ch7 gene disorders and pedigree analysis

7.0 Gene disorders and pedigree analysis

Prepared by Pratheep SandrasaigaranLecturer at Manipal International University

By the end of this chapter you should be able to:

• Understand laws of probability that help to explain genetic events.

• Do pedigree analysis for a given disorder.

• Recognize the gene disorders

Prepared by Pratheep Sandrasaigaran

Diagram adopted from Internet Sources

7.1 laws of probability in genetics


laws of probability in genetics


Product law

Sum law

Binomial theorem

Chi-Square (χ2)

a. Product law

• Probabilities range from 0.0 to 1.0.

• The probability of two or more events occurring simultaneously is equal to the product of their individual probabilities.

• Two or more events are independent of one another if the outcome of each one does not affect the outcome of any of the others under consideration



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a. Product law

• To illustrate the product law, consider the possible outcome from tossing two coins ($1, $2 coin).

• There are four possible outcomes:• (PH:NH) = (1/2)(1/2) = ¼• (PT:NH) = (1/2)(1/2) = ¼• (PH:NT) = (1/2)(1/2) = ¼• (PT:NT) = (1/2)(1/2) = ¼

• The probability of obtaining a head or a tail for either coin is ½ and the outcome of one coin does not influence the outcome of the other coin.



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b. Sum law

• Suppose we want to calculate the probability of possible outcomes of two events that are independent of one another but can occur in more than one way, what do you do?

• What is the probability of tossing the two coins ($1, $2) and observe one head and one tail?

• Here, we do not care on whether the $1 coin comes out as head or the $2 coin comes out as head provided the other coin has an alternative outcome.


b. Sum law

• There are two ways in which such outcomes can be observed, each with a probability of ¼.

• The sum law states that the probability of obtaining any single outcome, where that outcome can be achieved by two or more events, is equal to the sum of the individual probabilities of all such events.


b. Sum law

• p = (¼) + (¼) = ½

• One-half of all two-coin tosses are predicted to give such outcome.

• These probability laws will be used to study transmission genetics and for solving genetic problems.


c. Binomial theorem

• Probability calculation using the binomial theorem can be used to evaluate cases where there are alternative ways to achieve a combination of events.

• Consider the following question, what is the probability that in a family with five children, two will be males and three will be females?

• This question is complex as each birth is an independent event and different birth orders can achieve the same outcome.


c. Binomial theorem

• The expression of the binomial theorem is

(a + b)n = 1

• a and b are the respective probabilities of the two alternative outcomes.

• n is the number of trials.


c. Binomial theorem


Adopted from Concepts of Genetics, Klug, Cummings, Spencer, Palladino, 2012

c. Binomial theorem• To expand on any binomial, the various exponents of

a and b are determined using the pattern

(a + b)n = an, an-1b, an-2b2, an-3b3, …, bn

• What is the expansion of (a + b)7?

• As the value of n increases and the expanded binomial becomes more complex

• Pascal’s triangle is a useful way to determine the numerical coefficient of each term in the expanded equation


c. Binomial theorem• What is the probability that in a family with five

children, two will be males and three will be females?

• Assign initial probabilities to each outcome:

a = male = ½

b = female = ½

n = 5

• Write out the expanded binomial for the value of n = 5,

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5


c. Binomial theorem• Each term represents a possible outcome, with

exponent of a representing the number of males and the exponent of b representing the number of females.

• Therefore, the term describing the outcome of two males and three females – the expression of the probability (p) is

p = 10a2b3

= 10(1/2)2(1/2)3

= 10(1/2)5

= 10/32 = 5/16


c. Binomial theorem• If you prefer not to use Pascal’s triangle, a formula

can be used to determine the numerical coefficient for any set of exponents

n! / (s!t!)

n = the total number of events

s = the number of times outcome a occurs

t = the number of times outcome b occurs

Therefore, n = s + t

• Note that in factorials, 0! = 1.


c. Binomial theorem• Find the probability in a family

with seven children, five will be males and two females

• In this case, n = 7, s = 5, and t = 2; five events having outcome a and two events having outcome b.

• Of families with seven children, on the average, 21/128 are predicted to have five males and two females


d. Chi-Square (χ2) Analysis• The outcomes of independent assortment and

fertilization, like coin tossing, are subject to random fluctuations from their predicted occurrences as a result of chance deviation.

