Genetics Exam 1
A cell has four pairs of chromosomes. Assuming that crossing over does not occur, what is the probability that a gamete will contain all of the paternal chromosomes? If n equals the number of chromosomes in a set, which of the following expressions can be used to calculate the probability that a gamete will receive all of the paternal chromosomes: (1/2)n, (1/2)n-1, or n1/2?
(1/2)n = (1/2)4 = 1/16 or 6.25%
For the following pedigrees, describe what you think is the most likely inheritance pattern(dominant versus recessive)
(a) It behaves like a recessive trait because unaffected parents sometimes produce affected offspring. In such cases, the unaffected parents are heterozygous carriers. (b) It behaves like a dominant trait. An affected offspring always has an affected parent. However, recessive inheritance cannot be ruled out
With regard to genotypes, what is a true-breeding organism?
A true-breeding organism is a homozygote that has two copies of the same allele.
What is a genetic cross?
A genetic cross involves breeding two different individuals
What is the cause of Down Syndrome?
An extra chromosome (specifically an extra copy of chromosome 21) causes Down syndrome.
In cocker spaniel, solid coat color is dominant over spotted coat color. If two heterozygous dogs were crossed to each other, what would be the probability of the following combinations of offspring? A. A litter of five pups, four with solid fur and one with spotted fur. B. A first litter of six pups, four with solid fur and two with spotted fur, and then a second litter of five pups, all with solid fur. C. A first litter of five pups, the firstborn with solid fur, and then among the next four, three with solid fur and one with spotted fur, and then a second litter of seven pups in which the firstborn is spotted, the second born is spotted, and the remaining five are composed of four solid and one spotted animal. D. A litter of six pups, the firstborn with solid fur, the second born spotted, and among the remaining four pups, two with spotted fur and two with solid fur.
First construct a Punnett square. The chances are 75% of producing a solid pup and 25% of producing a spotted pup. A litter of five pups, four with solid fur and one with spotted fur A. Use the binomial expansion equation where n = 5, x = 4, p = 0.75, q = 0.25. The answer is 0.396 = 39.6% of the time. A first litter of six pups, fou with solid fur and two with spotted ufr, and then a second litter of five pups B. You can use the binomial expansion equation for each litter. For the first litter, n = 6, x = 4, p = 0.75, q = 0.25; for the second litter, n = 5, x = 5, p = 0.75, q = 0.25. Because the litters are in a specified order, we use the product rule and multiply the probability of the first litter times the probability of the second litter. The answer is 0.070, or 7.0%. A first litter of five pups, the firstborn with solid fur, and then among the next four, three with solid fur and one with spotted fur, and then a second litter of seven pups in which the firstborn is spotted, the second born is spotted, and the remaining five are composed of four solid and one spotted animal. C. To calculate the probability of the first litter, we use the product rule and multiply the probability of the first pup (0.75) times the probability of the remaining four. We use the binomial expansion equation to calculate the probability of the remaining four, where n = 4, x = 3, p = 0.75, q = 0.25. The probability of the first litter is 0.316. To calculate the probability of the second litter, we use the product rule and multiply the probability of the first pup (0.25) times the probability of the second pup (0.25) times the probability of the remaining five. To calculate the probability of the remaining five, we use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The probability of the second litter is 0.025. To get the probability of these two litters occurring in this order, we use the product rule and multiply the probability of the first litter (0.316) times the probability of the second litter (0.025). The answer is 0.008, or 0.8%. A litter of six pups, the firstborn with solid fur, the second born spotted, and among the remaining four pups, two with spotted fur and two with solid fur D. Because this is a specified order, we use the product rule and multiply the probability of the firstborn (0.75) times the probability of the second born (0.25) times the probability of the remaining four. We use the binomial expansion equation to calculate the probability of the remaining four pups, where n = 4, x = 2, p = 0.75, q = 0.25. The answer is 0.040, or 4.0%.
In a cross between a heterozygous tall pea plant and a dwarf plant, predict the ratios of the offspring's genotypes and phenotypes.
Genotypes: 1 Tt :1 tt Phenotypes: 1 tall : 1 dwarf
In people with maple syrup urine disease, the body is unable to metabolize the amino acids leucine, isoleucine, and valine. One of the symptoms is that the urine smells like maple syrup. An unaffected couple produced six children in the following order: unaffected daughter, affected daughter, unaffected son, unaffected son, affected son, and unaffected son. The youngest unaffected son and an unaffected woman have three children in the following order: affected daughter, unaffected daughter, and unaffected son. Draw a pedigree that describes this family. What type of inheritance (dominant or recessive) would you propose to explain maple syrup urine disease?
It is recessive inheritance. The pedigree is shown here. Affected individuals are shown with filled symbols. The mode of inheritance appears to be recessive. Unaffected parents (who must be heterozygous) produce affected children.
The arrangement of homologs during metaphase of meiosis I is a random process. In your own words, explain what this means.
It means that the maternally derived and paternally derived chromosomes are randomly aligned along the metaphase plate during metaphase I. Refer to Figure 3.16.
An individual has the genotype Aa Bb Cc and makes an abnormal gamete with the genotype AaBc. Does this gamete violate the law of independent assortment or the law of segregation (or both)? Explain your answer.
It violates the law of segregation because two copies of one gene are in the gamete. The two alleles for the A gene did not segregate from each other.
With regard to sister chromatids, which phase of mitosis is the organization phase, and which is the separation phase?
Metaphase is the organization phase and anaphase is the separation phase.
Would it be possible to deduce the law of independent assortment from a single-factor cross? Explain your answer.
No, the law of independent assortment applies to transmission patterns of two or more genes. In a single-factor cross, you are monitoring only the transmission pattern of a single gene.
Describe three advantages of using pea plants as an experimental organism
Pea plants are relatively small and hardy. They produce both pollen and eggs within the same flower. Because a keel covers the flower, self-fertilization is quite easy. In addition, cross-fertilization is possible by the simple manipulation of removing the anthers in an immature flower and later placing pollen from another plant. Finally, peas exist in several variants
Pick any example of a genetic technology and describe how it has directly affect your life.
There are many possible answers. Some common areas to discuss might involve the impact of genetics in the production of new medicines, the diagnosis of diseases, the production of new kinds of food, and the use of DNA fingerprinting to solve crimes
Let's suppose you conducted an experiment involving genetic crosses and calculated a chi square value of 1.005. There were four categories of offspring (i.e., the degrees of freedom equaled 3). Explain what the 1.005 value means. Your answer should include the phrase "80% of the time."
This means that a deviation value of 1.005 or greater (between the observed and expected data) would occur 80% of the time. In other words, it is fairly likely to obtain this value due to random sampling error. Therefore, we accept our hypothesis.
