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Pedigree Probability of Autosomal Recessive Trait

Pedigree Probability of Autosomal Recessive Trait


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Here is a pedigree:

The trait is autosomal recessive.

The question is: What is the probability that the bottom 2 people (4 and 5) have a child with the trait?

I tried doing2/3 * 2/3 * 2/3 *1/4and got2/27but this is wrong. I thought that the probability of III 1 being a carrier is 2/3. The probability that IV 4 is a carrier is also 2/3. The probability that III 6 is a carrier is 2/3 and IV 5 must be a carrier. So IV 4 and IV 5 must be carriers. The probability that they have a child with the trait is 1/4. But I am not getting the right answer.

The right answer is 1/6.


Starting with the left hand side of the diagram:

  • III:2 is definitely a carrier (Tt) as one parent (II:2) is affected (tt).
  • III:1 is also definitely a carrier (Tt) as when mating with III:2 they produce an affected (tt) offspring (IV:1)
  • This means that we can work out the possibilities for IV:4 as we know the parent genotypes. It follows the standard arrangement for two carrier parents giving the options of:
    • TT (1/4)
    • Tt (2/4 = 1/2)
    • tt (Normally 1/4 but in this case 0 as individual not marked as affected).
  • Therefore for this scenario, the probabilities for IV:4 are :
    • TT (1/3)
    • Tt (2/3)

Now if we look at the right hand side of the diagram.

  • IV:5 is definitely a carrier (Tt) as one of their parents (III:5) is affected.

This gives two possible Punnett squares to be examined:

|-------------------------------------------------------------------------------| | ♂ (IV:5) | | T t | | -------------------------------------------------------------------| | | | | | T | TT | Tt | | | | | | (IV:4) |-------------------------------------------------------------------| | ♀ | | | | T | TT | Tt | | | | | |-----------|-------------------------------------------------------------------|

This gives nil affected offspring so we can disregard this option for your question (as we are ONLY looking for scenarios which produce affected individuals).

Therefore the alternative is:

|-------------------------------------------------------------------------------| | ♂ (IV:5) | | T t | | -------------------------------------------------------------------| | | | | | T | TT | Tt | | | | | | (IV:4) |-------------------------------------------------------------------| | ♀ | | | | t | TT | tt | | | | | |-----------|-------------------------------------------------------------------|

Giving 1/4 affected offspring.

As mentioned above, in order to have affected offspring then IV:4 must be Tt. There is a2/3chance of this being the case. If this is the case, then there is a1/4chance of the child being tt.

Both conditions need to be true for this to happen so we multiply the fractions:

2/3 * 1/4 = 1/6

Pedigree Probability of Autosomal Recessive Trait - Biology


Calculating probability and risks in pedigree analysis:
Elementary principles


The pedigree shows the occurrence of an autosomal recessive trait, where the black squares have genotype aa . We wish to calculate the probability that IV-1 (shown as ? ) will be either affected ( aa ), or a carrier heterozygote ( Aa ).

(1) For IV-1 to be an affected recessive homozygote , s/he must inherit an a allele from the father and the mother. Given that II-1 must be aa , both great-grandparents ( I-1 and I-2 ) must be Aa . II-2 shows the dominant phenotype, and therefore has at least one A allele: the probability that the other is a is 1/2 . II-3 is from outside the affected pedigree and can be assumed to be AA . Like his father, III-1 shows the dominant phenotype, and therefore has at least one A . Then, the probability that III-1 is Aa is the probability that II-2 is heterozygous andpassed the a allele to III-1 : (1/2) x (1/2) = 1/4. The same reasoning leads to the conclusion that III-2 is heterozygous with a probability of 1/4. Thus, for IV-1 to be aa , both parents must be Aa , andthey must both pass the a allele to their offspring: 1/4 x 1/4 x 1/4 = 1/64

(2) Alternatively, for IV-1 to be a heterozygous , carrier, either s/he most inherit an a allele from the father, or from the mother. The chance of either parent being a heterozygote is 1/4, as calculated above. Then, the probability that both parents are heterozygotes, and the probability that two heterozygotes will have a heterozygous child, is 1/4 x 1/4 x 1/2 = 1/32.

(3) Finally, the probability that IV-1 is a dominant homozygote is 1 - 1/64 - 1/32 = (64 - 1 - 2)/64 = 61/64. This can also be calculated more tediously by summing the alternative probabilities at each of the steps above.

The calculations in this example involved a distinction between a priori and a posteriori probability, which are often presented incorrectly in elementary genetics textbooks . To take a simple case: the a priori probability of getting heads on a single toss of a penny is 1/2, since there are two equal possibilities, H or T . Then, given two pennies tossed at random, HH , HT , TH , and TT are all equally likely. The a prior i probability of getting at least one head is 3/4. The a priori probability that any combination with at least one head will have two tails (HT or TH vs HH ) is 2/3.

However, consider an experiment in which I have tossed two pennies. I show you that one is H , and ask, What is the probability that the other is also H ? The a posteriori probability is 1/2 : given the knowledge that one coin is H , the other is H or T with equal probability. In anticipation of the experiment, the a priori probability of HT given H- is 2/3 . In analyzing the results of any particular experiment, the added information changes probabilities a posteriori .

In the above example, we know that I-1 and I-2 are heterozygotes and II-2 shows the dominant phenotype. We therefore know a posteriori that he has inherited a dominant alleles from one parent, and the probability that he will inherit a recessive alleles from the other parent and be heterozygous is 1/2. It is incorrect to reason that, since 2/3 of all unaffected children (that is, all non- aa ) are heterozygotes a priori , his individual risk is also 2/3. Stated another way, by knowing the nature of one allele, we have lost one statistical degree of freedom .