• As the sample size increases, the average deviation from the expected results decreases.

• When we assume that data will fit a given ratio such as 1:1, 3:1, or 9:3:3:1, we establish what is called the null hypothesis (H0):

a. Be rejected or

b. Fail to be rejected.


d. Chi-Square (χ2) Analysis

• If it is rejected, the observed deviation from the expected result is judged not to be attributable to chance alone.

• In this case, the null hypothesis and the underlying assumptions leading to it must be reexamined.

• If the null hypothesis fails to be rejected, any observed deviations are attributed to chance.


d. Chi-Square (χ2) Analysis• The value for 2ᵪ is then used to estimate how

frequently the observed deviation can be expected to occur strictly as a result of chance.

o = Observed value for a given category

e = Expected value for that category

• (o - e) is the deviation (d)


d. Chi-Square (χ2) Analysis



How to interpret the χ2 value



How to interpret the χ2 value• 1st determine a value called the degrees of freedom

(df), which is equal to n – 1.

• n is the number of different categories into which the data are divided, in other words, the number of possible outcomes.

• For the 3:1 ratio, n = 2, so df = 1. For the 9:3:3:1 ratio, n = 4 and df = 3.

• Degrees of freedom must be taken into account because the greater the number of categories, the more deviation is expected as a result of chance.




χ2 values that lead to a p value of 0.05 or greater (darker blue areas) justify failure to reject the null hypothesis


What is a p value?

• It is simplest to think of the p value as a percentage.

• In the initial example, p = 0.26, which can be thought of as 26 percent.

• This indicates that if we repeat the same experiment many times, 26 percent of the trials would be expected to exhibit chance deviation as great as or greater than that seen in the initial trial.


What is a p value?

• Conversely, 74 percent of the repeats would show less deviation than initially observed as a result of chance.

• Does it prove (absolute) anything?

• Hence, a standard p value that is commonly applied is 0.05; reject or fail to reject the null hypothesis.


What is a p value?

• When it comes to chi-square analysis, a p value less than 0.05 means that the observed deviation in the set of results will be obtained by chance alone is less than 5% of the time.

• Such a p value shows that the difference between the observed and the expected results is substantial.

• Thus, the null hypothesis needs to be rejected.


What is a p value?

• p value of 0.26, the null hypothesis that independent assortment accounts for the results fails to be rejected. Therefore, the observed deviation can be reasonably attributed to chance.

• The null hypothesis is failed to be rejected (accepted), that is, where p value is more than 0.05


TEST YOUR KNOWLEDGE 1


1. What is the probability that in a family with four children, two will be males and two will be females?

2. If two parents, both heterozygous carriers of the autosomal recessive gene causing cystic fibrosis, have five children, what is the probability that exactly three will be normal?

3. In one dihybrid crosses, 317 round yellow, 110 round green, 101 wrinkled yellow and 32 wrinkled green F2 plants were observed. Analyze these data using the χ2 test to determine if they fit a 9:3:3:1 ratio?


4. In the laboratory, a genetics student crossed flies with normal long wings with flies expressing the dumpy mutation (truncated wings), which she believed was a recessive trait. In the F1 generation, all flies had long wings. The following results were obtained in the F2 generation:• 792 long-winged flies• 208 dumpy-winged flies• The student tested the hypothesis that the

dumpy wing is inherited as a recessive trait usingχ2 analysis of the F2 data. a. What ratio was hypothesized?b. Did the analysis support the hypothesis?


5. The following are F2 results of two of Mendel’s monohybrid crosses. For each cross, state a null hypothesis to be tested using analysis. Calculate theχ2 value and determine the p value fo both. Interpret the p values. Can the deviation in each case be attributed to chance or not? Which of the two crosses shows a greater amount of deviation?


6. In assessing data that fell into two phenotypic classes, a geneticist observed values of 250:150. She decided to perform a χ2 analysis by using the following two different null hypotheses: (a) the data fit a 3:1 ratio, and (b) the data fit a 1:1 ratio. Calculate the 2 values for each hypothesis. What can be concluded about each hypothesis?

7. The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the p value used to reject or not reject the hypothesis. In the case of the chi-square analysis of genetic crosses, would the use of a standard of p = 0.10 be more or less stringent about not rejecting the null hypothesis? Explain.


8. In a family of five children, what is the probability thata. all are males?b. three are males and two are females?c. two are males and three are females?