Your textbook describes how the detrimental symptoms associated with the disease phenylketonuria (PKU) are caused by a faulty gene. Even so, a change in diet can prevent these symtomps. Pick a trait in your favorite plant and explain how genetics and the environment may play important roles.
You could pick almost any trait. For example, flower color in petunias would be an interesting choice. Some petunias are red and others are purple. There must be different alleles in a flower color gene that affect this trait in petunias. In addition, the amounts of sunlight, fertilizer, and water also affect the intensity of flower color.
This picture shows a micrograph of chromosomes from a normal human cell. If you created this kind of image using a cell from a person with Down Syndrome, what would you expect to see?
You would see 47 chromosomes instead of 46. There would be three copies of chromosome 21 instead of two copies.
On rare occasions, an organism may have three copies of a chromosome and therefore has three copies of the genes on that chromosome (instead of the usual number of two copies). For such a rare organism, the alleles for each gene usually segregate so that a gamete will contain one or two copies of the gene. Let's suppose that a rare pea plant has three copies of the chromosome that carries the height gene. Its genotype is TTt. The plant is also heterozygous for the seed color gene, Yy, which is found on a different chromosome. With regard to both genes, how many types of gametes can this plant make, and in what proportions? (Assume that it is equally likely that a gamete will contain one or two copies of the height gene.)
2 TY, tY, 2 Ty, ty, TTY, TTy, 2 TtY, 2 Tty It may be tricky to think about, but you get 2 TY and 2 Ty because either of the two T alleles could combine with Y or y. Also, you get 2 TtY and 2 Tty because either of the two T alleles could combine with t and then combine with Y or y.
What is a DNA sequence?
A DNA sequence is a sequence of nucleotides. Each nucleotide may have one of four different bases (i.e., A, T, G, or C). When we speak of a DNA sequence, we focus on the sequence of bases.
At a molecular level, what is a gene? Where are genes located?
A chromosome is a very long polymer of DNA. A gene is a specific sequence of DNA within that polymer; the sequence of bases creates a gene and distinguishes it from other genes. Genes are located in chromosomes, which are found within living cells
What is a homolog? With regard to genes and alleles, how are homologs similar to and different from each other?
A homolog refers to one of the members of a chromosome pair. Homologs are usually the same size and carry the same types and order of genes. They may differ in that the genes they carry may be different alleles.
For the following events, specify whether they occur during mitosis, meiosis I, or meiosis II: Separation of conjoined chromatids within a pair of sister chromatids Pairing of homologous chromosomes Alignment of chromatids along the metaphase plate Attachment of sister chromatids to both poles
A. During mitosis and meiosis II B. During meiosis I C. During mitosis, meiosis I, and meiosis II D. During mitosis and meiosis II
A true-breeding tall plant was crossed to a dwarf plant. Tallness is a dominant trait. The F1 individuals were allowed to self-fertilize. What are the following probabilities for the F2 generation? The first plant is dwarf. The first plant is dwarf or tall. The first three plants are tall. For any seven plants, three are tall and four are dwarf. The first plant is tall, and then among the next four, two are tall and the other two are dwarf.
A. 1/4 B. 1, or 100% C. (3/4)(3/4)(3/4) = 27/64 = 0.42, or 42% D. Use the binomial expansion equation, where n = 7, p = 3/4, q = 1/4, x = 3: P = 0.058, or 5.8% E. The probability that the first plant is tall is 3/4. To calculate the probability that among the next four, any two will be tall, use the binomial expansion equation, where n = 4, p = 3/4, q = 1/4, and x = 2: P = 0.21 Then calculate the overall probability of these two events: (3/4)(0.21) = 0.16, or 16%
Corn has 10 chromosomes per set, and the sporophyte of the species is diploid. If you performed a karyotype, what is the total number of chromosomes you would expect to see in each of the following types of cells? A leaf cell The sperm nucleus of a pollen grain An endosperm cell after fertilization A root cell
A. 20 B. 10 C. 30 D. 20
A true-breeding pea plant with round and green seeds was crossed to a true-breeding plant with wrinkled and yellow seeds. Round and yellow seeds are the dominant traits. The F1 plants were allowed to self-fertilize. What are the following probabilities for the F2 generation? A. An F2 plant with wrinkled, yellow seeds. B. Three out of three F2 plants with round, yellow seeds. C. Five F2 plants in the following order: two have round, yellow seeds; one has round, green seeds; and two have wrinkled, green seeds. D. An F2 plant will not have round, yellow seeds
A. 3/16 B. (9/16)(9/16)(9/16) = 729/4096 = 0.18 C. (9/16)(9/16)(3/16)(1/16)(1/16) = 243/1,048,576 = 0.00023, or 0.023% D. Another way of looking at this is that the probability it will have round, yellow seeds is 9/16. Therefore, the probability that it will not is 1 - 9/16 = 7/16.
Three genes (A, B, and C) are found on three different chromosomes. For the following diploid genotypes, describe all of the possible gamete combinations. Aa Bb Cc AA Bb CC Aa BB Cc Aa bb cc
A. A B C, A B c, A b C, A b c, a B C, a b C, a B c, a b c B. A B C, A b C C. A B C, A B c, a B C, a B c D. A b c, a b c
Huntington disease is a rare dominant trait that causes neurodegeneration later in life. A man in his thirties, who already has three children, discovers that his mother has Huntington disease though his father is unaffected. What are the following probabilities? A. That the man in his thirties will develop Huntington disease. B. That his first child will develop Huntington disease. C. That one out of three of his children will develop Huntington disease
A. Construct a Punnett square. Because it is a rare disease, we would assume that the mother is a heterozygote and the father is normal. The chances are 50% that the man in his thirties will have the allele. B. Use the product rule: 0.5 (chance that the man has the allele) times 0.5 (chance that he will pass it to his offspring), which equals 0.25, or 25%. C. We use the binomial expansion equation. From part B, we calculated that the probability of an affected child is 0.25. Therefore the probability of an unaffected child is 0.75. For the binomial expansion equation, n = 3, x = 1, p = 0.25, q = 0.75. The answer is 0.422 or 42.2%
Let's suppose that a gene affecting pigmentation is found on the X chromosome (in mammals or insects) or the Z chromosome (in birds) but not on the Y or W chromosome. It is found on an autosome in bees. This gene exists in two alleles: D (dark) is dominant to d (light). What would be the phenotypic results of crosses between true-breeding dark females and true-breeding light males and of the reciprocal crosses involving true-breeding light females and true-breeding dark males for each of the following species? Refer back to Figure 3.17 for the mechanism of sex determination in these species. Birds Fruit flies Bees Humans
A. Dark males and light females; reciprocal: all dark offspring B. All dark offspring; reciprocal: dark females and light males C. All dark offspring; reciprocal: dark females and light males D. All dark offspring; reciprocal: dark females and light males
What would be the sex of a human with each of the following sets of sex chromosomes? XXX X (also described as X0) XYY XXY
A. Female; there are no Y chromosomes. B. Female; there are no Y chromosomes. C. Male; the Y chromosome determines maleness. D. Male; the Y chromosome determines maleness.
