Biology 111G Chapter 11 concept checks and end of chapter review

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 × aabbcc → AaBbCc b)AABbCc × AaBbCc → AAbbCC c)AaBbCc × AaBbCc → AaBbCc d)aaBbCC × AABbcc → AaBbCc

(a) 1; (b) ¹∕₃₂; (c) ⅛; (d) ½

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

Hemochromatosis 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) ¾ × ¾ × ¾ = ²⁷∕₆₄; (b) 1 - ²⁷∕₆₄ = ³⁷∕₆₄ (c) ¼ × ¼ × ¼ = ¹∕₆₄; (d) 1 - ¹∕₆₄ = ⁶³∕₆₄

In tigers, a recessive allele that is pleiotropic 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 of cross-eyed tigers will be white?

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

A pea plant heterozygous for inflated pods (Ii) is crossed with a plant homozygous for constricted pods (ii). Draw a Punnett square for this cross to predict genotypic and phenotypic ratios. Assume that pollen comes from the ii plant.

A cross of Ii × ii would yield offspring with a genotypic ratio of 1 Ii : 1 ii (2:2 is an equivalent answer) and a phenotypic ratio of 1 inflated : 1 constricted (2:2 is equivalent).

DRAW IT Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf?

According to the law of independent assortment, 25 plants (¹∕₁₆ of the offspring) are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.

When Mendel did crosses of true-breeding purple- and white-flowered pea plants, the white-flowered trait disappeared from the F1 generation but reappeared in the F2 generation. Use genetic terms to explain why that happened.

Alternative versions of genes, called alleles, are passed from parent to offspring during sexual reproduction. In a cross between purple- and white-flowered homozygous parents, the F1 offspring are all heterozygous, each inheriting a purple allele from one parent and a white allele from the other. Because the purple allele is dominant, it determines the phenotype of the F1 offspring to be purple, and the expression of the recessive white allele is masked. Only in the F2 generation is it possible for a white allele to exist in the homozygous state, which causes the white trait to be expressed.

Concept 11.4: Many human traits follow Mendelian patterns of inheritance

Analysis of family pedigrees can be used to deduce the possible genotypes of individuals and make predictions about future offspring. Such predictions are statistical probabilities rather than certainties. Many genetic disorders are inherited as simple recessive traits. Most affected (homozygous recessive) individuals are children of phenotypically normal, heterozygous carriers. The sickle-cell allele has probably persisted for evolutionary reasons: Heterozygotes have an advantage because one copy of the sickle-cell allele reduces both the frequency and severity of malaria attacks. Lethal dominant alleles are eliminated from the population if affected people die before reproducing. Nonlethal dominant alleles and lethal alleles that are expressed relatively late in life are inherited in a Mendelian way. Many human diseases are multifactorial—that is, they have both genetic and environmental components and do not follow simple Mendelian inheritance patterns. Using family histories, genetic counselors help couples determine the probability of their children having genetic disorders.

Tips for Genetics Problems 3

Determine what the problem is asking. If asked to do a cross, write it out in the form [genotype] × [genotype], using the alleles you've decided on.

Concept 11.3: Inheritance patterns are often more complex than predicted by simple Mendelian genetics

Extensions of Mendelian genetics for a single gene: Extensions of Mendelian genetics for two or more genes: The expression of a genotype can be affected by environmental influences. Polygenic characters that are also influenced by the environment are called multifactorial characters. An organism's overall phenotype reflects its overall genotype and unique environmental history. Even in more complex inheritance patterns, Mendel's fundamental laws still apply.

Tips for Genetics Problems 7

For pedigree problems, use the tips in Figure 11.14 and below to determine what kind of trait is involved. a)If parents without the trait have offspring with the trait, the trait must be recessive and both of the parents must be carriers. b)If the trait is seen in every generation, it is most likely dominant (see the next possibility, though). C)If both parents have the trait, then in order for it to be recessive, all offspring must show the trait. D)To determine the likely genotype of a certain individual in a pedigree, first label the genotypes of all the family members you can. Even if some of the genotypes are incomplete, label what you do know. For example, if an individual has the dominant phenotype, the genotype must be AA or Aa; you can write this as A-. Try different possibilities to see which fits the results. Use the rules of probability to calculate the probability of each possible genotype being the correct one.