Assume that the probability of a male child is equal to the probability of a female child.

8. In a family of eight children, where both parents are heterozygous for albinism, what mathematical expression predicts the probability that six are normal and two are albinos? (Use formula)


10. To assess Mendel’s law of segregation using tomatoes, a true-breeding tall variety (SS) is crossed with a true-breeding short variety (ss). The heterozygous F1 tall plants (Ss) were crossed to produce two sets of F2 data, as follows.

a. Using the χ2 test, analyze the results for both datasets. Calculate values and estimate the p values in both cases.

b. From the above analysis, what can you conclude about the importance of generating large datasets in experimental conditions?


7.2 Pedigree analysis


Pedigree and the analysis

• Pedigrees is a chart that display family relationships and depict which relatives have specific phenotypes and genotypes.

• What it may contain in a Pedigrees?• Molecular data• Test results• Haplotypes (genes or SNPs linked in

segments on a chromosome)• Genome-wide association study

information



• A pedigree consists of lines that connect shapes

• They also assigned with numbers• Roman numerals designate generations.• Arabic numerals or names indicate

individuals.

• Pedigree is not the same as the:• Genealogy in family tree • Genogram in social work




Adopted from Human Genetics concepts and Application 9th ed.


• Colored or shaded shapes indicate individuals who express a trait, and half-filled shapes are known carriers.

• A genetic counselor may sketch a pedigree while interviewing a client, then use a computer program and add test results that indicate genotypes


Pedigree and the analysis• Lacks an enzyme that manufacture the

pigment melanin and result in hair and skin are very pale.

• Can you interpret the pedigree shown here?

• Homozygous recessive individuals in the third (F2) generation having Albinism.

• Their parents are inferred to be heterozygotes (carriers).

• Grandparents?


Albinism (autosomal recessive)Adopted from Human Genetics

concepts and Application 9th ed.



• An autosomal dominant trait does not skip generations and can affect both sexes.

Adopted from Human Genetics concepts and Application 9th ed.

TEST YOUR KNOWLEDGE 2


• Deshawn has sickle cell disease, which is autosomal recessive.

• His unaffected parents, Kizzy and Ike, must each be heterozygotes.

• Deshawn’s sister, Taneesha, also healthy, is expecting her first child. Taneesha’s husband, Antoine, has no family history of sickle cell disease.

• Taneesha wants to know the risk that her child will inherit the mutant allele from her and be a carrier.


1. First, what is the risk that she is a carrier?

2. Second, what is the risk her child will inherit the mutant allele?

3. Taneesha has a 2 in 3 chance of being a carrier.

4. If so, the chance that she will transmit the mutant allele is 1 in 2, because she has two copies of the gene, and only one allele goes into each gamete.

5. likelihood of the second event—the child being a carrier—depends upon the first event.

6. Total probability = 2/3 × 1/2 = 1/3


• In 1883, Alexander Graham Bell reported on records of deaf individuals in the United States, and hypothesized that advances such as sign language and schools for the deaf would lead to increased incidence of the condition resulting from an increasing number of marriages between deaf individuals.

• More recently, researchers reported that the number of marriages between two deaf individuals in which all children will be deaf has increased five-fold over the past century, and that the incidence of hereditary deafness has also increased.

• What is the most likely mode of inheritance that applies to families in which all children are born deaf?


• Draw a pedigree to depict the following family:

– One couple has a son and a daughter with normal skin pigmentation.

– Another couple has one son and two daughters with normal skin pigmentation.

– The daughter from the first couple has three children with the son of the second couple.

– Their son and one daughter have albinism; their other daughter has normal skin pigmentation.


• Chands syndrome (MIM 214350) is autosomal recessive and causes very curly hair, underdeveloped nails, and abnormally shaped eyelids.

• In the following pedigree, which individuals must be carriers?


7.3 Genetic Disorders


Gene disorders

http://www.nature.com/scitable/topicpage/rare-genetic-disorders-learning-about-genetic-disease-979


Gene disorders

• Online Mendelian Inheritance in Man

• Database of human genes and human genetic disorders that are inherited in a Mendelian manner

• http://www.omim.org/


Individual Assignment

• For the given disorders in the previous table, write a creative illustrated summary (one A4 page).

• Your summery on the disease should include:• Description• History• Inheritance • Clinical Features• Diagnosis


ch7 gene disorders and pedigree analysis

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