When studying living cells in a laboratory, researchers sometimes use drugs as a way to cause cells to remain in a particular phase of the cell cycle. For example, aphidicolin inhibits DNA synthesis in eukaryotic cells and causes them to remain in the G1 phase because they cannot replicate their DNA. In what phase of the cell cycle—G1, S, G2, prophase, metaphase, anaphase, or telophase—would you expect somatic cells to stay if the following types of drug were added? A drug that inhibits microtubule formationPage 75 A drug that allows microtubules to form but prevents them from shortening A drug that inhibits cytokinesis A drug that prevents chromosomal condensation
A. G2 phase (it could not complete prophase) B. Metaphase (it could not enter anaphase) C. Telophase (it could not divide into two daughter cells) D. G2 phase (it could not enter prophase)
In fruit flies, curved wings are recessive to straight wings, and ebony body is recessive to gray body. A cross was made between true-breeding flies with curved wings and gray bodies and flies with straight wings and ebony bodies. The F1 offspring were then mated to flies with curved wings and ebony bodies to produce an F2 generation. Diagram the genotypes of this cross, starting with the parental generation and ending with the F2 generation. What are the predicted phenotypic ratios of the F2 generation? Let's suppose the following data were obtained for the F2 generation: 114 curved wings, ebony body 105 curved wings, gray body 111 straight wings, gray body 114 straight wings, ebony body Conduct a chi square analysis to determine if the experimental data are consistent with the expected outcome based on Mendel's laws.
A. If we let c+ represent normal wings and c represent curved wings, and e+ represent gray body and e represent ebony body: Parental Cross: cce+e+ × c+c+ee. F1 generation is heterozygous: c+ce+e An F1 offspring crossed to a fly with curved wings and ebony body is represented as: c+ce+e × ccee The F2 offspring have this 1:1:1:1 ratio: c+ce+e : c+cee : cc e+e : ccee B. The phenotypic ratio of the F2 flies would be 1:1:1:1: normal wings, gray body : normal wings, ebony bodies : curved wings, gray bodies : curved wings, ebony bodies C. From part B, we expect 1/4 of each category. There are a total of 444 offspring. The expected number of each category is 1/4 × 444, which equals 111. x^2= 0.49 With 3 degrees of freedom, a value of 0.49 or greater is likely to occur between 80% and 95% of the time. Therefore, we accept our hypothesis.
A human gene called the β-globin gene encodes a polypeptide that functions as a subunit of the protein known as hemoglobin. Hemoglobin is found within red blood cells; it carries oxygen. In human populations, the β-globin gene can be found as the common allele called the HbA allele, and it can also be found as the HbS allele. Individuals who have two copies of the HbS allele have the disease called sickle cell disease. Are the following examples descriptions of genetics at the molecular, cellular, organism, or population level? Examples: a. The HbS allele encodes a polypeptide that functions slightly differently from the polypeptide encoded by the HbA allele. b. If an individual has two copies of the HbSallele, that person's red blood cells take on a sickle cell shape. c. Individuals who have two copies of the HbA allele do not have sickle cell disease but are not resistant to malaria. d. Individuals with sickle cell disease have anemia because their red blood cells are easily destroyed by the body.
A. Molecular level. This is a description of a how an allele affects protein function. B. Cellular level. This is a description of how protein function affects cell structure. C. Population level. This is a description of how the two alleles affect members of a population. D. Organism level. This is a description of how the alleles affect the traits of an individual.
For pea plants with the following genotypes, list the possible gametes that the plant can make: TT Yy Rr Tt YY rr Tt Yy Rr tt Yy rr
A. T Y R, T y R, T Y r, T y r B. T Y r, t Y r C. T Y R, T Y r, T y R, T y r, t Y R, t Y r, t y R, t y r D. t Y r,t y r
Assuming that such a fly would be viable, what would be the sex of a fruit fly with the following chromosomal composition? One X chromosome and two sets of autosomes Two X chromosomes, one Y chromosome, and two sets of autosomes Two X chromosomes and four sets of autosomes Four X chromosomes, two Y chromosomes, and four sets of autosomes
A. The fly is a male because the ratio of X chromosomes to sets of autosomes is 1/2, or 0.5. B. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0. C. The fly is male because the ratio of X chromosomes to sets of autosomes is 0.5. D. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0.
Many organisms are studied by geneticists. Do you think each of the following species would be more likely to be studied by a transmission geneticist, a molecular geneticist, or a population geneticist? Explain your answer. Note: More than one answer may be possible for a given species. Dogs E. coli Fruit flies Leopards Corn
A. Transmission geneticists. Dog breeders are interested in how genetic crosses affect the traits of dogs. B. Molecular geneticists. This is a good model organism for studying genetics at the molecular level. C. Both transmission geneticists and molecular geneticists. Fruit flies are easy to cross and for studying the transmission of genes and traits from parents to offspring. Molecular geneticists have also studied many genes in fruit flies to see how they function at the molecular level. D. Population geneticists. Most wild animals and plants would be of interest to population geneticists. In the wild, you cannot make controlled crosses. But you can study genetic variation within a population and try to understand its relationship to the environment. E. Transmission geneticists. Agricultural breeders are interested in how genetic crosses affect the outcome of traits.
What are the primary interests of researchers working in the following fields of genetics?
A. Transmission genetics: How genes and traits are transmitted from parents to offspring. B. Molecular genetics: How the genetic material functions at the molecular and cellular levels. C. Population genetics: Why genetic variation exists in populations, and how it changes over the course of many generations.
In humans, the allele for brown eye color (B) is dominant to that for blue eye color (b). If two heterozygous parents produce children, what are the following probabilities? The first two children have blue eyes. A total of four children, two with blue eyes and the other two with brown eyes. The first child has blue eyes, and the next two have brown eyes.
