Learning Curve Activity Chapter 3
A ______ is an inherited that determines a charateristic
gene
is the set of alleles possessed by an individual organism.
genotype
If a chi-square value is calculated to be 0.455 with one degree of freedom, what is the probability that t he difference between the observed and expected numbers of offspring is due to chance?
0.5
A monohybrid cross between plants heterozygous for tall and short alleles is expected to produce two phenotypic classes of offspring. How many degrees of freedom can we use in the chi-square test for this cross?
1
What would be the probability of obtaining the phenotype AB from a cross AABB x aabb?
1
What would be the probability of obtaining the phenotype Ab from a cross Aabb x aabb?
1/2
The probability of rolling a single die and obtaining a 2 is 1/6. The probability of rolling a single die and obtaining a 5 is 1/6. What is the probability of rolling a single die and obtaining either a 2 or a 5? -1/36 -1/4 -1/2 -1/3 -1/6
1/3
The probability of rolling a single die and obtaining a 2 is 1/6. What is the probability of rolling two dice and obtaining a 2 on one and a 5 on the other? -1/36 -1/4 -1/2 -1/3 -1/6
1/36
What would be the phenotypic ratio produced by a cross between AaBb and aabb? -1:1:1:1 -1:1 -3:1 -9:3:3:1 -15:1
1:1:1:1
A dihybrid cross between heterozygous individuals is expected to produce four phenotypic classes of offspring. How many degrees of freedom can we use in the chi-square test for this cross?
3
8. (Problem 7, part 1) Which of the following crosses would likely produce a genotypic ratio of 1:2:1? B. Aa × aa C. AA × Aa E. No correct answer is listed. A. Aa × Aa D. aa × aa
A. Aa × Aa
36. (Problem 19b) Figure 3.7 shows the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced if a tall F1 progeny is backcrossed to the tall parent? C. 1/2 tall and 1/2 short B. 3/4 tall and 1/4 short D. 1/4 tall and 3/4 short E. All short A. All tall
A. All tall
11. (Problem 8, part 2) Why was the chromosome theory of heredity important? D. It helped Mendel to understand the results of his crosses. A. It explained the biological basis for Mendel's principles of heredity. E. It aided in the discovery of the concept of dominance. B. It determined whether DNA or protein was the genetic material. C. It described the difference between genotype and phenotype.
A. It explained the biological basis for Mendel's principles of heredity.
20. (Problem 13, part 2) Which of the following organisms does not have the characteristics that would make it suitable for studies of the principles of inheritance? C. Arabidopsis thaliana, a plant D. Caenorhabditis elegans, a nematode E. Drosophila melanogaster, a fruit fly B. Saccharomyces cerevisiae, a yeast A. Loxodonta africana, an elephant
A. Loxodonta africana, an elephant
2. (Problem 2) What is the difference between genotype and phenotype? B. Phenotype refers to the sets of alleles found within an individual, whereas genotype refers to the physical expression of a trait. C. Genotype refers to the physical expression of the genes or sets of alleles found within an individual, whereas phenotype refers to the physical manifestation of a trait. A. Phenotype refers to the physical manifestation of a trait, whereas genotype refers to the genes or sets of alleles found within an individual. E. Genotype and phenotype are essentially two terms for the same thing. D. Genotype refers to the physical location of a gene on a chromosome, whereas phenotype refers to the total number of alleles found within an individual.
A. Phenotype refers to the physical manifestation of a trait, whereas genotype refers to the genes or sets of alleles found within an individual.
26. (Problem 17a) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W.E. Castle and J.C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W.E. Castle and J.C. Phillips, 1909, Science 30:312-313). Which of the following statements explains the results of this cross? C. The eggs from the ovary of the black coat guinea pig were likely fertilized by a black-coated male guinea pig before the transfer. D. The researchers failed to remove the original ovaries of the white female before adding the ovaries from the black guinea pig. E. The white female was heterozygous for the black trait, which explains how the offspring inherited the trait for black coats. A. The color of the offspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth. B. The black coats likely arose from mutations that occurred from the transplanting of the ovaries.
A. The color of the offspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth.
12. (Problem 9) According to the principle of independent assortment, A. alleles at different loci separate independently of each other. E. two alleles of each gene are present in every cell. B. two alleles at the same locus separate independently of each other. C. alleles on the same chromosome stay together during gamete formation. D. alleles on different chromosomes stay together during meiosis.