[In the simplest case: the probability that the next child of I-1 and I-2 will be a boy is a priori 1/2: once the child is born, the probability a posterior is either 0 or 1].

Two further extensions of this idea. For this scenario, assume that a genetic test is available to distinguish AA from Aa , but II-5 is deceased and III-2 will not take the test.

(4) Suppose III-2 has a heterozygous sibling. How does this change the calculation IV-I 's risk? This would mean that II-5 must be a heterozygote with a probability of 1, not 1/2 as before. Then, the probability that III-2 is a heterozygote is 1/2, the probability that the father ( III-1 ) is a heterozygote remains 1/4, and the probability that IV-1 is aa is 1/2 x 1/4 x 1/4 = 1/32 .

(5) Suppose III-2 has one or more siblings who test as unaffected homozygotes ( AA ). How does this change the calculation of IV-1 's risk? Note that, whereas the birth of a heterozygous sibling proves that the mother ( II-5 ) is a heterozygote, the birth of unaffected homozygous offspring cannot prove that she is a homozygote. However, multiple births of unaffected siblings do decrease the probability that she is a heterozygote, as follows. The probability that a heterozygote will not pass the a allele to an offspring is 1/2. Then, the probability that she will not pass it to either of two offspring is (1/2)(1/2) = 0.5 2 = 1/4. The probability that she will pass it to none of three offspring is 0.5 3 = 1/8, to none of four is 1/16, and so on. Less than 0.1% of all families with ten children would have known with an a alleles, if II-5 were a heterozygote . In other words, this is strong a posteriori evidence that II-5 is a homozygote, which if true means that IV-1 cannot be affected. Of course, the birth of an eleventh child who is Aa immediately proves that II-5 is heterozygous, and returns IV-1 's risk calculation to 1/16, as in (4) above..


Pedigree Probability of Autosomal Recessive Trait - Biology

Pedigree Analysis: Six Modes of Inheritance

The inheritance of many monogenic (single-gene) traits can be determined
by the analysis of family trees ( pedigrees ) (IG1 15.22)
Pedigrees show relationships among individuals
Ex .: Queen Victoria's pedigree shows genetic inheritance of Hemophilia A

Six basic Modes of Inheritance
Distinguish autosomal vs sex-linked conditions
sex (X) - linked conditions affect predominantly males
autosomal conditions affect male and females equally

1. Autosomal recessive
ex .: Oculocutaneous Albinism (OMIM203100)
Batten Disease (OMIM204200): high frequency in Newfoundland
PTC tasting (OMIM1720) ( Homework )

Clues : Condition isusually rare (allele frequency is low)
Allele is usually present in heterozygous genotypes
Condition often occurs in matings of related individuals
consanguineous marriage - at least one ancestor in common
" inbreeding " -consanguineous marriages occur more frequently than expected
Conditions often " skips generations "

Clues : Allele is usually present in heterozygousgenotype
Condition must appear in every generation

Clues : Condition occurs predominantly in males
Affected males usually do not have affected offspring
Criss-Cross inheritance : affected males & females alternate among generations

4. X-linked dominant
ex .:
Hypophosphatemia (OMIM 307800) ( XLH )

Clues : Affected males always pass the condition to daughters
Affected females are usually heterozygous

ex .: Zinc-Finger Protein, Y-linked ( ZFY ) (OMIM490000)
Sex-determining Region Y ( SRY ) (OMIM480000)
Hairy Ears (OMIM425500) probably not Y -linked: a sex-limited trait

Clues : Condition occurs in all male descendants, and only males
Phenotypes are typical of males

6 . mtDNA-linked
ex.: The " Daughters of Eve " in Newfoundland
Matrilineal descendants of Queen Victoria

mitochondrial DNA (mtDNA) is a cytoplasmic (non-nuclear) genome
Clues : occurs in all descendants of an affected female
Affected males do not transmit the condition

[ Not covered in 2012 not on final exam]

Calculation of Risk in pedigree analysis involves simple probability theory
The probability of two independent events occurring together (" A and B ")
is the product of the separate probabilities
The probability of two alternative events occurring together (" A or B ")
is the sum of their separate probabilities
The a priori probability of events may be modified by a posteriori knowledge
calculation of the probability of single events is often influenced a posteriori

Homework: In the following pedigrees, which modes of inheritance can be ruled out ?
What individual(s) allow you to do so?
Which are possible ? Which is most likely ? Explain.
Write out the genotypes of each of the individuals.

Pedigree #1
Pedigree #2
Pedigree #3

MGA2 Chapter 5: Problems 2*, 10, 11, 13, 19, 20
iGen2 Chapter 12: Problems 42, 43, 44, 45, 46, 47


Pedigree Analysis: Meaning and Its Problem | Zoology

Pedigree is a genealogical table, chart or diagram representing the ancestral line of an in­dividual man or animal having a long span of life and low reproductive ability. Pedigree analysis is the process of determination of the exact mode of inheritance of a gene when sufficient family information is available.