A. Use the product rule: (1/4)(1/4)=1/16 B. Use the binomial expansion equation, where n = 4, p = ¼, q = ¾, x = 2: P = 0.21, or 21% C. Use the product rule: (1/4)(3/4)(3/4) = 0.14, or 14%
For all seven characters described in the data of Figure 2.5, Mendel allowed the F2 plants to self-fertilize. He found that when F2 plants with recessive traits were crossed to each other, they always bred true. However, when F2 plants with dominant traits were crossed, some bred true but others did not. A summary of Mendel's results is shown to the right.
According to Mendel's law of segregation, the genotypic ratio should be 1 homozygote dominant : 2 heterozygotes : 1 homozygote recessive. The data table considers only the plants with a dominant phenotype. The genotypic ratio should be 1 homozygote dominant : 2 heterozygotes. The homozygote dominants would be true-breeding while the heterozygotes would not be true-breeding. This 1:2 ratio is very close to what Mendel observed.
As in many animals, albino coat color is a recessive trait in guinea pigs. Researchers removed the ovaries from an albino female guinea pig and then transplanted ovaries from a true-breeding black guinea pig. They then mated this albino female (with the transplanted ovaries) to an albino male. The albino female produced three offspring. What were their coat colors? Explain the results
All three offspring had black fur. The ovaries from the albino female could only produce eggs with the dominant black allele (because they were obtained from a true-breeding black female). The actual phenotype of the albino mother does not matter. Therefore, all offspring would be heterozygotes (Bb) and have black fur.
Briefly explain how gene expression occurs at the molecular level.
At the molecular level, a gene (a sequence of DNA) is first transcribed into RNA. The genetic code within the RNA is used to synthesize a protein with a particular amino acid sequence. This second process is called translation.
Based on genes in pea plants that we have considered in this chapter, which statements is/are not correct? A. The gene causing tall plants is an allele of the gene causing dwarf plants. B. The gene causing tall plants is an allele of the gene causing purple flowers. C. The alleles causing tall plants and purple flowers are dominant
B. This statement is not correct because these are alleles of different genes
Marfan syndrome is a rare inherited human disorder characterized by unusually long limbs and digits plus defects in the heart (especially the aorta) and the eyes, among other symptoms. Following is a pedigree for this disorder. Affected individuals are shown with filled (black) symbols. What type of inheritance pattern do you think is the most likely?
Based on this pedigree, the disease is likely to be a dominant trait because an affected child always has an affected parent. In fact, it is a dominant disorder
Eukaryotic cells must sort their chromosomes during mitosis so that each daughter cell receives the correct number of chromosomes. Why don't bacteria need to sort their chromosomes?
Bacteria do not need to sort their chromosomes because they only have one type of chromosome. Though not discussed in the text, the attachment of the two copies of the chromosomes to the cell membrane prior to cell division also helps to ensure that each daughter cell receives one copy.
Ectrodactyly, also known as "lobster claw syndrome", is recessive disorder in humans. If a phenotypically unaffected couple produces a n affected offspring, what are the following probabilities? A. Both parents are heterozygotes. B. An offspring is a heterozygote. C. The next three offspring will be phenotypically unaffected. D. Any two out of the next three offspring will be phenotypically unaffected
Both parents are heterozygotes A. Barring a new mutation during gamete formation, the probability is 100% because the parents must be heterozygotes in order to produce a child with a recessive disorder. An offspring is heterozygote B. Construct a Punnett square. There is a 50% chance of heterozygous children. The next three offspring will be phenotypically unaffected C. Use the product rule. The chance of being phenotypically normal is 0.75 (i.e., 75%), so the answer is 0.75 × 0.75 × 0.75 = 0.422, which is 42.2%. Any two out of the next three offspring will be phenotypically unaffected D. Use the binomial expansion equation where n = 3, x = 2, p = 0.75, q = 0.25. The answer is 0.422, or 42.2%.
Honeybees are unusual in that male bees (drones) have only one copy of each gene, but female bees have two copies of their genes. This difference arises because drones develop from eggs that have not been fertilized by sperm cells. In bees, the trait of long wings is dominant over short wings, and the trait of black eyes is dominant over white eyes. If a drone with short wings and black eyes was mated to a queen bee that is heterozygous for both genes, what are the predicted genotypes and phenotypes of male and female offspring? What are the phenotypic ratios if we assume an equal number of male and female offspring?
C29. The drone is sB and the queen is SsBb. According to the laws of segregation and independent assortment, the male can make only sB gametes, while the queen can make SB, Sb, sB, and sb, in equal proportions. Therefore, male offspring will be SB, Sb, sB, and sb, and female offspring will be SsBB, SsBb, ssBB, and ssBb. The phenotypic ratios, assuming an equal number of males and females, will be: Males Females 1 normal wings/black eyes 2 normal wings, black eyes 1 normal wings/white eyes 2 short wings, black eyes 1 short wings/black eyes 1 short wings/white eyes
A woman with achondroplasia (a dominant form of dwarfism) and a phenotypically unaffected man have seven children, all of whom have achondroplasia. What is the probability of producing such a family if this woman is a heterozygote? What is the probability that the woman is a heterozygote if her eighth child does not have this disorder?
C34. Use the product rule. If the woman is heterozygous, there is a 50% chance of having an affected offspring: (0.5)7 = 0.0078, or 0.78%, of the time. This is a pretty small probability. If the woman has an eighth child who is unaffected, however, she has to be a heterozygote, because it is a dominant trait. She would have to pass a normal allele to an unaffected offspring. The answer is 100%.
How can you determine whether an organism is heterozygous or homozygous for a dominant trait?
Conduct a cross in which the unknown individual is bred to an individual that carries only recessive alleles for the gene in question.
A true-breeding plant with round and green seeds was crossed to a true-breeding plant with wrinkled and yellow seeds. The F1 plants were allowed to self-fertilize. What is the probability of obtaining the following plants in the F2 generation: two that have round, yellow seeds; one with round, green seeds; and two with wrinkled, green seeds?
Construct a Punnett square to determine the probability of these three phenotypes. The probabilities are 9/16 for round, yellow; 3/16 for round, green; and 1/16 for wrinkled, green. Use the multinomial expansion equation described question 4 in More Genetic TIPS, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, r = 1/16. The answer is 0.007, or 0.7%.
The period between meiosis I and meiosis II is called interphase II. Does DNA replication take place during interphase II?
DNA replication does not take place during interphase II. The chromosomes at the end of telophase I have already replicated (i.e., they are found in pairs of sister chromatids). During meiosis II, the sister chromatids separate from each other, yielding individual chromosomes.
In your own words, describe Mendel's Law of Segregation. Do not use the word segregation in your answer.
Diploid organisms contain two copies of each type of gene. When they make gametes, only one copy of each gene is found in a gamete. Two alleles cannot stay together within the same gamete.