A. alleles at different loci separate independently of each other.
Mendel's law of independent assortment depends on the events of meiosis because: A. different chromosomes will segregate independently during anaphase I. B. different chromosomes will segregate independently during anaphase II. C. sister chromatids will segregate independently during anaphase I. D. sister chromatids will segregate independently during anaphase II. E. None of the above.
A. different chromosomes will segregate independently during anaphase I.
The law of independent assortment describes the behavior of traits: A. only if genes are on different chromosomes. B. only if genes are on the same chromosome. C. only if the genes affect the same traits. D. B and C. E. None of the above.
A. only if genes are on different chromosomes.
Mendel observed that the number of peas that had an expected phenotype was rarely exactly what was predicted based on his laws. This can be explained by: A. random use of pollen without regard to its genotype. B. experimental errors in counting. C. failures of meiosis to occur properly. D. contamination by pollen from other sources. E. None of the above.
A. random use of pollen without regard to its genotype.
A 1:2:1 genotypic ratio in the progeny is produced by which cross? Please choose the correct answer from the following choices, and then select the submit answer button. AA x aa Aa x Aa aa x aa Aa x aa AA x Aa
Aa x Aa
Red (A) is dominant to white (a). Two red individuals are crossed and some white progeny are produced. What are the most likely genotypes of the two red parents of this cross? -aa x aa -Aa x aa -AA x aa -Aa x Aa -AA x Aa
Aa x Aa
Which parents produce a 3:1 phenotypic ratio in a simple genetic cross with dominance? -Aa x Aa -AA x aa -Aa x aa -AA x Aa -aa x aa
Aa x Aa
A 1:1 genotypic ratio in the progeny is usually produced by which cross? Please choose the correct answer from the following choices, and then select the submit answer button. AaBb x aabb AA x aa Aa x aa Aa x Aa aa x aa
Aa x aa
A 9:3:3:1 phenotypic ratio is produced by which cross? -AaBb x AaBb -AaBb x aabb -AaBb x Aabb -AAbb x aaBB -Aa x Aa
AaBb x AaBb
The key indicators for applying the _____ rule of probability are the word either and or.
Addition
When does the principle of independent assortment (Mendel's second law) occur in mitosis and/or meiosis? -Prophase I of meiosis -Anaphase of mitosis -Anaphase I of meiosis -Prophase of mitosis -Anaphase of mitosis and Anaphase I of meiosis
Anaphase I of meiosis
Where in mitosis and/or meiosis does the principle of segregation (Mendel's first law) occur? Prophase I of meiosis -Anaphase of mitosis -Anaphase I of meiosis -Prophase of mitosis -Anaphase of mitosis and Anaphase I of meiosis
Anaphase I of meiosis
9. (Problem 7, part 2) Which of the following crosses would likely produce a genotypic ratio of 1:1? A. Aa × Aa B. Aa × aa C. AA × aa E. No correct answer is listed. D. aa × aa
B. Aa × aa
Punnett squares are NOT: A. useful for predicting outcomes of crosses. B. filled in to represent meiosis. C. set up to represent the sexual life cycle. D. accurate representations of patterns of inheritance. E. Actually, they are all of these.
B. filled in to represent meiosis.
37. (Problem 20) Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains 1/2 white kittens and 1/2 black kittens. When the black kittens are interbred, they produce all black kittens. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait and why? A. White is the likely the recessive trait because it did not appear when the black kittens were interbred. C. White and black are likely both recessive traits because when Sam was crossed with a black cat, both white and black kitten offspring were produced. D. White is likely the recessive trait because when Sam was crossed with a white cat, one-half the offspring were white, and one-half the offspring were black. E. No correct answer is listed. B. Black is the recessive trait because the black kitten offspring only produced black offspring when interbred, suggesting they are all homozygous for black coat color.
B. Black is the recessive trait because the black kitten offspring only produced black offspring when interbred, suggesting they are all homozygous for black coat color.
44. (Problem 24) Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait? D. All of the above are crosses that would allow you determine if your rat terrier is heterozygous or homozygous. B. Cross your rat terrier with a hairless rat terrier. A. Cross your rat terrier with a rat terrier with hair from the same litter as your rat terrier. C. Cross your rat terrier with a rat terrier with hair from a different litter than your rat terrier. E. Cannot be determined
B. Cross your rat terrier with a hairless rat terrier.
16. (Problem 11, part 2) What does the probability associated with a chi-square value indicate about the results of a cross? A. It indicates the likelihood that the phenotypic ratio observed is accurate. B. It indicates the likelihood that random chance produced the deviations of the observed numbers from the expected numbers. D. It indicates the likelihood that the genetic cross has been carried out correctly. E. All of the above are indications suggested by the probability associated with the chi-square value. C. It indicates the likelihood that the correct genetic explanation for the results observed has been determined.
