Like Gram-positive bacteria, red blood cells have carbohydrates on the cell surface. The A, B, A/B, and O blood types are designations of the phenotypes expressed by the alleles that code for these cell surface carbohydrates.

1. Describe the relationship between regulation of expression and the differences among the A, B, A/B, and O types of blood cells.
2. Explain how the blood group phenotype does not display non-Mendelian inheritance, and describe the simplest alternative model that explains this deviation.
   
    
   Immune system T-cells recognize cell surface carbohydrates of bacterial and red blood cells. That these similarities have consequences for survival is indicated by the observation that individuals with blood type O are more susceptible to infection by Vibrio, the Gram-negative bacterium that causes cholera; in contrast, individuals with type A/B are more susceptible to infections from a broad range of E. coli variants, all of which are also Gram-negative.
3. Describe the likely reason for the increased susceptibility of individuals with type A/B blood in terms of the antigen-antibody model of specific immune response. Justify the selection of data that would allow a test of your reasoning.
4. The distribution of blood types is determined in a population. The results are displayed in the table. Fill in the frequency of each genotype. Don’t forget that the total should add up to 1.00 (100 percent).
    

   Type	Observed	Frequency
   A	501
   B	794
   A/B	236
   O	601
   Total, N	2,042

    
   Recall that the frequency of an allele can be determined if the genotype is known. If we take an example of a gene with two possible alleles, of which one is dominant and one is recessive, we need to calculate the number of each allele and divide by the total number of alleles (not genotypes). For example, let’s say that you have 50 homozygous dominant individuals and 50 homozygous recessive individuals. That means that you have 100 dominant alleles (two per homozygous dominant individual) and 100 recessive alleles (two per homozygous recessive individuals) and a total of 200 alleles. The frequency of the dominant allele is 100/200 = 0.5.
   •  Using the observed numbers of each genotype in the table, calculate the frequency of each allele as described above. Don’t forget that the total should add up to 1.00 (100 percent of the alleles).
      

     Genotype	Observed	Frequency
     IAIA	320
     IAio	181
     IBIB	349
     IBio	445
     IAIB	236
     ii	601
     Total, N	2,132

      

     Allele	Probability
     A
     B
     O

   • Using these probabilities, calculate the expected frequencies (E) of each blood type using the Hardy-Weinberg equation for three alleles given below.
     
     p = frequency of I^A
     q = frequency of I^B
     r = frequency of i
     p² + 2pq + q² + 2pr + 2qr + r²
     
      

     Genotype	Symbol	Frequency/Total individuals
     IAIA	p2	0.061/131
     IAIB	2pq	0.160/342
     IBIB	q2	0.104/222
     IAi	2pr	0.213/454
     IBi	2qr	0.277/591
     ii	r2	0.184/392

     Use the information in the table above to calculate expected numbers to add to this table.
      

     Type	Observed, O	Expected Frequency, E
     A	501
     B	794
     A/B	236
     O	601

   • Apply your understanding of the conceptual foundation of these equations by restating in words the relationship represented by E(A).
   • Apply a χ² test at the 95 percent confidence level and 3 degrees of freedom (number of traits minus one) to evaluate the claim that these data indicate Hardy-Weinberg equilibrium of the ABO system for this population. The definition of the statistic is provided on the AP Biology Exam.
     
      
     
      
     ${\chi }^2=\Sigma \frac{(O–E{)}^2}{E}$
     
      
     This table is also provided on the AP Biology Exam.
      

     p	1	2	3	4	5	6	7	8
     0.05	3.84	5.99	7.82	9.49	11.07	12.59	14.07	15.51
     0.01	6.64	9.32	11.34	13.28	15.09	16.81	18.48	20.09

     Degrees of Freedom
   • Hardy-Weinberg equilibrium is consistent with the assumption of no change in the distribution of alleles over time. Justify the selection of data that should be obtained to further test this assumption. Homologous genes coding for the carbohydrates that are present on the surfaces of red blood cells are found in amphibians and mammals but not in fish, implying a last common ancestor for the ABO gene system at least 20 million years ago.
     
      
     Homologous genes coding for the carbohydrates that are presented on the surfaces of red blood cells are found in amphibians and mammals but not in fish, implying a last common ancestor for the ABO gene system at least 20 million years ago.
5. The difference between genes coding for A and B is a single nucleotide replacement. Evaluate the likelihood that negative selection pressures have been active in the evolution of this system.

The immune system rejection of transplanted organs and the availability of organ donors are key factors in determining survival. Blood ABO compatibility is always a criterion in matching donors and recipients in adult patients and was once a consideration for infant patients. Period of time on the waitlist for a suitable donor is critical because the health of the patient degrades while waiting. Dipchand et al (American Journal of Transplantation, 10, 2010) made a comparison of survival rates for infants where the donor heart was ABO compatible and incompatible as shown in the graph.

Figure 31.21

Based on these data, justify the claim that expression of blood group immune response develops over time and that this provides a window of opportunity for transplantation.