Why is it necessary for the chromosomes to condense during mitosis and meiosis? What do you think might happen if the chromosomes were not condensed?
During interphase, the chromosomes are greatly extended. In this conformation, they might get tangled up with each other and not sort properly during meiosis and mitosis. The condensation process probably occurs so that the chromosomes easily align along the metaphase plate during metaphase without getting tangled up.
Describe the timing of meiosis I and II during human oogenesis
During oogenesis in humans, the cells are arrested in prophase I of meiosis for many years until selected primary oocytes progress through the rest of meiosis I and begin meiosis II. If fertilization occurs, meiosis II is completed.
A diploid species has four chromosomes per set for a total of eight chromosomes in its somatic cells. Draw such a cell as it would look in late prophase of meiosis II and prophase of mitosis. Discuss how prophase of meiosis II and prophase of mitosis differ from each other, and explain how the difference originates.
During prophase II, your drawing should show four replicated chromosomes (i.e., four structures that look like Xs). Each chromosome is one homolog. During prophase of mitosis, there should be eight replicated chromosomes (i.e., eight Xs). During prophase of mitosis, there are pairs of homologs. The main difference is that prophase II has a single copy of each of the four chromosomes, whereas prophase of mitosis has four pairs of homologs. At the end of meiosis I, each daughter cell has received only one copy of a homologous pair, not both. This is due to the alignment of homologs during metaphase I and their separation during anaphase I.
What is meant by the term genetic variation? Give two examples of genetic variation not discussed in Chapter 1. What causes genetic variation at the molecular?
Genetic variation involves the occurrence of genetic differences within members of the same species or different species. Within any population, variation may occur in the genetic material. Variation may occur in particular genes so that some individuals carry one allele and other individuals carry a different allele. An example would be differences in coat color among mammals. There also may be variation in chromosome structure and number. In plants, differences in chromosome number can affect disease resistance.
A cross is made between a pea plant that has constricted pods (a recessive trait; smooth is dominant) and is heterozygous for seed color (yellow is dominant to green) and a plant that is heterozygous for both pod texture and seed color. Construct a Punnett square that depicts this cross. What are the predicted outcomes of genotypes and phenotypes of the offspring?
Here c is the recessive allele for constricted pods; Y is the dominant allele for yellow color. The cross is ccYy × CcYy. Follow the directions for setting up a Punnett square, as described in Chapter 2. The genotypic ratio is 2 CcYY : 4 CcYy : 2 Ccyy : 2 ccYY : 4 ccYy : 2 ccyy. This 2:4:2:2:4:2 ratio can be reduced to a 1:2:1:1:2:1 ratio. The phenotypic ratio is 6 smooth pods, yellow seeds : 2 smooth pods, green seeds : 6 constricted pods, yellow seeds : 2 constricted pods, green seeds. This 6:2:6:2 ratio can be reduced to a 3:1:3:1 ratio.
A cross was made between a white male dog and two different black females. The first female gave birth to eight black pups, and the second female gave birth to four white and three black pups. What are the likely genotypes of the male parent and the two female parents? Explain whether you are uncertain about any of the genotypes.
If B is the black allele, and b is the white allele, the male is bb, the first female is probably BB, and the second female is Bb. We are uncertain of the genotype of the first female. She could be Bb, although it is unlikely because she didn't produce any white pups out of a litter of eight.
Identical twins are produced from the same sperm and egg, whereas fraternal twins are produced from separate sperm and separate egg cells. If two parents with brown eyes (dominant trait) produce one twin boy with blue eyes, what are the following possibilities A. If the other twin is identical, he will have blue eyes. B. If the other twin is fraternal, he or she will have blue eyes. C. If the other twin is fraternal, he or she will transmit the blue eye allele to his or her offspring. D. The parents are both heterozygotes
If the other twin is identical, he will have blue eyes A. 100% because they are genetically identical. If the other twins is fraternal, he or she will have blue eyes B. Construct a Punnett square. We know the parents are heterozygotes because they produced a blue-eyed child. The fraternal twin is not genetically identical, but it has the same parents as its twin. The answer is 25%. If the other twin is fraternal, he or she will transmit C. The probability that an offspring inherits the allele is 50% and the probability that this offspring will pass it on to his/her offspring is also 50%. We use the product rule: (0.5)(0.5) = 0.25, or 25%. The parents are both heterozygotes D. Barring a new mutation during gamete formation, the chance is 100% because they must be heterozygotes in order to produce a child with blue eyes.
For Mendel's data for the experiment in Figure 2.8, conduct a chi square analysis to determine if the data agree with Mendel's law of independent assortment.
If we construct a Punnett square according to Mendel's laws, we expect a 9:3:3:1 ratio. Because a total of 556 offspring were observed, the expected numbers of offspring with different phenotypes are 556 × 9/16 = 313 round, yellow 556 × 3/16 = 104 wrinkled, yellow 556 × 3/16 = 104 round, green 556 × 1/16 = 35 wrinkled, green If we plug the observed and expected values into the chi square equation, we get a value of 0.51. With four categories, our degrees of freedom equal n - 1, or 3. If we look up the value of 0.51 in the chi square table (see Table 2.1), we see that it falls between the P values of 0.80 and 0.95. This means that the probability is between 80% and 95% that any deviation between observed results and expected results was caused by random sampling error. Therefore, we accept the hypothesis. In other words, the results are consistent with the law of independent assortment.
A species is diploid and has three chromosomes per set. Make a drawing showing what the chromosomes would look like in the G1 and G2 phases of the cell cycle.
In G1, there should be six linear chromosomes. In G2, there should be 12 chromatids that are attached to each other as 6 pairs of sister chromatids.
From the point of view of crosses and data collection, what are the experimental differences between a single-factor and a two-factor cross?
In a single-factor cross, the experimenter is only concerned with the outcome of a single trait. In a two-factor cross, the experimenter follows the pattern of inheritance for two different traits
How does the attachment of kinetochore microtubules to the kinetochore differ in metaphase of meiosis I compared to metaphase of mitosis? Discuss what you think would happen if a sister chromatid was not attached to a kinetochore microtubule.
In metaphase I of meiosis, each pair of chromatids is attached to only one pole via the kinetochore microtubules. In metaphase of mitosis, there are two attachments (i.e., to both poles). If the attachment was lost, a chromosome would not migrate to a pole and might not become enclosed in a nuclear membrane after telophase. If left out in the cytosol, it would eventually be degraded.
What is the difference between cross-fertilization and self-fertilization?
In the case of plants, cross-fertilization occurs when the pollen and eggs come from different plants while in self-fertilization they come from the same plant.
