch. In the future be more precise.
A recessive allele t is responsible for a condition called distonia a man who has this condition marries a woman who doesnt one of their four children has the conditon what are the pssible genotypes?
Firstly, if the condition is recessive, both the man and the child with the condition must have the genotype tt. The mother must have the genotype Tt. This is because if she had TT, all of the children would be Tt and not have the condition. If she were tt, she would have the condition as well. Therefore if the father has tt and the mother has Tt, the other 3 children who do not have the condition must all have the genotype Tt. So: Mother - Tt Father - tt Affected child - tt Other children - Tt