By this, determination of genotypes and phenotypes of past, present as well as future generations the nature of inheritance, whether sex-linked or autosomal and dominant or recessive probable ratio of inheritance of a gene and segregation are pos­sible. To express the phenotype of individuals of a family for a few generations following symbols are used in the hypothetical pedigree of five generations (Fig. 55.1).

iii. Diamond (◊) a sib (sibling) of un­known (either) sex

iv. Darkened or shed symbols for the rarer genetic phenotype

v. Connection between circle and square (□—O), marriage line

vi. Horizontal line (O—□) below the parents connecting off-springs (siblings) is sib-ship line

vii. Generations, denoted by roman letter I, II, III, IV etc., on a side

viii. Off-springs in each generation are numbered from left side in order of birth

ix. Identical (monozygotic) twin (arise from splitting of a common zygote) represented by a common connection with sib-ship line (11-5 & 6)

x. Dizygotic twins with separate connection with sib-ship line (IV-6 & 7)’

xi. Numbers within the symbols (II-7 to 10) are sibs not individually listed

xii. The connection of sib-ship line and marriage line:

a. Inbred dominant with no genetic segregation

b. Out cross and no segregation

c. Test cross from which a recessive (male) results

d. Out cross by chance producing a recessive (female)

e. Inbred recessives produce only recessive (tt nontaster)

f. Inbred dominants in a cousin marriage, also a back-cross type, producing no phenotypic segregation but only dominant of the Tt, TT varieties (taster).

Problem I: Transmission of Autosomal Recessive Trait:

The figure 55.2 illustrates an hypothetical family pedigree for albinism * in which the albino individuals are designated by shading symbols.

Determine whether the gene is dominant or recessive and probable genotype of each individual.

In the given pedigree the albino daughters (II-6 & 7) from normal parents (I-3 & 4) and from albino father (I-1) and normal mother (I-2) normal children are born. Thus the albino character skip over the generation and appears in alternate generation. So the gene is recessive one. In case of a dominant trait, the affected individuals appear in all generations, there would be no generation skipping.

Here parents I-3 & 4 are the carrier and heterozygous for the recessive gene of albinism. The frequency of children that are affected is about one-fourth (1/4 th) of the total off-springs in lllrd generation born from two normal parents (II-4 & 5). This fact also supports that this gene is recessive.

A dominant trait is usually transmitted by mating between an affected parent, hetero­zygous and a homozygous parent to about half of the off-springs. As the gene is recessive (a) to the normal (+) the genotype of 1-1 is a = a, because the recessive trait can express itself only in homozygous condition. Affected daughters (II-6 & 7) appeared from normal parents (l-3 & 4).

So the genotypes of I-3 & 4 be =. Similar explanation is applicable for II-4 & 5. The probable genotype of all the individuals are as follows.

Problem II: Transmission of Autosomal Dominant Trait:

The figure (Fig. 55.4, 55.5) illustrates an hypothetical family pedigree for a rare trait chondrodystrophy*.

Determine whether this gene is dominant or recessive and show the genotype of each individual.

In the given pedigree, chondrodystrophy appeared in each and every generation where any one of the parent is affected by this trait. That means, the inherited trait did not skip any generation. Whereas no chondrodystrophy observed by marriage between two normal individuals of which one of them having chondrodistrophic history in their past generation (III-7 & 8, III-9 & 10, III-12 & 13, etc.). So the trait is dominant.

In this pedigree it is assumed to be caused by a gene present in the lst generation (I-2). If the trait was recessive, then all the normal individuals of llnd generation (II-2,3, 5 & 8) were heterozygous carrier. So there was a chance of appearance of chondrodistrophic off-springs from the marriage of such heterozygous individuals with normal one. But no such information is evident from this pedigree.

Let the chondrodistrophic gene be ‘c’. As it is a dominant trait over normal ‘†’, so the genotype of I-2 must be C/†, because normal off-springs were born from him. If he was homozygous for this dominant trait then all his children must be chondrodistrophic.

The probable genotype of the individuals are as follows:

Problem III: Transmission of Sex-linked Recessive Trait:

The following pedigree (Fig. 55.6, 55.7) represent the phenotype of a family where red- green colour blind individuals are indicated by shade.

Determine the nature of trait and explain.

From the given pedigree it is observed that the colour blind children in llnd generation (II-1, 6 & 8) are born from normal parents (I-1, 2 & 3). So the gene responsible for colour blindness is a recessive one. As only 50% of male children are affected by this trait, the trait is carried by the sex chromosome (sex-linked).

If it be autosomal then there is a probability to get affected daughters. If it be ‘Y’- linked then there is probability of appearance of the colour blindness to all the sons in llnd generation. But these are not evident from the pedigree and the crisscross type inheritance in the pedigree clearly indicate that the trait is a sex-linked recessive.

Here mother (I-2) is the carrier for the gene and is heterozygous c/†. So all her sons received their X-chromosome from her and Y-chromo some from father (I-1 & 3). For this reason 50% of her sons received the affected gene and rest 50% received normal gene resulting into 1: 1 ratio of colour blind and normal sons.

At the same time, as fathers are normal, they have one normal X-chromosome and all their daughters received one X-chromosome, result normal eye.

The probable genotype of the family pedigree displayed is as follows, taking colour blind trait as ‘c’.

Problem IV: Transmission of Y-linked Dominant Trait:

In the pedigree (Figs. 55.8, 55.9) the shaded symbols are represent as a rare trait.

State whether you believe it as caused by sex-linked or autosomal and dominant and recessive gene.

From the given pedigree it is observed that the gene responsible for the affected trait appeared in each generation. At the same time no affected children were born from marriage of normal male and female (I-3 & 4). So the gene is dominant one.

Another important event is that the trait is always associated with male but never with, females in all the three generations. This finding indicates that the gene is Y-linked, since sons receive their Y-chromosome from father and X-chromosome from mother.

