Chapter 14: Test Your Understanding
25% will be cross-eyed; all of the cross-eyed offspring will also be white.
In tigers, a recessive allele causes 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 will be white?
1⁄2 of their children could have extra digits.
A man has six fingers on each hang 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?
The man has Ao blood type and the woman had Bo blood type. The frequency of other genotypes is Oo 25% of the time, AB 25% of the time, Ao 25% of the time and Bo 25% of the time.
A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these three individuals? What genotypes, and in what frequencies, would you expect in future offspring from this marriage?
A) 1⁄64, B) 1⁄64, C) 1⁄8, D) 1⁄32
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? (a) homozygous for the three dominant traits (b) homozygous for three recessive traits (c) heterozygous for all three characters (d) homozygous for axial and tall, heterozygous for seed shape.
1/16
Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis.)
Matings of the original mutant cat with true-breeding noncurl cats will produce both curl and noncurl F1 offspring if the curl allele is dominant, but only noncurl offspring if the curl allele is recessive. You would obtain some true-breeding offspring homozygous for the curl allele from matings between the F1 cats resulting from the original curl × noncurl crosses whether the curl trait is dominant or recessive. You know that cats are true-breeding when curl × curl matings produce only curl offspring. As it turns out, the allele that causes curled ears is dominant.
In 1981, a stray black cat with unusual rounded, curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the curl cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you obtain true-breeding curl cats? How could you be sure they are true-breeding?
The dominant allele I is epistatic to the P/p locus, and thus the genotypic 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.
In 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 homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the offspring?
1/9
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 reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.
A) 3⁄4 × 3⁄4 × 3⁄4 = 27⁄64, B) 1 — 27⁄64 = 37⁄64, C) 1⁄4 × 1⁄4 × 1⁄4 = 1⁄64, D) 1 — 1⁄64=63 ⁄64
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) 1⁄256, B) 1⁄16, C) 1⁄256, D) 1⁄64, E) 1/128
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
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 also is Aa, since he has an affected child (Carla) with his heterozygous wife Ann. Sandra, Tina, and Christopher can each have the AA or Aa genotype.
The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the colored circles and squares, are 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?
A) AABBCC × aabbcc AaBbCc B) AABbCc × AaBbCc AAbbCC C) AaBbCc × AaBbCc AaBbCc D) aaBbCC × AABbcc AaBbCc Answer: A) 1, B) 1⁄32, C) 1⁄8, D) 1⁄2
What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.)
