# Science Practice Challenge Questions

### 12.2Characteristics and Traits

39.

The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et al, JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99% of humans with European origins. Both the wild type and mutant display codominance.

1. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype.
F1 m m
+
+
Table 12.6
F2

Table 12.7
2. Draw sister chromatids at anaphase II for both parents in the F1 generation and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square.
3. Explain which of Mendel's laws is violated by codominance.
4. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant:
F2
Phenotype Observed Expected
H 1206
I 2238
L 1124
Table 12.8

Complete the table. Explain the values expected in terms of the genotype of the offspring.
5. Using a c2 statistic at the 95% confidence level, evaluate the claim that the wild-type and mutant forms of SCLO24A5 are codominant. The definition of the statistic $Χc2=∑(Oi−Ei)2EiΧc2=∑(Oi−Ei)2Ei$ where X is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The Chi-square statistic table is provided in 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
Table 12.9
40.

Adrenoleukodystrophy (ALD) is a genetic disorder in which lipids with very high molecular weights are not metabolized and accumulate within cells. Accumulation of these fats in the brain damages the myelin that surrounds nerves. This progressive disease has two causes: an autosomal recessive allele, which causes neonatal ALD, and a mutation in the ABCD1 gene located on the X-chromosome. A controversial treatment is the use of Lorenzo’s oil, which is expensive; despite this treatment, neurological degradation persists in many patients. Gene therapy as a potential treatment is currently in trials but is also very costly.

An infant patient exhibits symptoms of neonatal ALD, which are difficult to distinguish from the X-linked form of the disease. The infant’s physician consults electronic health records to construct a pedigree showing family members who also presented symptoms similar to ALD. The pedigree is shown in this diagram. The infant patient is circled. Symbols for males (o) and females (m) are filled when symptoms are present.

Figure 12.26
1. Using the pedigree, explain which form of ALD (neonatal or X-linked) is present in the infant.
2. Sharing of digital records among health providers is one method proposed to improve the quality and reduce the cost of health care in the U.S. The privacy of electronic health records is a concern. Pose three questions that must be addressed in developing policies that balance the costs of treatments and diagnoses, patient quality of life, and risks to individual privacy.
41.

Two genes, A and B, are located adjacent to each other (linked) on the same chromosome. In the original cross (P0), one parent is homozygous dominant for both traits (AB), whereas the other parent is recessive (ab).

Characteristic Alleles Chromosome
Seed color yellow (I) / green (i) 1
Seed coat & flowers colored (A) / white (a) 1
Mature pods smooth (V) / wrinkled (v) 4
Flower stalk from leaf axils (Fa) / umbellate at top of plant (fa) 4
Height > 1 m (Le) / ~0.5 m (le) 4
Unripe pods green (Gp) / yellow (gp) 5
Mature seeds smooth (R) / wrinkled (r) 7
Table 12.10
1. Describe the distribution of genotypes and phenotypes in F1.
2. Describe the distribution of genotypes and phenotypes when F1 is crossed with the ab parent.
3. Describe the distribution of genotypes and phenotypes when F1 is crossed with the AB parent.
4. Explain the observed non-Mendelian results in terms of the violation of the laws governing Mendelian genetics.

### 12.3Laws of Inheritance

42.

Gregor Mendel’s 1865 paper described experiments on the inheritance of seven characteristics of Pisum sativum shown in the first column in the table below. Many years later, based on his reported outcomes and analysis of the inheritance of a single characteristic, Mendel developed the concepts of genes, their alleles, and dominance. These concepts are defined in the second column of the table using conventional symbols for the dominant allele for each characteristic. Even later, the location of each of these genes on one of the seven chromosomes in P. sativum were determined, as shown in the third column.

A. Before the acceptance of what Mendel called “factors” as the discrete units of inheritance, the accepted model was that the traits of progeny were “blended” traits of the parents. Evaluate the evidence provided by Mendel’s experiments in disproving the blending theory of inheritance.

B. Mendel published experimental data and analysis for two experiments involving the inheritance of more than a single characteristic. He examined two-character inheritance of seed shape and seed color. He also reported three-character inheritance of seed shape, seed color, and flower color. Evaluate the evidence provided by the multiple-character experiments. Identify which of the following laws of inheritance depend upon these multiple-character experiments for support:

1. During gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene.
2. Genes for different traits can segregate independently during the formation of gametes.
3. Some alleles are dominant; whereas others are recessive. An organism with at least one dominant allele will display the effect of the dominant allele.
4. All three laws can be inferred from the single-character experiments.

C. As shown in the table above, some chromosomes contain the gene for more than one of the seven characteristics Mendel studied, for example, seed color and flowers. The table below shows, with filled cells above the dashed diagonal line, the combinations of characteristics for which Mendel reported results. In the cells below the dotted diagonal line, identify with an X each cell where deviations from the law or laws identified in part B might be expected.

Figure 12.27

D. Explain the reasons for the expected deviations for those combinations of characteristics identified in part C.

E. In one of the experiments reported by Mendel, deviations from the law identified in part B might be expected. Explain how the outcomes of this experiment were consistent with Mendel’s laws.

43.

A dihybrid cross involves two traits. A cross of parental types AaBb and AaBb can be represented with a Punnett square:

Figure 12.28

This representation clearly organizes all of the possible genotypes and reveals the 9:3:3:1 distribution of phenotypes and a 4×4 grid of 16 cells. Expressed as a fraction of the 16 possible genotypes of the offspring, the phenotypic ratio describes the probability of each phenotype among the offspring:

3 (AA, Aa, aA) × 3 (BB, bB, Bb)/16 = 9/16
3 (AA, Aa, aA) × 1 (bb) /16 = 3/16
1 (aa) × 3 (BB, bB, Bb) = 3/16
1 (aa) × 1 (bb) = 1/16

A. Using the probability method, calculate the likelihood of these phenotypes from each dihybrid cross:

• recessive in the gene with alleles A and a from the cross AaBb × aabb
• dominant in both genes from the cross AaBb × aabb
• recessive in both genes from the cross AaBb × aabb
• recessive in either gene from the cross AaBb × aabb

A Punnett square representation of a trihybrid cross, such as the self-cross of AaBbCc, is more cumbersome because there are eight columns and rows (2×2×2 ways to choose parental genotypes) and 64 cells. A less tedious representation is to calculate the number of each type of genotype in the offspring directly by counting the unique permutations of the letters representing the alleles. For example, the probability of the cross AaBbCc × AaBbCc is 3 (AA, Aa, aA) × 3 (BB, Bb, bB) × 3 (CC, Cc, cC)/64 = 27/64.

B. Using the probability method, calculate the likelihood of these phenotypes from each trihybrid cross:

• recessive in all traits from the cross AaBbCc × aabbcc
• recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc
• dominant in the gene with alleles A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc

C. The probability method is an easy way to calculate the likelihood of each particular phenotype, but it doesn’t simultaneously display the probability of all possible phenotypes. The forked line representation described in the text allows the entire phenotypic distribution to be displayed. Using the forked line method, calculate the probabilities in a cross between AABBCc and Aabbcc parents:

• all traits are recessive: aabbcc
• traits are dominant at each loci, A?B?C?
• traits are dominant at two genes and recessive at the third
• traits are dominant at one gene and recessive at the other two
44.

Construct a representation showing the connection between the process of meiosis and the transmission six possible phenotypes from parents to F2 offspring. The phenotypes are labeled A, a, B, b and C, c. Expression of each phenotype is controlled by a separate Mendelian gene. Your representation should show the proportion of every possible combination of phenotypes (e.g., ABC, AbC, etc.) that will be present in the F2 offspring.