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MaRussiya [10]
4 years ago
6

Why would it be important to know the possible genotypic and phenotypic ratios of different genetic crosses?

Biology
2 answers:
OleMash [197]4 years ago
6 0
It’s important to know the possible genotypic and phenotypic ratios of different genetic crosses because the the phenotype shows what the offspring will physically look like what the genotype is what genes the offspring carries. phenotype doesn’t always show what alleles the offspring carries because if the offspring is heterozygous for the gene only the dominant allele shows and the recessive allele won’t be visible. the genotype can see what alleles the offspring carries, both dominant and recessive. knowing the genotype helps to know what alleles are passed on. if one of the parents have a genetic mutation that is passed on, the phenotype helps see what ration of offsprings will have the mutation visible and the genotype will help see what ratio of offsprings will pass on the allele for the mutation

probably isn’t useful. my brain is currently burnt
aleksandr82 [10.1K]4 years ago
3 0
It helps to understand the results of the offsprings.
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Some types of muscular dystrophy are X-linked recessive disorders. A mother who is a carrier for muscular dystrophy has children
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D gene is an X-chromosome-linked gene. a) 100% will have the disease, XdY // 0% will be carriers. b) 0% will have the disease // 100% will be carriers, XDXd.

<h3>What are the séx-linked genes?</h3>

These are genes that can be found only in the X-chromosome. The inheritance pattern expressed by these genes varies from the one expressed by genes located in autosomal chromosomes.

In the exposed exampe,

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- woman ⇒ carrier for muscular dystrophy ⇒ heter0zyg0us ⇒ XDXd

- man ⇒ doesn’t have muscular dystrophy ⇒ X-chromosome with the dominant allele ⇒ XDY

- XD ⇒ chromosome rarrying the dominant allele

- Xd ⇒ chromosomes carrying the recessive allele

Cross: Man x Woman

Parentals)   XDY     x      XDXd

Gametes) XD   Y           XD   Xd

Punnett square)       XD           Xd

                      XD   XDXd       XDXd

                      Y      XdY           XdY

F1) 1/2 = 50% of the progeny will express the normal phenotype,  XD-

    1/2 = 50% of the progeny will have muscular dystrophy, Xd-

    100% of the girls will be normal and carriers,  XDXd

    100% of the boys have muscular dystrophy, XdY

a. What percentage of their sons will have the disease, and what percentage of their sons will be carriers?

Considering only boys,

  • 100% of them will have the disease, XdY
  • 0% will be carriers.

b. What percentage of their daughters will have the disease, and what percentage of their daughters will be carriers?

Considering only girls,

  • 0% of them will have the disease,
  • 100% will be carriers, XDXd

You will elarn more about séx-linked genes at

brainly.com/question/15570

brainly.com/question/786413

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