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10. A rare condition known as adermatoglyphia leads to such smooth fingertips that the individual has no fingerprints. It has been dubbed the "immigration delay disease" because sufferers have such a hard time entering foreign countries Recently the cause has been traced to a point mutation in the very first nucleotide of an intron The allele with this mutation (F) is dominant to the wild type allele (f). The condition also leads to less hand sweat than the average person and researchers think that the gene might help skin cells fold over one another early in fetal development. A woman with adermatoglyphia marries a man with adermatoglyphia (assume both are heterozygous). Give two reasons why it might be difficult to calculate the probability they will have a child with fingerprints. Consider the genotypes and phenotypes of all potential children and the possible contribution of the gene to early fetal development? See article and abstract posted on Blackboard or on the web as follows: (http:// Amer. J. Hum. Genet. 89, 302-307, 2011. You do not need to consult this reference but you should read a commentary on it from AAAS posted on Black board with it. Note that this problem is as much to teach you about the ambiguities of some problems as statistics

Answer 1

According to the information, both the parents are heterozygous for the genetic disease, hence have the genotype Ff and Ff each. Although a Punnett square can be easily used to study the genotype of their kids, some difficulties might arise due to following reasons:

1. Since the disease arises due to random mutation, non-conservative mutation in this case might undergo reversion or extension thus leading to change in genotype, hence phenotype.
2. Spontaneous loss of mutated part of any of the chromosome or occurence of any spotaneous mutation can silence the gene, hence make it difficult to express.

However, if these problems are absent in this case, the Punnett square can be formulated for this couple as below:

	F	f
F	FF	Ff
f	Ff	ff

Genotype ratio: FF: Ff : ff :: 1:2:1

Phenotype ratio: Affected: normal:: 3:1

Here, it must be carefully noted that the heterozygous children i.e. Ff are actually normal for the condition but do bear the affected allele as well. Thus, if this affected allele is expressed in these individuals, they will have reduced ability to properly develop the finger skin and hence their developmental process would be altered from normal children.