B. It indicates the likelihood that random chance produced the deviations of the observed numbers from the expected numbers.
29. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type A Female parent= blood type B Kittens= 4 kittens with blood type A, 3 with blood type B B. Male I(A)i(B) × Female i(B)i(B) A. Male I(A)I(A) × Female i(B)i(B) D. Either A or B are possible parental genotypes. C. Male I(A)I(A) × Female I(A)i(B) E. No possible parental genotype is listed.
B. Male I(A)i(B) × Female i(B)i(B)
34. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type A Female parent= blood type B Kittens= 4 kittens with blood type A, 1 with blood type B C. Male i(B)i(B) × Female I(A)i(B) D. Male i(B)i(B) × Female i(B)i(B) A. Male I(A)I(A) × Female i(B)i(B) B. Male I(A)i(B) × Female i(B)i(B) E. No possible parental genotype is listed.
B. Male I(A)i(B) × Female i(B)i(B)
18. (Problem 12 part 2) The inheritance of red hair was discussed in the introduction to this chapter. At times in the past, red hair in humans was thought to be a recessive trait and at other times, it was thought to be a dominant trait. What features of heritance would red hair be expected to exhibit as a dominant trait? D. A mating between two red-haired parents would never produce children with non-red hair. A. Red hair would often appear in children whose parents lacked red hair. B. Red hair would appear only in children who had at least one parent with red hair. E. No correct answer is listed. C. A cross between two non-red-hair parents would produce some red-hair offspring.
B. Red hair would appear only in children who had at least one parent with red hair.
3. (Problem 3) What is the principle of segregation? Why is it important? E. The principle of segregation proposes that once genetic material is blended, it cannot be separated. It explains how the genotypic ratios of offspring are produced. B. The principle of segregation states that a diploid organism possesses two alleles for any one particular trait and that these alleles separate during the formation of gametes. It explains how the genotypic ratios in the haploid gametes are produced. D. The principle of segregation states that genes are located on chromosomes. It explains how genes are passed on from parent to offspring. C. The principle of segregation states that genes for different characteristics that are at different loci segregate independently of one another. It explains the phenotypic ratios seen in the diploid parents. A. The principle of segregation proposes that offspring are the result of blended genetic material from the parent and the genetic factors are not discrete units. It explains how the genotypic ratios in the diploid gametes are produced.
B. The principle of segregation states that a diploid organism possesses two alleles for any one particular trait and that these alleles separate during the formation of gametes. It explains how the genotypic ratios in the haploid gametes are produced.
27. (Problem 17b) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W.E. Castle and J.C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W.E. Castle and J.C. Phillips. 1909, Science 30:312-313). Give the genotype of the offspring from this cross. A. WW C. ww D. Both A and B are likely genotypes. B. Ww E. Both A and C are likely genotypes.
B. Ww
40. (Problem 22a part 2) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria, and her mother has normal metabolism. What is the genotype of Sally's brother and Sally's father? B. aa A. AA C. Aa D. Both A and C E. Both B and C
B. aa
Mendel's first law is described by all of the following EXCEPT: A. it is known as the law of segregation. B. it is known as the law of independent assortment. C. it states that we get two copies of every gene. D. it states that some alleles are dominant to others. E. it states that alleles assort randomly.
B. it is known as the law of independent assortment.
In applying a chi-square test, the square of the deviation is divided by the expected results. This means that: A. only positive numbers will be involved. B. larger sample sizes will produce more accurate results. C. smaller sample sizes will produce more accurate results. D. A and B. E. A and C.
B. larger sample sizes will produce more accurate results.
According to the multiplication rule of probability, the chance of crossing two pea plants that are heterozygous for height and have none of four offspring showing the recessive short phenotype is: A. ¼ × ¼ × ¼ × ¼. B. ¾ × ¾ × ¾ × ¾. C. (¼ × ¼ × ¼ × ¼) + (¾ × ¾ × ¾ × ¾). D. ¼ + ¼ + ¼ + ¼. E. None of the above.
