Written by Brian White. Copyright 1995
Another way to present the results of crosses is a pedigree. Pedigrees
are usually used when few individuals and many generations are involved -
this is often the case with human genetic data.
There is a standard set of pedigree symbols used to denote the phenotype
of individuals in a family.
Because of the small number of progeny, pedigrees can often be
ambiguous - more than one model may explain the observed pattern of
inheritance. For this reason, we often assume that the genetic abnormality
is rare in the population. This means that, in a given family, very
few (usually one) of the first generation individuals carry the abnormality
and very few (usually none) individuals who "marry" into the family
carry the abnormality. This assumption may help to rule out certain models.
Here is a sample human pedigree and one way to analyze it. Use A or a
to represent the abnormality if it is dominant or recessive, respectively.
Try the possible models one by one.
- - if the abnormality were Y-linked:
Y - normal Y chromosome
Y* - Y chromosome carrying abnormality (it doesn't matter if it is
dominant or recessive, since males are haploid for the Y chromosome)
It would appear only in males, as is observed. However, 1 or 3 would
also have to have Y* and be abnormal in order to pass it on to his sons.
This does not fit the data so it is not Y-linked.
- - if the abnormality were due to a dominant mutation in a gene on an
autosome (autosomal dominant):
A - dominant abnormal allele
a - recessive normal allele
Then either 2, or both 1 and 3 would have to show the abnormality for
it to be present in the children. This also does not fit he data.
- - if the abnormality were due to an autosomal recessive mutation:
a - recessive abnormal allele
A - dominant normal allele
In order for the second generation to have affected individuals (aa),
both parents of each family must be carriers. That is:
1, 2 , and 3 must be Aa
You would the predict that their children would have a 25% chance of
being affected (aa). The observed frequencies are 33% and 50% in the two
families which is not statistically
significant for this small sample size. Therefore, this model fits the data.
However, this model assumes that three non-blood relatives all carry
the abnormal gene (1, 2, and 3). If we assume that the abnormality is rare,
then the chance that three randomly selected individuals will have the
abnormality is very small. Therefore, this model is a possible explanation
of the data, but it is not the most likely. You would have to look at more
children in this family do determine if the chance meeting you have proposed
actually took place.
At this point, it is reasonable to try other possible models to see if
they fit the data better.
- if it is sex-linked dominant, 2 would have to be abnormal or both 1 and 3
would have to be abnormal to pass the abnormality to their children. This
does not fit the data.
- if it is sex-linked recessive:
X - X chromosome with recessive abnormal allele
X - X chromosome with dominant normal allele
1 and 3 must be XY - in order to be normal
2 must be XX -
in order that the abnormality run in the
family, someone has to carry it.
We would predict that the resulting daughters would be:
50% XX unaffected
50% XX affected
No affected daughters are observed, but the sample size is too
small to be significant.
(Note that statistical arguments are of limited value in pedigree
analysis because of the small sample sizes involved.)
We would predict that the resulting sons would be:
50% XY unaffected
50% XY affected
Both affected and unaffected sons are observed
Therefore, the data fit this model as well. This is also a more likely
and reasonable model than autosomal recessive since it requires only one
individual (2) to have the genetic abnormality. You could distinguish
between sex-linked recessive and autosomal recessive by looking at more
children in this family to see if any affected daughters of 1 and 2 or 2
and 3 appear. If so, the abnormality must be autosomal recessive. (Why?)