he first, and most important, thing to remember about autosomal recessive inheritance is that most, if not all, affected individuals have parents with normal phenotypes.
Why? Suppose the disease affects one in ten thousand live births, a good order of magnitude estimate for most autosomal recessive diseases. That would make the heterozygote frequency in the population one in fifty (see population genetics for calculations). The likelihood of two affected persons mating would be 1/10,000 x 1/10,000 or 1/100,000,000. By chance alone there might be two such matings in the Unites States, but no more than 2. The likelihood of an affected and a heterozygote mating would be 1/10,000 x 1/50 x 2(since either parent could be the affected) or 1/250,000. The likelihood of two heterozygotes (heterozygotes are usually called "carriers") mating is 1/50 x 1/50 or 1/2500, more than 99% of all possible matings. The Punnett Square for autosomal recessive diseases with an affected child in the family almost always looks like the following:
Where the father and mother are both Dd (dd is the recessive affected individual, Dd the heterozygous carrier individual, and DD the homozygous normal individual). The Punnet Square shows the origin of the famous Mendelian ration of 3/4 normal to 1/4 affected. For most autosomal recessive diseases, but not all, the heterozygote cannot be distinguished from the normal homozygote. In the normal phenotype categories of offspring in the above Punnett Square (Dd and DD produce the same normal phenotype), please note that two of the three are heterozygotes (carriers); one of the three is homozygous normal.
Within the normal siblings of an affected individual the probability of being a carrier is 2/3.
There are five hallmarks of autosomal recessive inheritance:
The above pedigree illustrates four of the five hallmarks of autosomal recessive inheritance. I-1 and I-2 are unrelated, yet they produced an affected offspring (affected offspring have normal parents). By chance, they both must have been carriers. Even though II-2 is affected, she produced no affected offspring (trait appears in siblings, not parents or offspring). By far the most probable genotype for an individual from outside the family (II-1) is homozygous normal. III-1, III-2 and III-3 are all obligate carriers (heterozygotes), since they are not affected but could only have inherited the recessive gene from II-2 II-3, II-5, and II-6 each have a 2/3 chance of being a carrier and a 1/3 chance of being homozygous normal. They are not affected, but they come from a carrier x carrier mating. II-4 and II-7 have a high probability of being homozygous normal since they are from outside the family. III-4, III-5, III-6, III-7, III-8, and III-9 all have a 1/3 chance of being carriers and a 2/3 chance of being homozygous normal. One parent of each is probably homozygous normal, the other has a 2/3 chance of being a carrier and a 1 in 2 chance of passing on the recessive allele if they were a carrier.
When consanguinity is involved, i.e., matings between related individuals, in the production of an affected child the assignment of probabilities changes, especially in the rarer autosomal recessive diseases. Since these relatively rare autosomal recessive diseases would have disease frequencies of 1/10,000 live births or less, the carrier frequency in the general population would not exceed 1/50. Normal individuals from the general population would have a probability of at least 49 to 1 of being homozygous. Consanguinity introduces the possibility of one founding parent being a carrier, with the recessive allele being passed through carrier offspring and meeting itself to produce an affected homozygous offspring some generations later. When an affected child is produced as the result of a consanguineous mating, those individuals in the direct line of descent are most probably carriers and those from outside the family are most probably normal homozygotes. In the following pedigree, V-1 is affected with an autosomal recessive disease. Her parents are second cousins, a legal marriage in most states. IV-1 and IV-2 must both be carriers since they produced an affected child. (The child must have received a recessive allele from each of her parents.) III-2 is an obligate carrier. Her father was affected, and hence, a homozygote for the recessive allele. III-5 must also be heterozygotes since IV-2 had to get her recessive allele from one of her parent, and the chance of III-6 being a carrier is less than 1 in 50. I-1 and I-2 must both have been carriers since they produced an affected offspring, II-1.
Normally one never considers the possibility of two unrelated individuals both being carriers unless there is evidence to the contrary. Here I-1 and I-2 are the exception to the rule. There is evidence that both must be carriers.
Before he had any children, II-5 had a 2/3 chance of being a carrier and a 1/3 chance of being homozygous normal (The normal siblings of affected rule explained above). But III-5 had to get her recessive allele from someone, and her other parent, II-6 had at most a 1/50 chance before her children were born. One can compare the two probabilities and calculate that in at least 100 out of 103 times II-5 will be the carrier. This is so close to 1 that for practical purposes one can say he is the carrier. In rare autosomal recessive diseases, when consanguinity is involved, those individuals in the direct line of descent within the family are considered to be carriers and those individuals from outside the family are considered homozygous normal unless there is evidence to the contrary.
What can we say about the carrier probabilities of the individuals within the family that are not in the direct line of descent to the affected child? In the above pedigree, III-1 must be a carrier. She is not affected, but she must have received a recessive allele from her father (II-1) who is homozygous recessive. II-3 and II-4 each have a 2/3 chance of being a carrier since they are phenotypically normal and come from a heterozygote x heterozygote mating. III-6 has a one in two chance of being a carrier. His father is a carrier (see above calculations) and his mother is from outside the family.
In human genetics counseling it doesn't help your patient to understand and make decisions when you are quoting statistics to the second or third decimal point. 49/100 or 0.488 are close to one half. One out of two, or one half, is a number your patient is more likely to understand and remember. You may get into trouble for using certainty (zero or 100%), or even ridiculous numbers like 1/1,000,000, when something doesn't go as you predicted, but not for rounding off to the nearest fraction. Always make your counseling calculations as accurately as possible, but make sure they are presented in a form that is something your patient will comprehend.
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