Ch. 17 review Sections 1, 4, and 5

The inheritance of the ABO blood group in humans is illustrated in the table. Check all that apply.

(C) The gene for blood type shows a complete dominance inheritance pattern. (C) The gene for blood type shows an incomplete dominance inheritance pattern. (C) The gene for blood type shows a codominance inheritance pattern. (C) The gene for blood type has only two different alleles. (C) The gene for blood type has multiple (> 2) alleles. (X) The phenotype for the ABO blood type reflects the types of antibodies produced in the blood stream. (X) A person with type O blood will have compatibility with blood from a type A donor. Hint: Antibodies attack antigens or foreign molecules that are not naturally produced by the individual.

In order to determine all possible allele combinations, fill in the missing genotypes in the Punnett square.

BR Br bR br BR BBRR BBRr BbRR BbRr Br BBRr BBrr BbRr Bbrr bR BbRR BbRr bbRR bbRr br BbRr Bbrr bbRr bbrr

Gregor Mendel

Carried out his experiments in an abbey garden. Was a monk. Carried out his experiments on heredity by doing thousands of genetic crosses on a common garden plant. Discovered/founded the basic principles of hereditary.

In humans, long eyelashes (L) are dominant to short eyelashes (l). On the left are three boxes, each containing one possible genotype for eyelash length. Match the terms/phrases on the right with their correct genotype(s); some labels may be used more than once, and some labels may not be used at all.

LL: All offspring will have long eyelashes when crossed with Ll genotype. Homozygous dominant. Will have long eyelashes. Ll: Heterozygous. Will have long eyelashes. ll: Homozygous recessive. Will have short eyelashes.

Indicate if the conditions described would result in the development of a human male or a female individual.

Male: One of each X and Y chromosomes. An individual with only 1 X and Y chromosome w/an SRY gene mutation. 4 X chromosomes and a Y chromosome 2 X chromosomes and a Y chromosome Female: XX chromosomes Only one X chromosome. One X chromosome and a Y chromosome with the SRY gene deleted. An individual with only one X chromosome and a Y chromosome w/an SRY gene mutation.

In humans, the allele for freckles is dominant (F) and the allele for no freckles is recessive (f). An individual who is heterozygous for freckles would have the genotype

Ff

Consider a newly discovered organism that is bioluminescent. Scientists studying this animal have discovered that there are two enzymes necessary for the animal to glow, encoded by the bright and red genes. Furthermore, scientists have confirmed that for bioluminescence to be displayed, an individual must have at least one copy of the dominant allele for each gene, (B and R). What are the correct genotypes and phenotypes?

Glowing: BBRR, BBRr, BbRr, BbRR Non-glowing: bbrr, Bbrr, BBrr, bbRR, bbRr

Patterns of inheritance: Codominance and Incomplete Dominance Match the descriptions with the correct pattern of inheritance: monohybrid cross A, B, or C.

Monohybrid cross A (Tall x Dwarf model): Simple mendelian inheritance Dominant/recessive as an inhertiance pattern Monohybrid cross B (Red x White cross model): Shows incomplete dominance as an inheritance pattern Some individuals in F2 show a blend of 2 parental phenotypes Monohybrid cross C (ABO model): Shows Codominance as an inheritance pattern Some individuals in F2 exhibit both parental phenotypes in the same flower

Mendel's Principle of Segregation states that

two alleles of a gene separate during gamete formation such that every gamete receives only one allele.

In pea plants, yellow seeds (Y) are dominant to green seeds (y) and round seeds (R) are dominant to wrinkled seeds (r). The genes for seed color and seed shape are on different chromosomes. Two true-breeding parents, one with yellow round peas and the other with green wrinkled peas, are crossed to produce a hybrid (heterozygous) F1. Two F1 individuals are crossed to give an F2; this is depicted in the Punnett square below. Place the correct genotypes in the Punnett square and the place the correct phenotypic ratios next to their appropriate phenotype on the right.

P: YYRR - YR | yyrr - yr (Self-fertilization) F1: YR x yr = YyRr F2: YyRr x YyRr Phenotypic ratios: Yellow, round: Yellow, wrinkled: Green, round: Green, Wrinkled. 9:3:3:1

In pea plants, a single gene controls pea texture. Smooth (S) peas are dominant over wrinkled (s) peas. A plant with smooth peas is crossed with a plant with wrinkled peas. Of the progeny, 252 are smooth and 247 are wrinkled. What are the genotypes of the parent plants?