Here II-2 & 5 children receive their Y-chromosome from parent I-2, who is affected. Similarly, the affected gene is inherited from II-2 to III-3 and II-5 to III-4 & 6 respectively. If the gene be X-linked dominant then all the daughters (II-3 & 4) would be affected, as they would receive one X-chromosome from their father. But in the given pedigree it is not evident. So the gene is Y-linked dominant.

Problem V. Muscular Dystrophy:

Muscular dis-trophy is a rare human disease. The given pedigree (Figs. 55.10, 55.11) represent the phenotypes of a certain family with shaded symbols as affected individuals.

Determine the mode of inheritance of the gene.

In the given pedigree, the affected trait appeared in all generations in which one of the parents (I-2) was affected. At the same time, no affected child was born from marriage of normal individuals (I-3 & 4). So the gene is dominant.

It is found that the trait has expressed with sex during transmission from generation to generation. In generation III all the daughters (III-1, 2, 4, 6, 7, 9 & 10) are muscular distrophied as their father is affected. But no male children are affected.

So it is probable that the gene is X-linked, for which all female children of generation III received the affected X- chromosome from their father (II-2 & 6), who received that gene from their affected mother (I-2). The gene is transmitted from mother to son and son to daughter in a crisscross way. So the gene is X-linked dominant.

1. In the pedigree (Figs. 55.12 and 55.13) affected individuals are shaded.

i. Indicate the trait is dominant or recessive.

ii. Mention the trait is either autosomal or sex-linked.

iii. Determine the genotype of each individual in the pedigree and indicate the choice of genotypes if you think more than one genotype is possible, for an individual.

2. Describe the following pedigree (Figs. 55.14 & 15). Analyse the mode of inheritance and nature of trait. The shaded individuals are affected. Find out the genotype of the shaded individual.

3. The figure (Fig. 55.16) represents four family pedigrees for a trait in hu­mans. Shaded symbols bear trait. For each pedigree (A, B, C, D), state by answering ‘yes’ or ‘no’ in the appropriate blank space whether transmission of the trait can be accounted for on the basis of each of the listed simple modes of inheritance.


Autosomal Dominant and Recessive Inheritance

Many important and well-understood genetic diseases are the result of a mutation in a single gene. The online edition of McKusick’s Mendelian Inheritance in Man ( www-ncbi-nlm-nih-gov.easyaccess1.lib.cuhk.edu.hk/Omim/ ) lists nearly 16,000 single genes and more than 8000 single-gene or monogenic traits defined thus far in humans. Of these 24,000 genes and traits, about 23,000 are located on autosomes, nearly 1,300 are located on the X chromosome, and 60 are located on the Y chromosome. The identification of genes that cause monogenic traits has led to new and exciting insights, not only in genetics, but also in the basic pathophysiology of disease.

In this chapter we focus on single-gene disorders caused by mutations on the autosomes. (Single-gene disorders caused by mutations on the sex chromosomes are the subject of Chapter 5 .) We discuss the patterns of inheritance of these diseases in families and factors that complicate these patterns. We also discuss the risks of transmitting single-gene diseases to one’s offspring, because this is usually an important concern for at-risk couples.

Basic Concepts of Formal Genetics

Gregor Mendel’s Contributions

Monogenic traits are also known as mendelian traits, after Gregor Mendel, the 19th-century Austrian monk who deduced several important genetic principles from his well-designed experiments with garden peas. Mendel studied seven traits in the pea, each of which is determined by a single gene. These traits included attributes such as height (tall versus short plants) and seed shape (smooth versus wrinkled). The variation in each of these traits is caused by the presence of different alleles at individual loci.

Several important principles emerged from Mendel’s work. The first, the principle of dominant and recessive inheritance, was discussed in Chapter 3 . Mendel also discovered the principle of segregation, which states that sexually reproducing organisms possess genes that occur in pairs and that only one member of this pair is transmitted to the offspring (i.e., it segregates). The prevalent thinking during Mendel’s time was that hereditary factors from the two parents are blended in the offspring. In contrast, the principle of segregation states that genes remain intact and distinct. An allele for “smooth” seed shape can be transmitted to an offspring in the next generation, which can in turn transmit the same allele to its own offspring. If genes were somehow blended in offspring instead of remaining distinct, it would be impossible to trace genetic inheritance from one generation to the next. Thus the principle of segregation was a key development in modern genetics.

Mendel’s principle of independent assortment was another significant contribution to genetics. This principle states that genes at different loci are transmitted independently. Consider the two loci mentioned previously. One locus can have either the “smooth” or the “wrinkled” allele, and the other can have either the “tall” or the “short” allele. In a reproductive event a parent transmits one allele from each locus to its offspring. The principle of independent assortment dictates that the transmission of a specific allele at one locus (“smooth” or “wrinkled”) has no effect on which allele is transmitted at the other locus (“tall” or “short”).

The principle of segregation describes the behavior of chromosomes in meiosis. The genes on chromosomes segregate during meiosis, and they are transmitted as distinct entities from one generation to the next. When Mendel performed his critical experiments, he had no direct knowledge of chromosomes, meiosis, or genes (indeed, the last term was not coined until 1909, long after Mendel’s death). Although his work was published in 1865 and cited occasionally, its fundamental significance was unrecognized for several decades. Yet Mendel’s research, which was eventually replicated by other researchers at the turn of the 20th century, forms the foundation of much of modern genetics.

Mendel’s key contributions were the principles of dominance and recessiveness, segregation, and independent assortment.

