HOW LIKELY ARE YOU TO HAVE CYSTIC FIBROSIS?

A very simple analysis can be conducted to understand how likely a person is to have Cystic Fibrosis. The only piece of information required here is the genotype (the pair of alleles) of the parents.

 

The likeliness of someone having Cystic Fibrosis can be analyzed by a simple tool known as the Punnett grid.

Let us consider the Dominant allele to be “C” and its corresponding recessive allele to be “c”.

We know that Cystic fibrosis is caused by the recessive allele. So, this allele should be present for the person to have the disease or be a carrier.

 

Five scenarios can be analyzed here:

HOW LIKELY ARE YOU TO HAVE CYSTIC FIBROSIS?

A very simple analysis can be conducted to understand how likely a person is to have Cystic Fibrosis. The only piece of information required here is the genotype (the pair of alleles) of the parents.

 

The likeliness of someone having Cystic Fibrosis can be analyzed by a simple tool known as the Punnett grid.

Let us consider the Dominant allele to be “C” and its corresponding recessive allele to be “c”.

We know that Cystic fibrosis is caused by the recessive allele. So, this allele should be present for the person to have the disease or be a carrier.

 

Five scenarios can be analyzed here:

  1. BOTH PARENTS WITH DOMINANT HOMOZYGOUS GENOTYPE:

Here, the parents will both be of genotype- CC

Hence, the following Punnett grid can be made:

​Here we can see that the only possible genotype for all the offspring is CC and the only possible phenotype is normal. This is mainly because of the lack of the recessive allele. Hence, any offspring of parents with these genotypes will be normal.

2. BOTH PARENTS WITH RECESSIVE HOMOZYGOUS GENOTYPE:

Here, the parents will both be of genotype- cc

Hence, the following Punnett grid can be made:

Here we can see that the only possible genotype for all the offspring is cc and the only possible phenotype is the disease cystic fibrosis. This is mainly because of the presence of only the recessive allele. Hence, any offspring of parents with these genotypes will have cystic fibrosis.

Embryonic Stem Cells

3. ONE PARENT WITH HETEROZYGOUS GENOTYPE AND ONE PARENT WITH DOMINANT HOMOZYGOUS GENOTYPE:

Here, one parent will be of genotype- Cc

And the other parent will be of genotype- CC

Hence, the following Punnett grid can be made:

  • Here we can obtain two different genotypes and one phenotype. 

  • The genotypic ratio of CC:Cc is 1:1. The only possible phenotype is normal. 

  • The offspring with genotype CC lack the recessive allele and hence are normal. 

  • On the other hand, due to the presence of a singular recessive allele in the offspring with genotype Cc, they are carriers of the disease but are not directly affected by it. This means that if crossed with a certain genotype, it is possible for these offspring to pass on their recessive allele and hence they are carriers. As the disease is caused by a recessive allele, which is never expressed in the presence of a dominant allele, all the offspring with genotype Cc are normal.

4. ONE PARENT WITH HETEROZYGOUS GENOTYPE AND ONE PARENT WITH RECESSIVE HOMOZYGOUS GENOTYPE:

Here, one parent will be of genotype- Cc

And the other parent will be of genotype- cc

Hence, the following Punnett grid can be made:

  • Here we can obtain two different genotypes and two different phenotypes. 

  • The genotypic ratio of Cc:cc is 1:1.

  • The phenotypic ratio of normal:cystic fibrosis is 1:1. 

  • The offspring with genotype cc lack the dominant allele and hence have cystic fibrosis as the recessive allele is only expressed in homozygous form. 

  • On the other hand, due to the presence of a singular dominant allele in the offspring with genotype Cc, they are carriers of the disease but are not directly affected by it. This means that if crossed with a certain genotype, it is possible for these offspring to pass on their recessive allele and hence they are carriers. As the disease is caused by a recessive allele, which is never expressed in the presence of a dominant allele, all the offspring with genotype Cc are normal.

5. BOTH PARENTS WITH HETEROZYGOUS GENOTYPES:

Here, both parents will be of genotype- Cc

Hence, the following Punnett grid can be made:

  • Here we can obtain three different genotypes and two different phenotypes. 

  • The genotypic ratio of Cc:Cc:cc is 1:2:1.

  • The phenotypic ratio of normal:cystic fibrosis is 3:1. 

  • The offspring with genotype cc lack the dominant allele and hence have cystic fibrosis as the recessive allele is only expressed in homozygous form. 

  • The offspring with genotype CC are normal. This is mainly because of the lack of the recessive allele.

  • On the other hand, due to the presence of a singular dominant allele in the offspring with genotype Cc, they are carriers of the disease but are not directly affected by it. This means that if crossed with a certain genotype, it is possible for these offspring to pass on their recessive allele and hence they are carriers. As the disease is caused by a recessive allele, which is never expressed in the presence of a dominant allele, all the offspring with genotype Cc are normal.

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