Concept 11.1: Mendel used the scientific approach to identify two laws of inheritance

Gregor Mendel formulated a theory of inheritance based on experiments with garden peas, proposing that parents pass on to their offspring discrete genes that retain their identity through generations. This theory includes two "laws." The law of segregation states that genes have alternative forms, or alleles. In a diploid organism, the two alleles of a gene segregate (separate) during meiosis and gamete formation; each sperm or egg carries only one allele of each pair. This law explains the 3:1 ratio of F2 phenotypes observed when monohybrids self-pollinate. Each organism inherits one allele for each gene from each parent. In heterozygotes, the two alleles are different: expression of the dominant allele masks the phenotypic effect of the recessive allele. Homozygotes have identical alleles of a given gene and are true-breeding. The law of independent assortment states that the pair of alleles for a given gene segregates into gametes independently of the pair of alleles for any other gene. In a cross between dihybrids (individuals heterozygous for two genes), the offspring have four phenotypes in a 9:3:3:1 ratio.

If a man with type AB blood marries a woman with type O, what blood types would you expect in their children? What fraction would you expect of each type?

Half of the children would be expected to have type A blood and half type B blood.

Tips for Genetics Problems 6

If the problem gives you the phenotypic ratios of offspring but not the genotypes of the parents in a given cross, the phenotypes can help you deduce the parents' unknown genotypes. a)for example, if ½ of the offspring have the recessive phenotype and ½ the dominant, you know that the cross was between a heterozygote and a homozygous recessive. b)If the ratio is 3:1, the cross was between two heterozygotes. c)If two genes are involved and you see a 9:3:3:1 ratio in the offspring, you know that each parent is heterozygous for both genes. Caution: Don't assume that the reported numbers will exactly equal the predicted ratios. For example, if there are 13 offspring with the dominant trait and 11 with the recessive, assume that the ratio is one dominant to one recessive.

MAKE CONNECTIONS In Table 11.1, note the phenotypic ratio of the dominant to recessive trait in the F2 generation for the monohybrid cross involving flower color. Then determine the phenotypic ratio for the offspring of the second-generation couple in Figure 11.14b. What accounts for the difference in the two ratios?

In the monohybrid cross involving flower color, the ratio is 3.15 purple : 1 white, while in the human family in the pedigree, the ratio in the third generation is 1 can taste PTC : 1 cannot taste PTC. The difference is due to the small sample size (two offspring) in the human family. If the secondgeneration couple in this pedigree were able to have 929 offspring as in the pea plant cross, the ratio would likely be closer to 3:1. (Note that none of the pea plant crosses in Table 11.1 yielded exactly a 3:1 ratio.)

Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms?

Incomplete dominance describes the relationship between two alleles of a single gene, whereas epistasis relates to the genetic relationship between two genes (and the respective alleles of each).

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?

Man IAi; woman IBi; child ii. Genotypes for future children are predicted to be ¼ IAIB, ¼ IAi, ¼ IBi, ¼ ii.

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?

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. If dominant, you wouldn't be able to tell true-breeding, homozygous offspring from heterozygotes without further crosses whether the curl trait is dominant or recessive. You know that cats are true-breeding when curl × curl matings produce only curl offspring for several generations. As it turns out, the allele that causes curled ears is dominant.

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?

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 either the AA or Aa genotype.

MAKE CONNECTIONS In some pea plant crosses, the plants are self-pollinated. Explain whether self-pollination is considered asexual or sexual reproduction

Self-pollination is sexual reproduction because meiosis is involved in forming gametes, which unite during fertilization. As a result, the offspring in self-pollination are genetically different from the parent.

Which genetic relationships listed in the first column of the two tables are demonstrated by the inheritance pattern of the ABO blood group alleles? For each genetic relationship, explain why this inheritance pattern is or is not an example.

The ABO blood group is an example of multiple alleles because this single gene has more than two alleles (I A, I B, and i). Two of the alleles, I A and I B, exhibit codominance, since both carbohydrates (A and B) are present when these two alleles exist together in a genotype. I A and I B each exhibit complete dominance over the i allele. This situation is not an example of incomplete dominance because each allele affects the phenotype in a distinguishable way, so the result is not intermediate between the two phenotypes. Because this situation involves a single gene, it is not an example of epistasis or polygenic inheritance.

WHAT IF? A rooster with gray feathers and a hen of the same phenotype produce 15 gray, 6 black, and 8 white chicks. What is the simplest explanation for the inheritance of these colors in chickens? What phenotypes would you expect in the offspring of a cross between a gray rooster and a black hen?