The Concept of Phenotype

The term genotype has been defined as an individual’s genetic constitution at a locus. The phenotype is what is actually observed physically or clinically. Genotypes do not uniquely correspond to phenotypes. Individuals with two different genotypes, a dominant homozygote and a heterozygote, can have the same phenotype. An example is cystic fibrosis ( Clinical Commentary 4.1 ), an autosomal recessive condition in which only the recessive homozygote is affected. Conversely, the same genotype can produce different phenotypes in different environments. An example is the recessive disease phenylketonuria (PKU, see Chapter 7 ), which is seen in approximately 1 of every 10,000 European-ancestry births. Mutations at the locus encoding the metabolic enzyme phenylalanine hydroxylase render the homozygote unable to metabolize the amino acid phenylalanine. Although babies with PKU are unaffected at birth, their metabolic deficiency produces a buildup of phenylalanine and its toxic metabolites. This process is highly destructive to the central nervous system, and it eventually produces severe mental impairment. It has been estimated that babies with untreated PKU lose, on average, 1 to 2 IQ points per week during the first year of life. Thus the PKU genotype can produce a severe disease phenotype. However, it is straightforward to screen for PKU at birth (see Chapter 13 ), and damage to the brain can be avoided by initiating a low-phenylalanine diet within 1 month after birth. The child still has the PKU genotype, but the phenotype has been profoundly altered by environmental modification.

Cystic fibrosis (CF) is one of the most common single-gene disorders in North America, affecting approximately 1 in 2000 to 1 in 4000 European-American newborns. The prevalence among African-Americans is about 1 in 15,000 births, and it is less than 1 in 30,000 among Asian-Americans. Approximately 30,000 Americans and 70,000 people worldwide are estimated to have this disease.

CF was first identified as a distinct disease entity in 1938 and was termed “cystic fibrosis of the pancreas.” This refers to the fibrotic lesions that develop in the pancreas, one of the principal organs affected by this disorder ( Fig. 4.1 ). Approximately 85% of CF patients have pancreatic insufficiency, in which the pancreas is unable to secrete digestive enzymes, contributing to chronic malabsorption of nutrients. The intestinal tract is also affected, and approximately 15% to 20% of newborns with CF have meconium ileus (thickened, obstructive intestinal matter). The sweat glands of CF patients are abnormal, resulting in high levels of chloride in the sweat. This is the basis for the sweat chloride test commonly used in the diagnosis of this disease. More than 95% of males with CF are sterile due to absence or obstruction of the vas deferens.

The major cause of morbidity and mortality in CF patients is pulmonary disease. Patients with CF have lower airway inflammation and chronic bronchial infection, progressing to end-stage lung disease characterized by extensive airway damage and fibrosis of lung tissue. Airway obstruction and lung injury are thought to be caused by a dehydrated airway surface and reduced clearance, resulting in thick airway mucus. This is associated with infection by bacteria such as Staphylococcus aureus and Pseudomonas aeruginosa. The combination of airway obstruction, inflammation, and infection leads to destruction of the airways and lung tissue, resulting eventually in death from pulmonary disease in more than 90% of CF patients.

As a result of improved nutrition, airway clearance techniques, and antibiotic therapies, the survival rate of CF patients has improved substantially during the past three decades. Median survival time in the United States is now nearly 40 years. CF has highly variable expression, with some patients experiencing only mild respiratory difficulty and nearly normal survival. Others have much more severe respiratory problems and may survive less than two decades.

CF is caused by mutations in a gene, CFTR, ∗ that encodes the cystic fibrosis transmembrane conductance regulator ( Fig. 4.2 ). CFTR encodes cyclic AMP-regulated chloride ion channels that span the membranes of specialized epithelial cells, such as those that line the bowel and lung. In addition, CFTR is involved in regulating the transport of sodium ions across epithelial cell membranes. The role of CFTR in sodium and chloride transport helps us to understand the multiple effects of mutations at the CF locus. Defective ion transport results in salt imbalances, depleting the airway of water, and producing the thick, obstructive secretions seen in the lungs. The pancreas is also obstructed by thick secretions, leading to fibrosis and pancreatic insufficiency. The chloride ion transport defect explains the abnormally high concentration of chloride in the sweat secretions of CF patients chloride cannot be reabsorbed from the lumen of the sweat duct.

DNA sequence analysis has revealed nearly 2000 different mutations at the CFTR locus. The most common of these, labeled F508del, is a three-base deletion that results in the loss of a phenylalanine residue (F) at position 508 of the CFTR protein. F508del accounts for nearly 70% of all CF mutations. This mutation, along with several dozen other relatively common ones, is assayed in the genetic diagnosis of CF (see Chapter 13 ).

Identification of the specific mutation or mutations that are responsible for CF in a patient can help to predict the severity of the disease. For example, the most severe classes of mutations (of which F508del is an example) result in a complete lack of chloride ion channel production or in channels that cannot migrate to the cell membrane. Patients homozygous for these mutations nearly always have pancreatic insufficiency. In contrast, other mutations (e.g., R117H, a missense mutation) result in ion channels that do proceed to the cell membrane but respond poorly to cyclic AMP and consequently do not remain open as long as they should. The phenotype is thus milder, and patients who have this mutation are less likely to have pancreatic insufficiency. Some males with mild CFTR mutations have only congenital bilateral absence of the vas deferens (CBAVD) but little, if any, lung or gastrointestinal disease. The correlation between genotype and phenotype is far from perfect, however, indicating that modifier loci and environmental factors must also influence expression of the disease (see text). In general there is a reasonably good correlation between genotype and pancreatic function and a more variable relationship between genotype and pulmonary function.

The ability to identify CFTR mutations has led to surveys of persons who have one (heterozygous) or two (homozygous) CFTR mutations, but who do not have cystic fibrosis. They have increased risks for a number of disease conditions, including CBAVD, bronchiectasis (chronic dilatation of the bronchi and abnormal mucus production), and pancreatitis (pancreatic inflammation).