Ss x ss yields: Ss and ss OR 1:1 ratio of Smooth to Wrinkled plants.

Tay-Sachs is an autosomal recessive condition that results from a mutation in a gene encoding the lysosomal enzyme, hexA, that normally degrades a specific type of lipid in brain cells. Failure to degrade the lipid results in degradation of brain neurons. Match the statements about Tay Sachs and the hex A enzyme to the appropriate genotype. T = dominant allele; t = recessive allele

TT: Does not have Tay Sachs Makes normal amount of HexA enzyme Tt: Makes 1/2 the normal amount of HexA enzyme, but enough to degrade lipids. Does not have Tay Sachs, but is a carrier. tt: Has Tay Sachs. Makes no HexA enzyme.

Role of Environment (in plant height)

Temperature influences the growth of plants. The environment likely impacts the expression of genes important to plant height. Despite having the identical genetics the phenotype of organisms show continuous variation with respect to the environment.

In garden peas, the allele for tall plants (T) is dominant over the allele for dwarf plants (t). A true-breeding tall plant is bred with a true-breeding dwarf plant to produce heterozygous tall plants. The F1 heterozygotes then self-fertilize to yield F2 offspring. Complete the Punnett square for this monohybrid cross using the F1 gametes with the appropriate genotype and placing the correct F2 genotype/phenotype combination in the Punnett square.

Tt x Tt yields: TT, Tt, tt OR 3:1 ratio of Tall to Dwarf plants.

Match the organisms with the correct mechanism of sex determination.

XY: Humans. The male is heterozygous for sex chromosomes. ZW: The male is homozygous for sex chromosomes. XO: Insects that are hemizygous for sex chromosomes. Environmental: In alligators, males are produced at intermediate temperature, females at low temperature, and a mix of both at high temperatures. Haploidiploid: Male bees have half the number of chromosomes relative to female bees.

Red-green colorblindness in humans is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. Complete the Punnett square to answer the following questions: What is the probability that they will have a color-blind daughter? (X = wild type allele, Xc = color blind allele)

XcY x XXc Genotypes: XXc, XcXc, XcY, and XY Probability of having a colorblind daughter: 1/4

Red-green colorblindness in humans is caused by a sex-linked recessive allele. A colorblind man (XcY) marries a woman with normal vision whose father was colorblind (XXc). The couple has a son. What is the probability that he is colorblind?

XcY x XXc Genotypes: XXc, XcXc, XcY, and XY Probability of having a colorblind son: 1/2

Match types of Molecular Basis of Inheritance: A. Pattern baldness in humans occurs from an autosomal allele that is dominant in males and recessive in females. B. A women who is heterozygous for the recessive hemophilia gene does not exhibit the disease. Her sons, but not daughters, inherit the disease. C. Individuals who are homozygous for the normal gene that encodes an enzyme for metabolizing phenylalanine have low levels of this amino acid, while individuals heterozygous for the mutated gene have intermediate levels of phenylalanine, and individuals homozygous for the mutated gene have very high levels of phenylalanine, and therefore exhibit phenylketonuria. D. An inheritance pattern that occurs when the heterozygous individual expresses both alleles simultaneously. For example, a human carrying the A and B alleles for the ABO antigens of red blood cells produces both the A and the B antigens (has an AB blood type). E. A person with Tay-Sachs disease is homozygous for a mutated form of the hexosaminidase A gene (Tay-Sachs allele). This results in little production of the functional hexosaminidase A enzyme, leading to poor lipid metabolism, excess brain lipid deposition and eventual death. Individuals who are heterozygous for the Tay-sachs allele, produce enough of the hexosaminidase enzyme for proper lipid metabolism, and hence do no exhibit the disease.

