THE RECESSIVE GENE AND INFERTILITY

It is generally thought (so I think) that a recessive gene that
causes infertility when received from both parents, will die out
naturally.  (With "infertility" can be included any disease that
causes a person to die before reproductive age or even
homosexuality, if it were ever proved to be caused by a recessive
gene.)

Having done the mathematics I disagree and I will show why and
what proportion of the population will continue to be carriers of
the gene.

We are dealing here with the simple idealized case where there
are no other effects of the gene other than to cause infertility
in persons with a copy of the gene from both parents and that has
no effect on carriers (who have the gene from one or other of
their parents but not both).

Let x be the proportion of the fertile population (abbreviated to
simply "population") that does not carry the gene at all.  We
will call these persons "healthy" and use the symbol H for them.
The "carriers" are those who carry a single copy of the gene and
these will be denoted by C.  The "diseased" (D) are those who
carry two copies of the gene, one from each parent.  The diseased
group are not counted as part of the population as they will not
reproduce.  Thus the proportion of the population who are
carriers is 1-x.

If we denote n for a normal gene and g for the recessive gene:

H people have nn genes and are healthy, proportion x
C people have ng genes are carriers, proportion 1-x
D people have gg genes and do not survive to produce offspring.

When the next generation are born, we can divide them according
to the rules of genetics with the assumption that the parents are
taken at random from the two sets: H and C.

Parents can be CC, CH or HH with probabilities (1-x)2, 2x(1-x)
and x2 respectively.

The CC group will produce offspring of type gg, ng and nn with
probabilities (1-x)2/4, (1-x)2/2 & (1-x)2/4 respectively.

The CH group will produce offspring of type gg, ng and nn with
probabilities zero, 3x(1-x)/2 & x(1-x)/2 respectively.

The HH group will only produce nn offspring so their probabilities
will be x2.

The nn groups are now summed and we have

from CC parents     (1-x)2/4
from CH parents     x(1-x)/2
from HH parents     x2

These three quantities add up to (1+3x2)/4 for the nn group to
produce H offspring.

The sum of the ng groups are also needed as will be seen:

from CC parents     (1-x)2/2
from CH parents     3x(1-x)/2
from HH parents     none

These add up to (1+x-2x2)/2

But the population size has decreased owing to the non-survival
of the D group with gg genes.  The reduction is the ratio of 1 to
the sum of the nn + ng groups or

        1/( (1+3x2)/4  +  (1+x-2x2)/2  )
      = 4/(3+2x-x2)

So we now multiply the proportion in the nn group by the
reduction in population ratio and get

        (1+3x2) / (3+2x-x2)

which we set equal to x in order to see if such a proportion is
stable.  It may not be stable but if we solve the equation first
we can test if it is stable or not afterwards.  The resulting
cubic equation is

        x3+x2-3x+1 = 0

But we know that x=1 must be a solution for if there were no g
genes in the population, no-one would ever be of type gn or gg.
So we divide the cubic by 1-x to get:

        x2+2x-1 = 0

which has solutions x1 = 0.414.. and x2 = -2.414..

the 0.414.. fraction being sqrt(2)-1.

A numerical test shows that the x = 0.414 solution is stable
because any variation will produce a next generation closer to
this figure.  The x=1 solution, however is not stable in the
usual sense; that is to say, any small reduction from 1.00 due to
a mutation, say, will get amplified and eventually end up at
0.414 again.

But there is a factor not considered in the mathematics and that
is the taboo on brother-sister and first cousin pairings.  These
taboos will help to remove the odd mutation.  The taboos do not
affect the general conclusion because we have assumed a large
population from which pairings for reproduction are entirely
random.  The smaller the population and the more restricted the
range of pairings due, for example, to other considerations such
as geographical or religious barriers, the more likely it is that
the ratio x will vary from the calculated value.

CONCLUSIONS

1. Contrary to popular belief, a stable population of carriers of
a recessive gene that causes infertility can exist at a figure of
59.6% of the fertile population (41.4% not carrying the gene),
assuming a large population with randomly chosen parents for each
generation of offspring.

2. Whilst no such gene exists in a population, the ratio of
healthy people in the population can remain high, any such rogue
gene being stamped out by the traditional taboos on the marriage
of close relatives, but once such a gene gets a hold (initiated by
a mutation or acquired from another population group) the ratio of
healthy individuals in the population will decrease to the stable
value approaching 41.4%.

Footnote: In the case of the gene that causes sickle cell anemia
in West Africa, the C group get a better chance of survival than
the H group because they have a degree of immunity from malaria.
This increases the proportion of the population who are carriers
and the number of children born who suffer the gene from both
parents.  A cure for malaria would reduce the suffering from
sickle cell anemia but not eradicate it.

(c) David Broughton, March 2004

Comments are invited from Geneticists.