MCQ Detailed View

Diabetes, in many cases, can be inherited when the responsible gene is passed from parents to their child. If both parents are heterozygous normal, it means that each parent carries one normal allele (N) and one recessive diabetic allele (d). They do not have diabetes themselves because the normal allele dominates, but they can still pass the recessive allele to their offspring.To calculate the probability of having a diabetic baby, we use a Punnett square to predict the genetic combinations of the offspring:

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  Each parent has two alleles: N (normal) and d (diabetic).

  
  


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  Possible combinations for the child are:

  
  
  
  
  1. 
     

     N + N = NN (normal, no diabetes)

     
     
  
  
  2. 
     

     N + d = Nd (carrier, no diabetes)

     
     
  
  
  3. 
     

     d + N = dN (carrier, no diabetes)

     
     
  
  
  4. 
     

     d + d = dd (diabetic, disease expressed)

     
     
  
  
  
  

From these four possible combinations, only one (dd) results in a diabetic baby. This equals 1 out of 4, or ¼ (25%) chance.

The other options are incorrect because:

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  Zero: This would only be true if neither parent carried the diabetic allele.

  
  


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  ½: A 50% chance occurs only in specific dominant-recessive crosses, not in this case.

  
  


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  ¾: This would mean 75% diabetic children, which is not possible here.

  
  

This principle of inheritance is based on Mendel’s laws of genetics, where recessive traits only appear when both parents pass on the recessive allele. Many genetic diseases, including some forms of diabetes, follow this autosomal recessive inheritance pattern.

Thus, when both parents are heterozygous normal carriers of the diabetes gene, the chance of having a diabetic baby is ¼ or 25%.