A: Sex-influenced inheritance - is an autosomal trait that is expressed differently, either in frequency or degree, in males and females. B: X-linked inheritance - refers to the inheritance of genes located on the X chromosome; the sex of an individual will play a role. C: Incomplete dominance - is a form of intermediate inheritance in which neither allele is completely expressed in a heterozygote; the phenotype of the heterozygote is intermediate between homozygous parents. D: Codominance - is a form of inheritance in which both alleles are completely expressed in a heterozygote. E: Simple Mendelian inheritance - refers to inheritance of alleles that show a clear dominant/recessive relationship; one allele is completely expressed in heterozygotes. In a recessive condition, both alleles must be defective for an individual to have a disease phenotype.

Phenotype is not often determined by one gene (i.e., metabolic rate is determined by the combination of multiple genes). In addition to an organism's genotype, the environment can influence phenotype. A rabbit breeder has established that rabbit body weight is determined by the combination of three genes as well as ambient temperature during the first four weeks after birth. The three genes involved in the determination of body weight are gene A (fat deposition), gene B (appetite), and gene C (gastric motility). Each gene can occur in heavy and light alleles (indicated by upper case and lower case letters, respectively). If a rabbit is heterozygous for each gene (Aa, Bb, Cc), the rabbit is an intermediate weight. If a rabbit is homozygous for all of the light genes (aa, bb, cc), the rabbit is lightweight. If a rabbit is homozygous for all of the heavy genes (AA, BB, CC), the rabbit is heavyweight. Additionally, rabbits reared at low temperatures during early development will deposit additional fat, making them heavier, whereas rabbits reared at high temperatures will not deposit as much adipose tissue. Arrange the following genotype and ambient temperature combinations from heaviest to lightest weight rabbits.

AABBCC, AaBBCc, AABbCc, AaBbCc, aabbcc In degrees F: -4, -4, 70, 70, 70

Why is muscular dystrophy more common in boys? Muscular dystrophy (MD) is a group of disorders that involve muscle weakness and loss of muscle tissue. Duchenne muscular dystrophy (DMD) is a rapidly-worsening form of MD. DMD is caused by a defect in the dystrophin gene. The dystrophin protein is a vital part of a protein complex involved in muscle support and structure. DMD is much more common in boys than in girls (about 1/3500 boys have DMD, girls rarely have DMD).

The gene for dystrophin must be on the X chromosome. If a female has a defective dystrophin allele, she will likely have a normal allele to act as a "back up". Males are hemizygous, so they will have DMD even if they have only one defective allele. DMD is inherited as an X-linked recessive trait. A woman who is a carrier for DMD will pass the defective allele to half her sons, and those sons will have DMD.

In pea plants, yellow seeds (Y) are dominant to green seeds (y) and round seeds (R) are dominant to wrinkled seeds (r). The genes for seed color and seed shape are on different chromosomes. Two true-breeding parents, one with yellow round peas and the other with green wrinkled peas, are crossed. Which of the following is/are accurate regarding the resulting F1 plants? Check all that apply.

The genotype of the F1 plants is YyRr. All of the F1 plants have yellow round seeds. The F1 plants make YR, Yr, yR, and yr gametes in equal proportions.

In pea plants, yellow seeds (Y) are dominant to green seeds (y) and round seeds (R) are dominant to wrinkled seeds (r). The genes for seed color and seed shape are on different chromosomes. Two true-breeding parents, one with yellow round peas and the other with green wrinkled peas, are crossed. What are the genotypes of the parents, and what kind of gametes will they produce?

The genotype of the plant with yellow round seeds will be YYRR and produce YR gametes, while the genotype of the plant with green wrinkled seeds will be yyrr and produce yr gametes.

In humans, the allele for freckles is dominant (F) and the allele for no freckles is recessive (f). An individual who is heterozygous for freckles would have which of the phenotypes

They would have freckles.

Chickens that have the dominant allele of the frizzle gene (F_) have feathers that curl outward. Chickens that are ff have feathers that lie flat against their bodies. Chickens with F alleles—along with producing defective feathers—have higher metabolic rates, abnormal body temperatures, high blood flow capacity, and low egg-laying rates. This is an example of

pleiotropy. - mutation in single gene can have multiple effects on an organism's phenotype

Recessive alleles in Mendelian inheritance patterns

will have no noticeable effects on an organism's phenotype if a dominant allele is present. will have an effect on an organism's phenotype if the individual is homozygous for the recessive allele. can be either on the maternal chromosome or the paternal chromosome.