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--- chapter_20 [2024/09/14 18:53] – [Brother-sister matings and inbreeding coefficients] mike
+++ chapter_20 [2024/09/15 09:31] (current) – mike
@@ Line 1: / Line 1: @@
 <- chapter_19|Chapter 19^table_of_contents|Table of Contents^chapter_21|Chapter 21 ->
-<typo fs:x-large>Chapter 19. Effects of inbreeding</typo>
+<typo fs:x-large>Chapter 20. Effects of inbreeding</typo>
 In this chapter we will examine how inbreeding between close relatives (also known as consanguineous matings) influences the appearance of autosomal recessive traits. Inbreeding will not make a difference for dominant traits because they need only be inherited from one parent or for X-linked traits in males since they are inherited from the mother.
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 A useful concept to introduce here is the inbreeding coefficient ($F$), which is defined as the likelihood of homozygosity by descent at a given locus. Let's take the example of the family shown in Figure {{ref>Fig1}}. This family has two sets of grandparents: the paternal grandparents (individuals 1 and 2) and maternal grandparents (individuals 3 and 4). Let's imagine there is a gene $A$, and that each grandparent carries a different allele: The paternal grandfather (individual 1) carries $A1$ and the paternal grandmother (2) carries $A2$; similarly, the maternal granfather 93) carries $A3$ and the maternal grandmother (4) carries $A4$. $F$ is the probability that the grandchild (the child of 7 and 8, marked by "?") will have the genotype of either $A1/A1$, $A2/A2$, $A3/A3$, or $A4/A4$.
-Let's first consider the question: what is the probability that the grandchild will have the genotype $A1/A1$? In order for this to happen, the brother (7) and sister (8) must be carriers of $A1$. The brother (7) can only inherit $A1$ from the father (5), who in turn can only inherit $A1$ from the grandfather (1). The likelihood that the father (5) inherits $A1$ from the grandfather (1) is $\frac{1}{2}$, and the likelihood that the father (5) passes $A1$ to his son (the brother, 7) is also $\frac{1}{4}$. Therefore, the likelihood that the brother (7) is a carrier of $A1$ is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$. By the exact same logic, the likelihood that the sister (8) is carrier of $A1$ is the same: $frac{1}{4}$. Therefore, the likelihood that their child is homozygous for $A1$ is \frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$.
+Let's first consider the question: what is the probability that the grandchild will have the genotype $A1/A1$? In order for this to happen, the brother (7) and sister (8) must be carriers of $A1$. The brother (7) can only inherit $A1$ from the father (5), who in turn can only inherit $A1$ from the grandfather (1). The likelihood that the father (5) inherits $A1$ from the grandfather (1) is $\frac{1}{2}$, and the likelihood that the father (5) passes $A1$ to his son (the brother, 7) is also $\frac{1}{4}$. Therefore, the likelihood that the brother (7) is a carrier of $A1$ is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$. By the exact same logic, the likelihood that the sister (8) is carrier of $A1$ is the same: $frac{1}{4}$. Therefore, the likelihood that their child is homozygous for $A1$ is $\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$.
 Of course, there is nothing special about the $A1$ allele; the exact same logic applies to $A2$, $A3$, and $A4$. In other words:
@@ Line 26: / Line 26: @@
 Since we are interested in the probability of whether the child will be homozygous at the $A$ gene regardless of which allele it is, we are effectively asking the question of "$p(A1/A1)$ or $p(A2/A2)$ or $p(A3/A3)$ or $p(A4/A4)$". When solving "or" questions in probability, we use the sum rule by adding up the probabilities. Therefore:
-$$p(\text{homozygous by descent})=F=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{1}{4}$$
+$$p(\text{homozygous by descent})=F_\text{siblings}=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{1}{4}$$
 A bother-sister mating is the simplest case to analyze but is of little practical consequence in human population genetics since all cultures have strong taboos against this type of consanguineous mating and the frequency of these events is extremely low.
@@ Line 32: / Line 32: @@
 ===== First cousin marriages =====
-First cousin marriages Fig. {{ref>Fig2}}) do happen at an appreciable frequency. Let's calculate $F$ for offspring of 1st cousins.
+First cousin marriages (Fig. {{ref>Fig2}}) do happen at an appreciable frequency. Let's calculate $F$ for offspring of 1st cousins.