By enhancing our understanding of the pathophysiology of CF, identification of CFTR has opened the possibility of new treatments for this disease. Examples include administration of drugs that cause ribosomes to read through the premature stop codons that account for approximately 7% of CFTR mutations. Other drugs can increase the activity of chloride channels in patients with class III or IV mutations. The first FDA-approved drug for CF treatment, ivacaftor, increases CFTR channel activity in response to ATP. A second FDA-approved drug, lumacaftor, can be used in combination with ivacaftor and has been shown to significantly improve lung function in CF patients homozygous for the common F508del mutation. Gene therapy, in which the normal CFTR gene is placed in viral or other vectors that are then introduced to the patient’s airway (see Chapter 13 ), is also being actively investigated. This strategy, however, has encountered difficulties because viral vectors can induce an inflammatory immune response.

∗ Conventionally the symbol for a gene, such as CFTR , is shown in italics, and the symbol for the protein product is not.

This example shows that the phenotype is the result of the interaction of genotype and environmental factors. It should be emphasized that “environment” can include the genetic environment (i.e., genes at other loci whose products can interact with a specific gene or its product).

The phenotype, which is physically observable, results from the interaction of genotype and environment.

Basic Pedigree Structure

The pedigree is one of the most commonly used tools in medical genetics. It illustrates the relationships among family members, and it shows which family members are affected or unaffected by a genetic disease. Typically an arrow denotes the proband, the first person in whom the disease is diagnosed in the pedigree. The proband is sometimes also referred to as the index case or propositus (proposita for a female). Fig. 4.3 describes the features of pedigree notation.

When discussing relatives in families, one often refers to degrees of relationship. First-degree relatives are those who are related at the parent–offspring or sibling (brother and sister) level. Second-degree relatives are those who are removed by one additional generational step (e.g., grandparents and their grandchildren, uncles or aunts and their nieces or nephews). Continuing this logic, third-degree relatives would include, for example, one’s first cousins and great-grandchildren.

Autosomal Dominant Inheritance

Characteristics of Autosomal Dominant Inheritance

Autosomal dominant diseases are seen in roughly 1 of every 200 individuals (see Table 1.3 in Chapter 1 ). Individually each autosomal dominant disease is rather rare in populations, with the most common ones having gene frequencies of about 0.001. For this reason matings between two individuals who are both affected by the same autosomal dominant disease are uncommon. Most often affected offspring are produced by the union of an unaffected parent with an affected heterozygote. The Punnett square in Fig. 4.4 illustrates such a mating. The affected parent can pass either a disease allele or a normal allele to his or her children. Each event has a probability of 0.5. Thus on the average, half of the children will be heterozygotes and will express the disease, and half will be unaffected homozygotes.

Postaxial polydactyly, the presence of an extra digit next to the fifth digit ( Fig. 4.5 ), can be inherited as an autosomal dominant trait. An idealized pedigree for this disease, shown in Fig. 4.6 , illustrates several important characteristics of autosomal dominant inheritance. First, the two sexes exhibit the trait in approximately equal ratios, and males and females are equally likely to transmit the trait to their offspring. This is because postaxial polydactyly is an autosomal disease (as opposed to a disease caused by an X chromosome mutation, in which these ratios typically differ). Second, there is no skipping of generations if an individual has polydactyly, one parent must also have it. This leads to a vertical transmission pattern in which the disease phenotype is usually seen in one generation after another. If neither parent has the trait, none of the children will have it. Third, father-to-son transmission of the disease gene is observed. Although father-to-son transmission is not required to establish autosomal dominant inheritance, its presence in a pedigree rules out some other modes of inheritance (particularly X-linked inheritance see Chapter 5 ). Finally, as we have already seen, an affected heterozygote transmits the disease-causing allele to approximately half of their children. However, because gamete transmission, like coin tossing, is subject to chance fluctuations, it is possible that all or none of the children of an affected parent will have the trait. When large numbers of matings of this type are studied, the proportion of affected children closely approximates 1/2.

Autosomal dominant inheritance is characterized by vertical transmission of the disease phenotype, a lack of skipped generations, and roughly equal numbers of affected males and females. Father-to-son transmission may be observed.

Recurrence Risks

Parents at risk for producing children with a genetic disease typically want to know the risk, or probability that their future children will be affected. This probability is termed the recurrence risk. If one parent is affected by an autosomal dominant disease and the other is unaffected, the recurrence risk for each child is 1/2 (assuming that the affected parent is a heterozygote, which is nearly always the case). It is important to keep in mind that each birth is an independent event, as in the coin-tossing examples. Thus even if the parents have already had a child with the disease, their recurrence risk remains 1/2. Even if they have had several children, all affected (or all unaffected) by the disease, the law of independence dictates that the probability that their next child will have the disease is still 1/2. Although this concept seems intuitively obvious, it is commonly misunderstood by the lay population. Further aspects of communicating risks to families are discussed in Chapter 15 .

The recurrence risk for an autosomal dominant disorder is 50%. Because of independence, this risk remains constant no matter how many affected or unaffected children are born.

Autosomal Recessive Inheritance

Like autosomal dominant diseases, autosomal recessive diseases are individually fairly rare in populations. As shown previously, heterozygous carriers for recessive disease alleles are much more common than affected homozygotes. Consequently the parents of individuals affected with autosomal recessive diseases are usually both heterozygous carriers. As the Punnett square in Fig. 4.7 demonstrates, one-fourth of the offspring of two heterozygotes will be unaffected homozygotes, half will be phenotypically unaffected heterozygous carriers, and one-fourth will be homozygotes affected with the disease (on average).

