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A term female infant receives phototherapy for hyperbilirubinemia 48 hours after
birth. A blood specimen is sent to the laboratory to determine her blood type. The
mother's blood type is A, and the husband states that his blood type is also A. The
prevalence of blood type A in the population is 0.40 and of type AB is 0.04.
Of the following, the probability that the infant's blood type will be A is CLOSEST
to:
0.75
0.80
0.85
0.90
0.95
You selected
, the correct answer is
.
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The ABO blood type is determined by a single dominant allele on chromosome 9.
In the vignette, each parent could be homozygotic or heterozygotic for the A blood
type. Using the Hardy-Weinberg principle, the computations below will show that
the child has a 0.82 probability of having blood type A. Correcting for the small
chance of nonpaternity will decrease that probability to 0.81.
According to the Hardy-Weinberg principle, at population equilibrium, the
prevalence of an allele in one generation will determine the prevalence in the next
generation. Specifically, if blood type A is the dominant allele and a is the
recessive, and P and q are their respective gene frequencies in the population,
then:
(P+q) x (P+q) = P 2 + 2Pq + q 2 = 1,
where P 2 becomes the prevalence of people in the next generation with genotype
AA,2Pq becomes the prevalence of genotype Aa, and q 2 becomes the prevalence of
genotype aa. The sum of all the probabilities is 1. People with genotypes Aa and
AA would show the phenotype of blood type A. Some of the people in the Aa
genotype group actually would have AO genotypes, but for the purposes of the
calculations they can be combined in the Aa group without loss of accuracy. The a
allele as discussed here includes alleles coding for the B and O phenotypes. People
with genotype aa would show a blood type other than A, such as O or B. Table 1 is
the crossing table.
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Education Module Learner
Table 1.
P
P
P2
q
Pq
q
Pq
q2
From the vignette, the portion of the population showing blood type A or AB is
0.40 + 0.04 = 0.44.
In the table above, this corresponds to the sum of P 2, Pq, and Pq.
The portion not showing blood type A or AB is 1 - 0.44 = 0.56, corresponding to q 2
in the table. Taking the square root gives q = 0.75. The gene frequency of the
blood type A allele is then p = 1 - 0.75 = 0.25. Now we can fill in the rest of the
population crossing table, Table 2.
Table 2.
A allele = 0.25
A allele = 0.25
AA genotype: P 2 = 0.06
Aa genotype: Pq = 0.19
Aa genotype: Pq = 0.19
aa genotype: q 2 = 0.56
a allele = 0.75
a allele = 0.75
The portion of the population having the homozygous blood type AA genotype is
0.06, and the portion having the heterozygous blood type Aa is 0.38. The portion
of all those showing the phenotypic blood type A who are genetic homozygotes
(AA) is
0.06 / (0.06 + 0.38) = 0.14.
The portion of blood type A persons who are heterozygous is 1 - 0.14 = 0.86.
In the vignette, the chances of the mother passing an A allele to her child is the
probability of her being an AA homozygote, plus half the chance of her being an Aa
heterozygote, or
0.14 + (0.86/2) = 0.57.
The chance of her passing an a allele is 1 - 0.57 = 0.43. The same numbers can be
derived for the husband. A crossing table is now made for the child, Table 3.
Table 3.
Husband's A allele = 0.57
Husband's a allele = 0.43
Mom's A allele = 0.57
AA genotype: 0.32
Aa genotype: 0.25
Mom's a allele = 0.43
Aa genotype: 0.25
aa genotype: 0.18
The probability of the child having a phenotypic blood type A is the sum of the
probabilities of the AA and Aa genotypes, or
0.32 + (2 x 0.25) = 0.82.
http://emb.aap.org/courseprodv2/Index.asp[4/4/2012 3:29:28 PM]
Education Module Learner
The chance of nonpaternity in large studies is approximately 0.04. If the husband
is not the father, the chance of the child receiving an A allele from the undeclared
father is that of the general population, 0.25. Using this new datum, Table 3, the
child's crossing table, can be rewritten as Table 4.
Father's A allele = 0.25
Father's a allele =
Table 4.
0.75
Mom's A allele =
0.57
Mom's a allele = 0.43
AA genotype: 0.14
Aa genotype: 0.43
Aa genotype: 0.11
aa genotype: 0.32
In the case of nonpaternity, the probability of the child having a phenotypic blood
type A is again the sum of the probabilities of the AA and Aa genotypes, or
0.14 + 0.11 + 0.43 = 0.69.
The final probability of the child expressing blood type A is the sum of the
weighted probabilities of the true paternity and the nonpaternity cases.
(0.82 x 0.96) + (0.69 x 0.04) = 0.81.
Interestingly, a rare condition allows parents of any blood type to produce
offspring of blood type O. The H antigen is a precursor of the A and B blood-type
glycoproteins. The H allele assorts independently from the A and B alleles, and has
a rare h allele that in the homozygous state hh prevents production of A or B
antigens. This homozygous (hh) condition is called the Bombay phenotype and is
expressed as blood type O. This is a rare event, estimated as 1 in 10,000 among
Bombay Indians and 1 in 1 million among Europeans. If both parents are
heterozygous (Hh) they have a 1-in-4 chance for each pregnancy to produce a
homozygous (hh) child showing blood type O, even if the parents are each blood
type AB or homozygous for AA or BB. The low frequency of the Bombay phenotype
allows it to be ignored in the calculations in the vignette.
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References:
American Red Cross. Blood types. Available
at: http://www.pleasegiveblood.org/education/blood_types.php. Accessed February
3, 2006
Blood Type Facts. Bloodbook.com Web site. Available at: http://www.bloodbook.com/typefacts.html. Accessed February 3, 2006
Keats BJ, Sherman SL. Population genetics. In: Rimoin DL, Connor JM, Pyeritz RE,
Korf BR. Emery and Rimoin's Principles and Practice of Medical Genetics. 4 th ed.
London, England: Churchill Livingston; 2002:425-438
Watkins-Chow W, McKusick VA. Fucosyltransferase 1; FUT1 Hh Bombay
phenotype. Available at the Online Mendelian Inheritance in Man (OMIM) Web site:
http://www.ncbi.nlm.nih.gov/entrez/dispomim.cgi?id=211100. Accessed
September 4, 2005
http://emb.aap.org/courseprodv2/Index.asp[4/4/2012 3:29:28 PM]
Education Module Learner
Content specification(s):
Understand the inheritance pattern of the major blood groups
Be able to calculate the gene frequency of a disease inherited on a single gene by
knowing the population incidence of that disease
http://emb.aap.org/courseprodv2/Index.asp[4/4/2012 3:29:28 PM]