A true-breeding tall pea plant was crossed to a true-breeding dwarf plant. What is the probability that an F1 individual will be truebreeding? What is the probability that an F1 individual will be a true-breeding tall plant?
It is impossible for the F1 individuals to be true-breeding because they are all heterozygotes.
Describe the cellular differences between male and female gametes.
Male gametes are usually small and mobile. Animal and some plant male gametes contain flagella, which make them motile. The mobility of the male gamete makes it likely that it will come in contact with the female gamete. Female gametes are usually much larger and contain nutrients to help the growth of the embryo after fertilization occurs.
Why did Mendel's work refute the idea of blending inheritance?
Mendel's work showed that genetic determinants are inherited in a dominant/recessive manner. This was readily apparent in many of his crosses. For example, when he crossed two true-breeding plants for a trait such as height (i.e., tall versus dwarf), all the F1 plants were tall. This was not consistent with blending. Perhaps more striking was the result obtained in the F2 generation: 3/4 of the offspring were tall and 1/4 were short. In other words, the F2 generation displayed phenotypes that were like the parental generation. There did not appear to be a blending to create an intermediate phenotype. Instead, the genetic determinants did not seem to change from one generation to the next.
A eukaryotic cell is diploid and contains 10 chromosomes (5 in each set). In mitosis and meiosis, how many daughter cells will be produced, and how many chromosomes will each one contain?
Mitosis—two diploid cells containing 10 chromosomes each (two complete sets). Meiosis—four haploid cells containing 5 chromosomes each (one complete set)
Describe the significance of nonparental with regard to the law of independent assortment. In other words, explain how the appearance nonparental refutes a linkage hypothesis.
Offspring with a recombinant (nonparental) phenotype are consistent with the idea of independent assortment. If two different traits were always transmitted together as unit, it would not be possible to get recombinant phenotypic combinations. For example, if a true-breeding parent had two dominant traits and was crossed to a true-breeding parent having the two recessive traits, the F2 generation could not have offspring with one recessive and one dominant phenotype. However, because independent assortment can occur, it is possible for F2 offspring to have one dominant and one recessive trait.
In Morgan's experiments, which result do you think is the most convincing piece of evidence pointing to X-linkage of the eye color gene? Explain your answer.
Perhaps the most convincing observation was that all of the white-eyed flies of the F2 generation were males. This suggests a link between sex determination and the inheritance of this trait. Because sex determination in fruit flies is determined by the number of X chromosomes, this outcome suggests a relationship between the inheritance of the X chromosome and the inheritance of this trait.
List several ways in which telophase appears to be the reverse of prophase and prometaphase.
Prophase/Prometaphase Telophase Nuclear membrane dissociates. Nuclear membrane re-forms. Mitotic spindle forms. Mitotic spindle disassembles. Chromosomes condense. Chromosomes decondense. Chromosomes attach to spindle. Chromosomes detach from the spindle
What is a sister chromatid? Are sister chromatids genetically similar or identical? Explain
Sister chromatids are identical copies derived from the replication of a parental chromosome. They remain attached to each other at the centromere. They are genetically identical, barring rare mutations and crossing over with homologous chromosomes.
A pea plant that is heterozygous regarding seed color(yellow is dominant to green) can self-fertilize. What are the predicted outcomes of genotypes and phenotypes of the offspring?
The genotypes are 1 YY : 2 Yy : 1 yy. The phenotypes are 3 yellow : 1 green.
The fungus Melampsora lini causes a disease known as flax rust. Different strains of M. lini cause varying degrees of the rust disease. Conversely, different strains of flax are resistant or sensitive to the various varieties of rust. The Bombay variety of flax is resistant to M. lini-strain 22 but sensitive to M. lini-strain 24. A strain of flax called 770B is just the opposite; it is resistant to strain 24 but sensitive to strain 22. When 770B was crossed to Bombay, all F1 individuals were resistant to both strain 22 and strain 24. When F1 individuals were self-fertilized, the following data were obtained: 43 resistant to strain 22 but sensitive to strain 24 9 sensitive to strain 22 and strain 24 32 sensitive to strain 22 but resistant to strain 24 110 resistant to strain 22 and strain 24 Explain the inheritance pattern for flax resistance and sensitivity to M. lini strains.
The data are consistent with two genes (let's call them gene 22 and gene 24) that exist in two alleles each, a susceptible allele and a resistant allele. The observed data approximate a 9:3:3:1 ratio. This is the expected ratio if two genes are involved, and if resistance is dominant to susceptibility
A tall pea plant with axial flowers was crossed to a dwarf plant with terminal flowers. Tallness and axial flowers are dominant traits. The following offspring were obtained: 27 tall, axial flowers; 23 tall, terminal flowers; 28 dwarf, axial flowers; and 25 dwarf, terminal flowers. What are the genotypes of the parents?
The dwarf parent with terminal flowers must be homozygous for both genes, because it is expressing these two recessive traits: ttaa, where t is the recessive dwarf allele, and a is the recessive allele for terminal flowers. The phenotype of the other parent is dominant for both traits. However, because this parent was able to produce dwarf offspring with axial flowers, it must have been heterozygous for both genes: TtAa.
Explain the technical differences between a cross-fertilization experiment versus a self-fertilization experiment.
The experimental difference depends on where the pollen comes from. In self-fertilization, the pollen and eggs come from the same plant. In cross-fertilization, they come from different plants.
What is the genetic code?
The genetic code is the way in which the sequence of bases in RNA is read to produce a sequence of amino acids within a protein.
Describe the difference between genotype and phenotype. Give 3 examples. Is it possible for two individuals to have the same phenotype but different genotypes?
The genotype is the type of genes that an individual inherits while the phenotype is the individual's observable traits. Tall pea plants, red hair in humans, and vestigial wings in fruit flies are phenotypes. Homozygous, TT, in pea plants; a heterozygous carrier of the cystic fibrosis allele; and homozygotes for the cystic fibrosis allele are descriptions of genotypes. It is possible to have different genotypes and the same phenotype. For example, a pea plant that is TT or Tt would both have a tall phenotype.
A pea plant that is dwarf with green, wrinkled seeds was crossed to a true-breeding plant that is tall with yellow, round seeds. The F1 generation was allowed to self-fertilize. What types of gametes, and in what proportions, would the F1 generation make? What would be the ratios of genotypes and phenotypes of the F2 generation?