Characteristics of Autosomal Recessive Inheritance

Fig. 4.8 is a pedigree showing the inheritance pattern of an autosomal recessive form of albinism that results from mutations in the gene that encodes tyrosinase, a tyrosine-metabolizing enzyme. ∗ The resulting tyrosinase deficiency creates a block in the metabolic pathway that normally leads to the synthesis of melanin pigment. Consequently the affected individual has very little pigment in the skin, hair, and eyes ( Fig. 4.9 ). Because melanin is also required for the normal development of the optic fibers, persons with albinism can also display nystagmus (rapid uncontrolled eye movement), strabismus (deviation of the eye from its normal axis), and reduced visual acuity. The pedigree demonstrates most of the important criteria for distinguishing autosomal recessive inheritance ( Table 4.1 ). First, unlike autosomal dominant diseases in which the disease phenotype is seen in one generation after another, autosomal recessive diseases are usually observed in one or more siblings, but not in earlier generations. Second, as in autosomal dominant inheritance, males and females are affected in equal proportions. Third, on average, one-fourth of the offspring of two heterozygous carriers will be affected with the disorder. Finally, consanguinity is present more often in pedigrees involving autosomal recessive diseases than in those involving other types of inheritance (see Fig. 4.8 ). The term consanguinity (Latin, “with blood”) refers to the mating of related persons. It is sometimes a factor in recessive disease because related persons are more likely to share the same disease-causing alleles. Consanguinity is discussed in greater detail later in this chapter.

∗ This form of albinism, termed tyrosinase-negative oculocutaneous albinism (OCA1), is distinguished from a second, milder form termed tyrosinase-positive oculocutaneous albinism (OCA2). OCA2 is typically caused by mutations in a gene on chromosome 15, whose protein product is thought to be involved in the transport and processing of tyrosine and tyrosinase.

Autosomal recessive inheritance is characterized by clustering of the disease phenotype among siblings, but the disease is not usually seen among parents or other ancestors. Equal numbers of affected males and females are usually seen, and consanguinity may be present.


Pedigree Probability of Autosomal Recessive Trait - Biology

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Glossary

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Determining Genotypes

Not all pedigre es use the half shaded convention to show carriers of a trait. One very useful purpose of pedigrees is to be able to identify the genotype, or possible genotypes of an individual, based on their family heritage.

Steps in Determining Genotype

  • In the information given, usually in a title, determine if the trait being discussed is dominant or recessive.
  • If the trait is dominant, th e n individuals with the trait will have their shapes coloured in, if the trait is recessive, th e n individuals with the trait will have unshaded circles or squares.
  • Locate the recessive individuals in the pedigree, and assign their genotype - two lower case alleles (ff).
  • All of the individuals with the dominant trait will have one capital letter, which expresses the trait. You can assign their genotype temporarily as capital letter, question mark to represent their alleles (F?)
  • To determine the second allele, you must examine the genotypes of the parents or offspring.
  • Each parent must give one allele. If one of the parents is homozygous recessive, ff, then the offspring must have a heterozygous, Ff, genotype.
  • If both parents are homozygous dominant, FF, then the offspring must also be homozygous dominant, FF.

Examine the example below, while listening to the audio, to see how genotypes can be determined.

To make any of the images larger, click on them. Or, download the word document below (Steps_In_Determining_Genotpe_Images.docx) and follow along on a printed copy. You could use this page to make your own notes at the bottom.

Pedigree for Cystic Fibrosis

AUDIO: Genotypes of CF Pedigree

By examining the parents or offspring, we can det ermine some , but not always all, of the genotypes of the individuals.

In this example, the mother in the first generation, only has a recessive allele to give. The offspring must have received the dominant allele, F, from their father. Since they must get one allele from their mother, all of their genotypes must be heterozygous, Ff.

Additional Example
Recall the tongue rolling example from the previous page. Can you determine the genotypes of the individuals in this family?


Autosomal recessive inheritance in pedigree and experiment, examples of traits in man

All individuals with the defect/disease in pedigrees (and in population) are homozygotes of recessive defective (deleterious, nonactive, affected, mutated etc.) alleles, aa. Two copies of a disease allele are needed for an individual to express the phenotype. The parents of an affected (suffering with an AR disease) individual are healthy (not affected by this AR disease) but are disease carriers. Typically, once a child in a family is born with AR trait/disease we may suppose parents as carriers (heterozygotes Aa) and there is a 25% (1/4) chance (probability, risk) that any next offspring will inherit two copies of the disease allele and will therefore have the disease phenotype. N.B.: There is a 50% (1/2) chance that the offspring of carrier parents will inherit one copy of the disease allele and will be a carrier, and there is a 25% probability the offspring will inherit healthy (normal, wild-type etc.) allele from both parents (will be a homozygote AA) and will not express the disease phenotype or be a disease carrier. This AA individual would not pose a risk for passing the disorder on to his/her offspring.

Many AR diseases (disorders) are seen more frequently in individuals of certain ethnic origins than others because these individuals are descendants of the same ancestors. However, because these common ancestors are generally more distantly related to these individuals, couples of the same ethnic background would generally have fewer genes in common than consanguineous couples.