The black and white alleles are incompletely dominant, with heterozygotes being gray in color. A cross between a gray rooster and a black hen should yield approximately equal numbers of gray and black offspring.

Both members of a couple know that they are carriers of the cystic fibrosis allele. None of their three children has cystic fibrosis, but any one of them might be a carrier. The couple would like to have a fourth child but are worried that he or she would very likely have the disease, since the first three do not. What would you tell the couple? Would it remove some more uncertainty in their prediction if they could find out whether the three children are carriers?

The chance of the fourth child having cystic fibrosis is ¼, as it was for each of the other children, because each birth is an independent event. We already know that both parents are carriers, so whether their first three children are carriers or not has no bearing on the probability that their next child will have the disease. The parents' genotypes provide the only relevant information.

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 homozygous 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 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.

Concept 11.2: Probability laws govern Mendelian inheritance

The multiplication rule states that the probability of two or more events occurring together is equal to the product of the individual probabilities of the independent single events. The addition rule states that the probability of an event that can occur in two or more independent, mutually exclusive ways is the sum of the individual probabilities. The rules of probability can be used to solve complex genetics problems. A dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneously. In calculating the chances of the various offspring genotypes from such crosses, each character is first considered separately and then the individual probabilities are multiplied.

List all gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape. How large a Punnett square is needed to predict the offspring of a self-pollination of this "trihybrid"?

The plant could make eight different gametes (YRI, YRi, YrI, Yri, yRI, yRi, yrI, and yri). To fit all the possible gametes in a self-pollination, a Punnett square would need 8 rows and 8 columns. It would have spaces for the 64 possible unions of gametes in the offspring.

Tips for Genetics Problems 4

To figure out the outcome of a cross, set up a Punnett square. a)Put the gametes of one parent at the top and those of the other on the left. To determine the allele(s) in each gamete for a given genotype, set up a systematic way to list all the possibilities. (Remember, each gamete has one allele of each gene.) Note that there are 2n possible types of gametes, where n is the number of gene loci that are heterozygous. For example, an individual with genotype AaBbCc would produce 23 = 8 types of gametes. Write the genotypes of the gametes in circles above the columns and to the left of the rows. b)Fill in the Punnett square as if each possible sperm were fertilizing each possible egg, making all of the possible offspring. In a cross of AaBbCc × AaBbCc, for example, the Punnett square would have 8 columns and 8 rows, so there are 64 different offspring; you would know the genotype of each and thus the phenotype. Count genotypes and phenotypes to obtain the genotypic and phenotypic ratios. Because the Punnett square is so large, this method is not the most efficient. Instead, see tip 5.

Tips for Genetics Problems 1

Write down symbols for the alleles. (These may be given in the problem.) When represented by single letters, the dominant allele is uppercase and the recessive allele is lowercase.

Tips for Genetics Problems 2

Write down the possible genotypes, as determined by the phenotype. If the phenotype is that of the dominant trait (for example, purple flowers), then the genotype is either homozygous dominant or heterozygous (PP or Pp, in this example). If the phenotype is that of the recessive trait, the genotype must be homozygous recessive (for example, pp). If the problem says "true-breeding," the genotype is homozygous.

Tips for Genetics Problems 5

You can use the rules of probability if the Punnett square would be too big. (For example, see the question at the end of the summary for Concept 11.2 and question 7.) You can consider each gene separately (see the section Solving Complex Genetics Problems with the Rules of Probability in Concept 11.2).

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 11.1. 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, e)heterozygous for seed shape

a) ¹∕₆₄; (b) ¹∕₆₄; (c) ¹∕₈; (d) ¹∕₃₂

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, nor their parents have cystic fibrosis.)

¹∕₆

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 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.

¹∕₉

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?

½

Beth and Tom each have a sibling with cystic fibrosis, but neither Beth nor Tom nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability if a test revealed that Tom is a carrier but Beth is not? Explain your answers.

½ (Since cystic fibrosis is caused by a recessive allele, Beth and Tom's siblings who have CF must be homozygous recessive. Therefore, each parent must be a carrier of the recessive allele. Since neither Beth nor Tom has CF, this means they each have a ⅔ chance of being a carrier. If they are both carriers, there is a ¼ chance that they will have a child with CF. ⅔ × ⅔ × ¼ = ¹∕₉); 0 (Both Beth and Tom would have to be carriers to produce a child with the disease.)