The genotype of the F1 plants is Tt Yy Rr. According to the laws of segregation and independent assortment, the alleles of each gene will segregate from each other, and the alleles of different genes will randomly assort into gametes. A Tt Yy Rr individual could make eight types of gametes: TYR, TyR, Tyr, TYr, tYR, tyR, tYr, and tyr, in equal proportions (i.e., 1/8 of each type of gamete). To determine genotypes and phenotypes, you could make a large Punnett square that would contain 64 boxes. You would need to line up the eight possible gametes across the top and along the side, and then fill in the 64 boxes. Alternatively, you could either the multiplication method or forked-line method described in Figure 2.11. The genotypes and phenotypes are as follows: 1 TT YY RR 2 TT Yy RR 2 TT YY Rr 2 Tt YY RR 4 TT Yy Rr 4 Tt Yy RR 4 Tt YY Rr 8 Tt Yy Rr = 27 tall, yellow, round 1 TT yy RR 2 Tt yy RR 2 TT yy Rr 4 Tt yy Rr = 9 tall, green, round 1 TT YY rr 2 TT Yy rr 2 Tt YY rr 4 Tt Yy rr = 9 tall, yellow, wrinkled 1 tt YY RR 2 tt Yy RR 2 tt YY Rr 4 tt Yy Rr = 9 dwarf, yellow, round 1 TT yy rr 2 Tt yy rr = 3 tall, green, wrinkled 1 tt yy RR 2 tt yy Rr = 3 dwarf, green, round 1 tt YY rr 2 tt Yy rr = 3 dwarf, yellow, wrinkled 1 tt yy rr = 1 dwarf, green, wrinkled
The arctic fox has 50 chromosomes (25 per set), and the common red fox has 38 chromosomes (19 per set). These species can interbreed to produce viable but infertile offspring. How many chromosomes would the offspring have? What problems do you think may occur during meiosis that would explain the offspring's infertility?
The hybrid offspring would have 44 chromosomes (i.e., 25 + 19). The reason for infertility is because each chromosome does not have a homologous partner. Therefore, the chromosomes cannot properly pair during metaphase I, and the gametes do not receive one copy of each homolog. Gametes will be missing certain chromosomes, which makes them infertile.
A cross was made between two strains of plants that are agriculturally important. One strain was disease-resistant but herbicide-sensitive; the other strain was disease-sensitive but herbicide-resistant. A plant breeder crossed the two plants and then allowed the F1 generation to self-fertilize. The following data were obtained: F1 generation: All offspring are disease-sensitive and herbicide-resistant F2 generation: 157 disease-sensitive, herbicide-resistant 57 disease-sensitive, herbicide-sensitive 54 disease-resistant, herbicide-resistant 20 disease-resistant, herbicide-sensitive Total: 288 Formulate a hypothesis that you think is consistent with the observed data. Test the goodness of fit between the data and your hypothesis using a chi square test. Explain what the chi square results mean.
The hypothesis is that disease sensitivity and herbicide resistance are dominant traits and they are governed by two genes that assort independently. According to this hypothesis, the F2 generation should yield a ratio of 9 disease sensitive, herbicide resistant : 3 disease sensitive, herbicide sensitive : 3 disease resistant, herbicide resistant : 1 disease resistant, herbicide sensitive. Because there are a total of 288 offspring produced, the expected numbers would be 9/16 × 288 = 162 disease sensitive, herbicide resistant 3/16 × 288 = 54 disease sensitive, herbicide sensitive 3/16 × 288 = 54 disease resistant, herbicide resistant 1/16 × 288 = 18 disease resistant, herbicide sensitive x^2= 0.54 If you look up this value in the chi square table under 3 degrees of freedom, the value lies between the 0.95 and 0.80 probability values. Therefore, you expect a value equal to or greater than 0.54, at least 80% of the time, due to random sampling error. Therefore, you would accept the hypothesis.
With regard to question C13, how would the phenomenon of crossing over affect the results? In other words, would the probability of a gamete inheriting only paternal chromosomes be higher or lower? Explain your answer
The probability of inheriting only paternal chromosomes would be much lower because pieces of maternal chromosomes would be mixed with the paternal chromosomes. Therefore, inheriting a chromosome that was completely paternally derived would be unlikely
Albinism, a condition characterized by a partial or total lack of skin pigment, is a recessive human trait. If a phenotypically unaffected couple produce an albino child, what is the probability that their next child will be albino?
The parents must be heterozygotes, so the probability is 1/4.
Explain why the products of meiosis may not be genetically identical, whereas the products of mitosis are.
The products of meiosis have only one copy of each type of chromosome. For example, one human gamete may contain the paternally derived copy of chromosome 11, whereas a different gamete may contain the maternally derived copy of chromosome 11. These two homologs may carry different alleles of the same genes and therefore are not identical. In contrast, mitosis produces genetically identical daughter cells that have both copies of all the pairs of homologous chromosomes.
Do you know the genotype of an individual with a recessive trait and/or a dominant trait? Explain
The recessive phenotype must be a homozygote. The dominant phenotype could be either homozygous or heterozygous.
Identify the key events during meiosis that result in a 50% reduction in the amount of genetic material per cell
The reduction occurs because there is a single DNA replication event but two cell divisions. Because of the nature of separation during anaphase I, each cell receives one copy of each type of chromosome.
With regard to biological evolution, which of the following statements is incorrect? Explain why. a. During its lifetime, an animal evolves to become better adapted to its environment. b. The process of biological evolution has produced species that are better adapted to their environments. c. When an animal is better adapted to its environment, the process of natural selection makes it more likely that the animal will reproduce.
The statement in part A is not correct. Individuals do not evolve. Populations evolve because certain individuals are more likely to survive and reproduce and pass their genes to succeeding generations.
Most genes encode proteins. Explain how the structure and function of proteins produce an organism's traits
The structure and function of proteins govern the structure and function of living cells. The cells of the body determine an organism's traits
What is meant by the term diploid? Which cells of the human body are diploid, and which cells are not?
The term diploid means that a cell has two copies of each type of chromosome. In humans, nearly all of the cells are diploid except for gametes (i.e., sperm and egg cells). Gametes usually have only one set of chromosomes.
Wooly hair is a rare dominant trait found in people of Scandinavian descent in which the hair resembles the wool of a sheep. A male with wooly hair, who has a mother with straight hair, moves to an island that is inhabited by people who are not of Scandinavian descent. Assuming that no other Scandinavians immigrate to the island, what is the probability that a great-grandchild of this male will have wooly hair? (Hint: You may want to draw a pedigree to help you figure this out.) If this wooly-haired male has eight great-grandchildren, what is the probability that one out of eight will have wooly hair?