AR inheritance in hybridization experiment [ edit | edit source ]

In experimental procedure, pure lines (e.g. inbred strains) are chosen showing difference in a character (purple versus white flowers, yellow versus green seeds/pies, colour versus albino coat/fur). Parental generation pairs are formed combining one individual from each line/strain. They are crossed in order to obtain F1 hybrids, these are phenotypically uniform, all expressing a single form of the trait (character). By intercrossing the F1 individuals, the F2 generation is obtained, where phenotypes segregate in ratio 3 : 1 (one vs. the other form of the trait). This 1/4 correspond to recessive phenotype (not seen in the uniform F1 generation) and, retrospectively, also allow to decide which line (strain) in P generation expressed the recessive form of the trait.

In a general population with uniform (wild) phenotype, an individual with an exceptional (newly appearing) form of the phenotype can be tested by hybridization experiment, too, if he/she/it expresses the dominant or recessive form of the trait.

If this new, different form of a trait is recessive (i.e. new phenotype individual is a homozygote nn), then this hybridization test (crossing with any individual with normal, common, wild phenotype in the population) represents de facto:

  1. most probably, a parental cross (wild phenotype individual is a homozygote NN) which leads to uniform (F1) hybrids expressing the wild phenotype, or
  2. rarely, a backcross (in case that the wild phenotype individual is a hidden, “silent” heterozygote Nn) resulting in phenotypic ratio 1 to 1 .

Experimental cross in a population (AR inheritance) – new, spontaneous mutant is recessive homozygote, and wild-type phenotype individuals are (most probably) dominant homozygotes


Backcross (experimental test-cross) in AR inheritance – rarely, wild-type phenotype individuals are heterozygotes (by the way, they had to be at least two of them, of opposite gender in preceding generation of the population it is the only way how a recessive homozygote could appear in the present generation of the population).


Inheriting a specific disease, condition, or trait depends on the type of chromosome that is affected. The two types are autosomal chromosomes and sex chromosomes. It also depends on whether the trait is dominant or recessive.

A mutation in a gene on one of the first 22 nonsex chromosomes can lead to an autosomal disorder.

Genes come in pairs. One gene in each pair comes from the mother, and the other gene comes from the father. Recessive inheritance means both genes in a pair must be abnormal to cause disease. People with only one defective gene in the pair are called carriers. These people are most often not affected with the condition. However, they can pass the abnormal gene to their children.

CHANCES OF INHERITING A TRAIT

If you are born to parents who both carry the same autosomal recessive gene, you have a 1 in 4 chance of inheriting the abnormal gene from both parents and developing the disease. You have a 50% (1 in 2) chance of inheriting one abnormal gene. This would make you a carrier.

In other words, for a child born to a couple who both carry the gene (but do not have signs of disease), the expected outcome for each pregnancy is:

  • A 25% chance that the child is born with two normal genes (normal)
  • A 50% chance that the child is born with one normal and one abnormal gene (carrier, without disease)
  • A 25% chance that the child is born with two abnormal genes (at risk for the disease)

Note: These outcomes do not mean that the children will definitely be carriers or be severely affected.


Introduction

Humans make terrible subjects for genetic studies. That’s because the primary experimental tool of genetics, the controlled cross, is simply not possible for humans. There is no such thing as a highly homozygous “true-breeding line” of humans, and even if there were, it would be impossible (or at the very least highly unethical) to compel two people to mate with each other to reveal the pattern of inheritance of a particular trait. Almost since the dawn of genetics at the turn of the 20th century, therefore, human geneticists have relied on a much less powerful (but at least feasible) method, the analysis of pedigree charts.

“Eugenicists employed pedigrees to bolster their claims that not only were physical traits strictly inherited as simple Mendelian characters, but so too were mental, emotional, and behavioral characteristics.”

Pedigree charts are branching diagrams that depict the appearance of a trait in a family (Resta, 1993). If the family is large enough, comprising several generations with multiple affected persons, it may be possible to infer how the trait is inherited, whether the responsible gene resides on the X chromosome or an autosome, and whether it is dominant or recessive.

Genealogical charts were drawn for centuries before they became so important in human genetics. At first, they were simply graphical representations of family relationships. The name is derived from the Middle French pié de grue, literally meaning “foot of crane,” because some early diagrams contained so many branching diagonal lines that they resembled a bird’s foot. One of the earliest surviving pedigrees showed the inheritance of hemophilia A, an X-linked recessive condition, in the royal families of Europe.

Pedigrees have grown in sophistication in recent decades, and may now incorporate information from genetic testing as specific as single-nucleotide differences in genes. But their basic construction has changed little since the first decades of the 20th century. What most students of genetics are unaware of, however, is that pedigree analysis was developed in this country by proponents of eugenics, the pseudoscientific social movement with the aim of controlling human heredity (Resta, 1993). Eugenicists employed pedigrees to bolster their claims that not only were physical traits strictly inherited as simple Mendelian characters, but so too were mental, emotional, and behavioral characteristics. As we will see, their analysis was deeply flawed, based on slipshod data collection, unsupported assumptions, and circular reasoning.


Pedigree Probability of Autosomal Recessive Trait - Biology

2. Two phenotypically normal people marry. Unknown to them, they have a common grandfather who suffered from a rare disorder that is inherited as an autosomal recessive trait. Assume that the people marrying into this pedigree are all DD , that there has been no earlier consanguinity, and that no affected (dd) individuals have occurred. What is the probability that the first child of this couple (a baby girl) will be affected? Draw the pedigree.

3. This pedigree is for the autosomal dominant trait achondroplasia, a rare form of dwarfism. The probability that III-1 and III-2 will have an affected child is

A. zero B. 1/4 C. 1/4 D. 1/16 E. 1/8

* If you need a little bit of help with the above, check out these problem solving tips.



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