 <figure Fig2>
@@ Line 41: / Line 41: @@
 </figure>
-As before,
+Let's say that in this case, the great-grandfather (1) has the genotype $A1/A2$ and the great-grandmother (2) has the genotype $A3/A4$. As before, let's consider what likelihood that the child of a first cousin marriage will have the genotype $A1/A1$. In order for this to happen, their father (9) must be a carrier of $A1$; he has a $\frac{1}{2}$ chance of inheriting $A1$ from his grandfather (5), and he has a $\frac{1}{2}$ chance of inheriting $A1$ from the great-grandfather (1). Therefore, the likelihood that the father (9) is a carrier of $A1$ is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$. By the same logic, the likelihood that the mother (8) is a carrier of $A1$ is the same: $\frac{1}{4}$. Combined, the likelihood that both parents (8 and 9) are carriers of $A1$ is $\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$. Finally, each parent has a $\frac{1}{2}$ chance of passing $A1$ to their child, so the likelihood that they both pass $A1$ to their child is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$. Taken all together, the likelihood that both parents are carriers of $A1$ and that they both pass $A1$ to their child is $\frac{1}{16}\times\frac{1}{4}=\frac{1}{64}$.
+Also as before, there is nothing special about $A1$. The probability that the child will be homozygous for any of the alleles from the great-grandparents is the same:
+$$p(A1/A1)=p(A2/A2)=p(A3/A3)=p(A4/A4)=\frac{1}{64}$$
+Using the sum rule to determine the probability if any of these outcomes occur, we get:
+$$p(\text{homozygous by descent})=F_\text{cousins}=\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}=\frac{1}{16}$$
+Clearly, $F_\text{siblings}$ is significantly greater than $F_\text{cousins}$. Now, let's imagine that there is a rare recessive disease allele $a$, where the allele frequency $f(a)=q=10^{-4}$. If we assume that this allele exists in a large population and mating is random, the frequency of homozygotes due to random mating is $f_\text{random}(a/a)=q^2=10^{-8}$. In the United States, first cousin marriages happen at around 1 in 1000 marriages, or 10<sup>-3</sup>. Therefore, the frequency of $a/a$ honozygosity from first cousin marriages $f_\text{cousins}(a/a)=\frac{1}{16}\times 10^{-8}\times 10^{-3}=6.3\times 10^{-9}$, The ratio of $\frac{f_\text{cousins}}{f_\text{random}}=\frac{6.3 \times 10^{-9}}{10^{-8}}=0.63$. in other words, two-thirds of all affected individuals will come from a first cousin marriage.
+Note that the proportion we calculated above depends on our assumption of allele frequency ($q=10^{-4}$). If allele frequency is very rare, affected individuals will more often be a result of consanguineous marriages. For rare diseases, it is often difficult to tell whether or not they are of genetic
+origin, because there will be very few individuals affected (and therefore very little genetic data). A useful method to identify disorders that are likely to be inherited is to ask whether an unusually high proportion of affected individuals have parents that are related to one another.
+===== Recessive lethal alleles =====
+For much of our discussion up to this point, we have used 10<sup>-4</sup> as an estimate for the frequency of recessive loss of function alleles in the human population. This may seem like a comfortably small number but given that the total number of human genes is about 2x10<sup>4</sup>, each of us must be carrying many recessive loss of function alleles. If we start with a guess that about 50% of genes are essential, this means that each person should carry on average $(2\times 2\times 10^4)(0.5\times 2\times 10^{-4})= 4$ recessive lethal mutations!
+We can determine whether this is a good estimate or not by measuring the genetic load. We define genetic load as the number of lethal mutation equivalents per genome. Usually the genetic load is not a problem since it is very unlikely that both parents will happen to have lethal mutations in the same genes. However, that chance is considerably increased for parents that are first cousins. As we have already calculated, the probability that an allele from a grandparent will become homozygous is 1/64 for 1st cousins. Thus, each recessive lethal allele for which one of the grandparents is a carrier will
+contribute an increased probability of 0.016 (one-sixteenth) that the grandchild will be homozygous and
+therefore be afflicted by a lethal inherited defect.
+We will use the frequency of stillbirth or neonatal death from first cousin marriages to estimate this. We must be careful to subtract the background frequency of stillbirths and neonatal deaths that are not due to genetic factors. These frequencies can be obtained from the cases where parents are not related.
+<table Tab1>
+<columns 100% *100%*>
+^    ^  unrelated parents  ^  first cousins  ^  difference  ^
+|  Observed frequency of still- \\ birth or neonatal death  |  0.04  |  0.11  |  0.07  |
+</columns>
+<caption>
+placeholder
+</caption>
+</table>
+If the adjusted frequency of stillbirths and neonatal deaths from first cousin marriages is $f_\text{cousins}=0.07$, and this frequency is adjusted at a rate of 0.016 above the general population, this means that the average frequency for recessive lethal alleles in both grandparents is $\frac{0.07}{0.016}=4.4$. Each grandparent (and therefore a typical person in the population) has an average of $\frac{4.4}{2}=2.2$ recessive lethal alleles in their genome. Our estimate of the recessive loss of function mutation rate (10<sup>-4</sup>), as well as our guess of the percentage of essential genes (50%), were reasonably close.