BSCI Genetics (Ch. 14) Practice Problems

A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Remember that extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits?

1/2

Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to see if they have the sick-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease

1/9

In tigers, a recessive allele of a particular gene causes both an absence of fur pigmentation (a white tiger) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage of cross-eyed tigers will be white?

25%, or ¼, will be cross-eyed; all (100%) of cross-eyed offspring will also be white.

The pedigree on the right traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the colored circles and squares, area unable to metabolize a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant allele or by a recessive allele? Fill in the genotypes of the individuals whose genotypes can be deduced. What genotypes are possible for each of the other individuals?

Recessive. All affected individuals (Arlene, Tom, Wilma, and Carla) are homozygous recessive aa. George is Aa, since some of his children with Arlene are affected. Sam, Ann, Daniel, and Alan are each Aa, since they are all unaffected children with one affected parent. Michael is also Aa, since he has an affected child (Carla) with his heterozygous wife Ann. Sandra, Tina, and Christopher can each have either Aa or AA genotype.

In maize (corn) plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant allele P causes purple kernel color, while the recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the offspring?

The dominant allele I is epistatic to the P/p locus, and thus the genotype ratio for the F1 generation will be 9 I_P_ (colorless): 3 I_pp (colorless): 3 iiP_ (purple): 1 iipp (red). Overall, the phenotypic ratio is 12 colorless: 3 purple: 1 red.

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.) a) AABBCC x aabbcc ---> AaBbCc b) AABbCc x AaBbCc ---> AAbbCC c) AaBbCc x AaBbCc ---> AaBbCc d) aaBbCC x AABbcc ---> AaBbCc

a.) 1 b.) 1/32 c.) 1/8 d.) 1/2

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a) aabbccdd b) AaBbCcDd c) AABBCCDD d) AaBBccDd e) AaBBCCdd

a.) 1/256 b.) 1/16 c.) 1/256 d.) 1/64 e.) 1/128

Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as indicated in Table 14.1. If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.) a) Homozygous for the three dominant traits b) Homozygous for the three recessive traits c) Heterozygous for all three characters d) Homozygous for axial and tall, heterozygous for seed shape

a.) 1/64 b.) 1/64 c.) 1/8 d.) 1/32

Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? a) All three children are of normal phenotype b) One or more of the three children have the disease c) All three children have the disease d) At least one child is phenotypically normal

a.) 3/4 x 3/4 x 3/4 = 27/64 b.) 1 - 27/64 = 37/64 c.) 1/4 x 1/4 x 1/4 = 1/64 d.) 1 - 1/64 = 63/64