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Sex Linkage (pedigree tree) (1 Viewer)

asdfqwerty

asdfqwerty

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you can tell if its sex-linked or not by looking at the male. If the male has the trait expressed and his future sons all have the traits expressed as well, then it means that its sex-linked because all the males express the trait due to only have one X chromosome and the females have the potential to be carriers as they have two X chromosomes so the trait is not expressed.
 
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SanjoyM

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asdfqwerty said:
you can tell if its sex-linked or not by looking at the male. If the male has the trait expressed and his future sons all have the traits expressed as well, then it means that its sex-linked because all the males express the trait due to only have one X chromosome and the females have the potential to be carriers as they have two X chromosomes so the trait is not expressed.
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Sorry, I disagree with this.
In order to be a male, a Y chromosome must be passed on from the parents. Since the father is the only parent with a Y chromosome, it must be from him. Thus, the x chromosome is received from the mother.

In the case of the sex-chromosome, the X chromosome is significantly larger than the Y. This means the X chromosome contains more genes, and thus governs more traits.

Hence, only the mother needs to be a carrier or affected for a male to be effected, regardless of whether or not the father is affected.

If a male inherits an affected X-chromosome, the Y chromosome does not have a corresponding allele to mask it as the Y chromosome is too small, so the male is haploid for the X allele.

For a sex linked recessive trait, if the mother is affected, all sons will be affected, regardless of whether or not the father is affected.
If a son is affected from the mating of two unaffected people, the mother was a carrier.

As featured, this also provides an explanation for why sex-linked traits appear less often in women. As women have two homologous X-chromosomes, the recessive allele only occurs on one of them. Thus, the corresponding dominant allele on the other X-chromosome masks the recessive allele, so the recessive phenotype does not manifest.
Hence, affected females are always HOMOZYGOUS for the recessive allele.

For a female to express the trait, the father must be affected and the mother should atleast be a carrier.

In addition to this, some traits are also governed by dominant alleles.
Sex-linked dominant traits:

- if an unaffected male and a homozygous affected female cross, all male and female offsprings will be affected.
- if an unaffected male and a heterozygous female cross, half of the male and female offspring will be affected
- if an affected male and an unaffected female cross, alll female offspring will be affected and NONE of the males offsprings will be affected

Hope this helps!! :)


 
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Power Rangers

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Dominant traits show their phenotype, and you see that 1 (the father) exhibits the trait. Male= XY. Female in I is a carrier (heterozygous) because 6 in II is heterozygous to produce a daughter that shows the trait (6 is homozygous recessive).
All the sons are affected in I, so if it's sex linkage, it means that the mother is a carrier (heterozygous). Notice how only the females are unaffected in II? 2 and 4 don't show the trait? So it ONLY AFFECTS THE MALE IN I.
And then looking at III, you see that 2 is a female and has the trait? It means that 4 in II is a carrier. For a female to exhibit a trait, she must be homozygous to exhibit the symptom. Sex linkage usually occurs on the X chromosome, so I have no idea why 4 in III doesn't show the trait. If you think about it, his mother has the trait in both her alleles (otherwise she wouldn't exhibit the phenotype) and the father has both the alleles too (otherwise 3 in iI wouldn't show the trait).


I don't think it's sex linkage.
 
jason2kool

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Power Rangers said:
Dominant traits show their phenotype, and you see that 1 (the father) exhibits the trait. Male= XY. Female in I is a carrier (heterozygous) because 6 in II is heterozygous to produce a daughter that shows the trait (6 is homozygous recessive).
All the sons are affected in I, so if it's sex linkage, it means that the mother is a carrier (heterozygous). Notice how only the females are unaffected in II? 2 and 4 don't show the trait? So it ONLY AFFECTS THE MALE IN I.
And then looking at III, you see that 2 is a female and has the trait? It means that 4 in II is a carrier. For a female to exhibit a trait, she must be homozygous to exhibit the symptom. Sex linkage usually occurs on the X chromosome, so I have no idea why 4 in III doesn't show the trait. If you think about it, his mother has the trait in both her alleles (otherwise she wouldn't exhibit the phenotype) and the father has both the alleles too (otherwise 3 in iI wouldn't show the trait).


I don't think it's sex linkage.
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thanks dude, this stuff is sooo confusing
 
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Enzym3

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Sec linked traits are present on the X or Y chromosome, however the Y-chromosome is small and the only disease gene that has every been even suggested for the Y-chromosome is one for a hairy ear condition. Therefore in human genetics when talking about sex linked we are talking about the X-chromosome.


Like autosomal conditions (some teachers will say non-sex linked) X linked traits come in both dominant e.g. Rett (one allele and you express the phenotype i.e. have the condition) and recessive e.g. Colour blindness and Hemophilia (you need both alleles to be the same to have the condition if you're a girl or 1 allele to have the condition if you're a guy)


For X-linked dominant traits:

Every affected person has an affected parents and
An affected father will always pass it on to all of his daughters and none of his sons


For X-linked recessive traits

Their are affected people without affected parents but
Every affected girl has an affected father and
Their are substantially more males affected than females



If people are keen I could probably knock up a dozen or so practice questions and upload them and give you guys the guide that I teach with
 
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