This document outlines the principles of genetic probability, pedigree analysis, and chi-square methodologies for geneticists. It covers laws of probability such as product and sum laws, the binomial theorem for evaluating genetic combinations, and techniques for analyzing pedigrees to understand genetic inheritance. Additionally, it discusses how to interpret chi-square values and p-values to test genetic hypotheses and the significance of observed deviations from expected ratios.

1. 7.0 Gene disorders and pedigree
analysis
Prepared by Pratheep Sandrasaigaran
Lecturer at Manipal International University

2. 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

3. 7.1 laws of probability in genetics
Prepared by Pratheep Sandrasaigaran

4. laws of probability in genetics
Prepared by Pratheep Sandrasaigaran
Product law
Sum law
Binomial theorem
Chi-Square (χ2
)

5. 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
Prepared by Pratheep Sandrasaigaran
Diagram adopted from Internet Sources

6. 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.
Prepared by Pratheep Sandrasaigaran
Diagram adopted from Internet
Sources

7. 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.
Prepared by Pratheep Sandrasaigaran

8. 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.
Prepared by Pratheep Sandrasaigaran

9. 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.
Prepared by Pratheep Sandrasaigaran

10. 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.
Prepared by Pratheep Sandrasaigaran

11. 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.
Prepared by Pratheep Sandrasaigaran

12. c. Binomial theorem
Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics, Klug, Cummings, Spencer, Palladino, 2012

13. 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-1
b, an-2
b2
, an-3
b3
, …, 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
Prepared by Pratheep Sandrasaigaran

14. Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics, Klug, Cummings, Spencer, Palladino, 2012

15. 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
+ 5a4
b + 10a3
b2
+ 10a2
b3
+ 5ab4
+ b5
Prepared by Pratheep Sandrasaigaran

16. 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 = 10a2
b3
= 10(1/2)2
(1/2)3
= 10(1/2)5
= 10/32
= 5/16
Prepared by Pratheep Sandrasaigaran

17. 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.
Prepared by Pratheep Sandrasaigaran

18. 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
Prepared by Pratheep Sandrasaigaran

19. 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.
Prepared by Pratheep Sandrasaigaran

20. 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.
Prepared by Pratheep Sandrasaigaran

21. 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)
Prepared by Pratheep Sandrasaigaran

22. d. Chi-Square (χ2
) Analysis
Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics, Klug, Cummings, Spencer, Palladino, 2012

23. How to interpret the χ2
value
Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics, Klug, Cummings, Spencer,
Palladino, 2012

24. 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.
Prepared by Pratheep Sandrasaigaran

25. How to interpret the χ2
value
Prepared by Pratheep Sandrasaigaran
χ2 values that lead to a p value of 0.05 or
greater (darker blue areas) justify failure to
reject the null hypothesis
Adopted from Concepts of Genetics, Klug, Cummings,
Spencer, Palladino, 2012

26. How to interpret the χ2
value
Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics, Klug, Cummings, Spencer, Palladino, 2012

27. 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.
Prepared by Pratheep Sandrasaigaran

28. 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.
Prepared by Pratheep Sandrasaigaran

29. 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.
Prepared by Pratheep Sandrasaigaran

30. 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
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31. TEST YOUR KNOWLEDGE 1
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32. 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?
Prepared by Pratheep Sandrasaigaran

33. 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?
Prepared by Pratheep Sandrasaigaran

34. 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?
Prepared by Pratheep Sandrasaigaran

35. 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.
Prepared by Pratheep Sandrasaigaran

36. 8. In a family of five children, what is the probability
that
a. 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)
Prepared by Pratheep Sandrasaigaran

37. 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?
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38. 7.2 Pedigree analysis
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39. 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
Prepared by Pratheep Sandrasaigaran

40. Pedigree and the analysis
• 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
Prepared by Pratheep Sandrasaigaran

41. Pedigree and the analysis
Prepared by Pratheep Sandrasaigaran
Adopted from Human Genetics
concepts and Application 9th ed.

42. Pedigree and the analysis
Prepared by Pratheep Sandrasaigaran
Adopted from Human Genetics
concepts and Application 9th ed.

43. Pedigree and the analysis
• 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
Prepared by Pratheep Sandrasaigaran

44. 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?
Prepared by Pratheep Sandrasaigaran
Albinism (autosomal recessive)
Adopted from Human Genetics
concepts and Application 9th ed.

45. Pedigree and the analysis
Prepared by Pratheep Sandrasaigaran
• An autosomal dominant trait does not skip generations and can affect both
sexes.
Adopted from Human Genetics concepts and Application 9th ed.

46. Prepared by Pratheep Sandrasaigaran
Adopted from Concepts of Genetics,
Klug, Cummings, Spencer, Palladino,
2012

47. TEST YOUR KNOWLEDGE 2
Prepared by Pratheep Sandrasaigaran

48. • 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.
Prepared by Pratheep Sandrasaigaran

49. 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
Prepared by Pratheep Sandrasaigaran

50. • 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?
Prepared by Pratheep Sandrasaigaran

51. • 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.
Prepared by Pratheep Sandrasaigaran

52. • 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?
Prepared by Pratheep Sandrasaigaran

53. 7.3 Genetic Disorders
Prepared by Pratheep Sandrasaigaran

54. Gene disorders
http://www.nature.com/scitable/topicpage/rare-genetic-disorders-learning-about-genetic-disease-979
Prepared by Pratheep Sandrasaigaran

55. Gene disorders
http://www.nature.com/scitable/topicpage/rare-genetic-disorders-learning-about-genetic-disease-979
Prepared by Pratheep Sandrasaigaran

56. 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/
Prepared by Pratheep Sandrasaigaran

57. 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
Prepared by Pratheep Sandrasaigaran