The wooly haired male is a heterozygote, because he has the trait and his mother did not. (He must have inherited the normal allele from his mother.) Therefore, he has a 50% chance of passing the wooly allele to his offspring; his offspring have a 50% of passing the allele to their offspring; and these grandchildren have a 50% chance of passing the allele to their offspring (the wooly haired man's great-grandchildren). Because this is an ordered sequence of independent events, we use the product rule: 0.5 × 0.5 × 0.5 = 0.125, or 12.5%. Because no other Scandinavians are on the island, the chance is 87.5% for the offspring being normal (because they could not inherit the wooly hair allele from anyone else). We use the binomial expansion equation to determine the likelihood that one out of eight great-grandchildren will have wooly hair, where n = 8, x = 1, p = 0.125, q = 0.875. The answer is 0.393, or 39.3%.
If a diploid cell contains six chromosomes (i.e., three per set), how many possible random arrangements of homologs could occur during metaphase of meiosis I?
There are three pairs of chromosomes. The number of different, random alignments equals 2n, where n equals the number of chromosomes per set. So the possible number of arrangements equals 23, which is 8.
A woman with an abnormally long chromosome 13 (and a normal homolog of chromosome 13) has children with a man with an abnormally short chromosome 11 (and a normal homolog of chromosome 11). What is the probability of producing an offspring that will have both a long chromosome 13 and a short chromosome 11? If such a child is produced, what is the probability that this child will eventually pass both abnormal chromosomes to one of his or her offspring?
There is a 1/2 chance that the mother will transmit her abnormal chromosome and a 1/2 chance that the father will. We use the product rule to calculate the chances of both events happening. So the answer is 1/2 × 1/2 = 1/4, or 25%. The probability that such a child will pass both chromosomes to an offspring is also 25% because that child had a 1/2 chance of passing either chromosome
The process of binary fission begins with a single mother cell and ends with two daughter cells. Would you expect the mother and daughter cells to be genetically identical? Explain why or why not.
They are genetically identical, barring rare mutations, because they receive identical copies of the genetic material from the mother cell.
What are the expected phenotypic ratios from the following cross: Tt Rr yy Aa × Tt rr YY Aa, where T = tall, t = dwarf, R = round, r = wrinkled, Y = yellow, y = green, A = axial, a = terminal; T, R, Y, and A are dominant alleles. Note: Consider using the multiplication method in answering this problem
This problem is a bit unwieldy, but we can solve it using the multiplication method. For height, the ratio is 3 tall : 1 dwarf. For seed texture, the ratio is 1 round : 1 wrinkled. For seed color, they are all yellow. For flower location, the ratio is 3 axial : 1 terminal. Thus, the product is (3 tall + 1 dwarf)(1 round + 1 wrinkled)(1 yellow)(3 axial + 1 terminal) Multiplying this out, the answer is 9 tall, round, yellow, axial 9 tall, wrinkled, yellow, axial 3 tall, round, yellow, terminal 3 tall, wrinkled, yellow, terminal 3 dwarf, round, yellow, axial 3 dwarf, wrinkled, yellow, axial 1 dwarf, round, yellow, terminal 1 dwarf, wrinkled, yellow, terminal
The technique known as DNA sequencing enables researchers to determine the DNA sequence of genes. Would this technique be used primarily by transmission geneticists, molecular geneticists, or population geneticists?
This would be used primarily by molecular geneticists, but it could also be used by transmission and population geneticists. The sequence of DNA is a molecular characteristic of DNA. In addition, as you will learn throughout this text, the sequence of DNA is interesting to transmission and population geneticists as well.
Nine-banded armadillos almost always give birth to four offspring that are genetically identical quadruplets. Explain how you think this happens.
To produce identical quadruplets, fertilization begins with one sperm and one egg cell. This fertilized egg then could divide twice by mitosis to produce four genetically identical cells. These four cells could then separate from each other to begin the lives of four distinct individuals. Another possibility is that mitosis could produce two cells that separate from each other. These two cells could then divide by mitosis to produce two pairs of cells, which also could separate to produce four individual cells.
At puberty, the testes contain a finite number of cells and produce an enormous number of sperm cells during the life span of a male. Explain why testes do not run out of spermatogonial cells.
To produce sperm, a spermatogonial cell first goes through mitosis to produce two cells. One of these remains a spermatogonial cell and the other progresses through meiosis. In this way, the testes continue to maintain a population of spermatogonial cells.
How long did it take Mendel to complete the experiment in Figure 2.5?
Two generations would take two growing seasons. About 1 and 1/2 years
. A recessive allele in mice results in an unusally long neck. Sometimes, during early embryonic development, the long neck causes the embryo to die. An experimenter began with a population of true-breeding normal mice and true-breeding mice with long necks. Crosses were made between these two populations to produce an F1 generation of mice with normal necks. The F1 mice were then mated to each other to obtain an F2 generation. For the mice that were born alive, the following data were obtained: 522 mice with normal necks 62 mice with long necks What percentage of homozygous mice (that would have had long necks if they had survived) died during embryonic development?
You would expect a ratio of 3 normal : 1 long neck. In other words, there should be 1/3 as many long-necked mice as normal mice. If you multiply 522 times 1/3, the expected value is 174. However, only 62 were observed. Therefore, it appears that 174 - 62, or 112, mice died during early embryonic development; 112 divided by 174 gives us the percentage that died, which equals 0.644, or 64.4%.
Explain the relationship between each of the following pairs of the genetic terms:
a. Gene and trait: A gene is a segment of DNA. For most genes, the expression of the gene results in the production of a functional protein. The functioning of proteins within living cells affects the traits of an organism. B. Gene and chromosome: A gene is a segment of DNA that usually encodes the information for the production of a specific protein. Genes are found within chromosomes. Many genes are found within a single chromosome. C. Allele and gene: An allele is an alternative version of a particular gene. For example, suppose a plant has a flower color gene. One allele could produce a white flower, while a different allele could produce an orange flower. The white allele and the orange allele are then two versions of the flower color gene. d. DNA sequence and amino acid sequence: A DNA sequence is a sequence of nucleotides. The information within a DNA sequence (which is transcribed into an RNA sequence) specifies the amino acid sequence within a protein.
The data with Figure 2.5 show the results of the F2 generation for seven of Mendel's experiments. Conduct a chi square analysis to determine if these data are consistent with the law of segregation. Follow the same basic chi square analysis used before. We expect a 3:1 ratio, or 3/4 of the dominant phenotype and 1/4 of the recessive phenotype. The observed and expected values are as follows (rounded to the nearest whole number):
χ2 = 2.15 *Due to rounding, the observed and expected values may not add up to precisely the same number. Because n = 14, there are 13 degrees of freedom. If we look up this value in the chi square table, we have to look between 10 and 15 degrees of freedom. In either case, we would expect the value of 2.15 or greater to occur more than 99% of the time. Therefore, we accept the hypothesis.
