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Biology for AP® Courses

Science Practice Challenge Questions

Biology for AP® CoursesScience Practice Challenge Questions

44.

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.

A. Describe the relationship between regulation of expression and the differences among the A, B, A/B, and O types of blood cells.

B. Explain how the blood group phenotype does not display non-Mendelian inheritance and 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 bacteria that causes cholera, and 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.

C. 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.

D. The distribution of blood types was determined in a population. The results are displayed in the table.

Type Observed Frequency
A 501
B 794
A/B 236
O 601
Total, N 2132
Table 31.1

Recall that the frequency of an allele can be determined if the genotype is known. For example, for a gene with two possible alleles the frequency of the dominant allele in a homozygous dominant population is just twice the number of individuals 2N divided by the number of alleles which is 2N also since each individual has 2 alleles at each gene. The frequency would be 1. For a population composed entirely of heterozygous individuals the frequency would be 0.5.

  • Using the usual mathematical relationship and the three-allele system calculate the frequency of each type and add that value to the table.

    From these frequencies the probabilities of the A, B, and O alleles in the population were determined as shown in the table below.



    Allele probability
    A 0.201
    B 0.265
    O 0.542
    Table 31.2
  • Using these probabilities calculate the expected frequencies, E, of each blood type using

    E(A)=( p A 2 +2 p A +2 p A p O )N E(B)=( p B 2 +2 p B +2 p B p O )N E(A/B)=2· p a · p B ·N E(O)= p O 2 ·N E(A)=( p A 2 +2 p A +2 p A p O )N E(B)=( p B 2 +2 p B +2 p B p O )N E(A/B)=2· p a · p B ·N E(O)= p O 2 ·N


    Add these expected frequencies to the table.

    Type Observed, O Expected Frequency, E
    A 501
    B 794
    A/B 236
    O 601
    Table 31.3
  • Apply your understanding of the conceptual foundation of these equations by restating in words the relationship represented by E(A).
  • Apply a χ 2 χ 2 test at the 95% 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

    χ 2 = (OE) 2 E χ 2 = (OE) 2 E


    and this table are provided on the AP Biology Exam.
    Degrees of Freedom
    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
    Table 31.4
  • 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 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.

E. 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.

45.

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.

This line graph has a title of Freedom from rejection. The vertical line reads Proportion of patients (%) and it has tick marks at 0, 20, 40, 60, 80, and 100. The horizontal axis is labeled Years since transplantation. It has tick marks at  0, 1, 2, 3, 4, 5, 6, 7, and 8. There is a red and a blue line that marks the graph. The first is a blue line and it is labeled Incompatible ABO the line begins at 100, and drops to 75 around the 1 year mark and stays mostly straight until the right year marker. The second line which is red is labeled compatible ABO. It begins at the 100 and drops to around the 55 around year one, and continues on straight around 50 until the 8 year mark.
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.

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