B. ¾ × ¾ × ¾ × ¾.
A dibybrid cross is performed and sixteen offspring are counted. Four phenotypes are expected for this cross. Interestingly, only three of the phenotypic classes are found among the offspring. If a chi-square value for this cross is calculated to be 1.1, what is the probability that the difference between the observed and expected numbers of offspring is due to chance? -Between 0.5 and 0.1 -Between 0.9 and 0.5 -Less than 0.05 -About 0.9
Between 0.9 and 0.5
Predict which of the following hypothetical organisms would be the most useful for studying heredity. A) A mammal that gives birth to one offspring every year and feeds exclusively on a diet of beetles and plant nectar B) A plant that reaches sexual maturity after 5 years of growth and produces abundant fruits with 5 seeds C) An insect that reaches sexual maturity in 9 days and can lay 1000 eggs per week D) A virus that infects blue whales E) A microbe that cannot tolerate oxygen and requires extreme pressure to grow
C
35. (Problem 19a) Figure 3.7 shows the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced if a tall F1 progeny is backcrossed to the short parent? A. All tall D. 1/4 tall and 3/4 short B. 3/4 tall and 1/4 short E. All short C. 1/2 tall and 1/2 short
C. 1/2 tall and 1/2 short
39. (Problem 22a part 1) Alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria, and her mother has normal metabolism. What is Sally's genotype? C. Aa D. Both A and C A. AA B. aa E. No correct answer is listed.
C. Aa
13. (Problem 10, part 1) In which phase of meiosis are the principles of segregation and independent assortment at work? A. Prophase I C. Anaphase I B. Prophase II D. Telophase I E. Telophase II
C. Anaphase I
31. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type B Female parent= blood type A Kittens= 8 with blood type A C. Male i(B)i(B) × Female I(A)I(A) E. No possible parental genotype is listed. A. Male I(A)I(A) × Female i(B)i(B) D. Male i(B)i(B) × Female I(A)i(B) B. Male I(A)i(B) × Female i(B)i(B)
C. Male i(B)i(B) × Female I(A)I(A)
5. (Problem 5) What is the concept of dominance? A. The concept of dominance states that when two different alleles are present in a genotype, both alleles can be observed equally in the phenotype. B. The concept of dominance states that recessive alleles are only visible in the heterozygous condition. C. The concept of dominance states that when two different alleles are present in a genotype, only the dominant allele is observed in the phenotype. E. None of these statements accurately explains the concept of dominance. D. The concept of dominance states that an organism can only exhibit two alleles at a time.
C. The concept of dominance states that when two different alleles are present in a genotype, only the dominant allele is observed in the phenotype.
24. (Problem 15) Figure 1.1 (below) shows three girls, one of whom has albinism. Could the three girls shown in the photograph be sisters? Why or why not? B. No, the girls could not be sisters. Hopi parents never had more than one child, so the child with albinism could not have had any sisters. C. Yes, the girls could be sisters. Albinism is inherited as an autosomal recessive trait, so the children could be sisters if the parents were heterozygous. E. Not enough information is given to know whether or not they could possibly be sisters. A. No, the girls could not be sisters. The frequency of albinism is very rare, only 1 in 20,000 persons, so it is not likely that they are sisters. D. Yes, the girls could be sisters. Albinism is known to show up more often in males than females, so these children could be sisters since most of them have normal pigmentation.
C. Yes, the girls could be sisters. Albinism is inherited as an autosomal recessive trait, so the children could be sisters if the parents were heterozygous.
The test cross: A. is used to determine the genotype of the recessive individual. B. uses a homozygous dominant in the cross. C. gives results either 100% dominant or 1:1 dominant: recessive. D. A and C. E. None of the above.
C. gives results either 100% dominant or 1:1 dominant: recessive.
If we apply a branch diagram to a dihybrid cross between organisms heterozygous for both genes, we can calculate: A. the chance of an individual being recessive for both is ¼. B. the chance of an individual being recessive for both is 1/8. C. the chance of an individual being recessive for both is 1/16. D. the chance of an individual being recessive for both is 9/16. E. None of the above is correct.
C. the chance of an individual being recessive for both is 1/16.
A heterozygote is an individual: A. with two different genes for a trait. B. with one locus that has two traits. C. with two different alleles of a gene. D. whose phenotype is recessive. E. None of the above.
C. with two different alleles of a gene.
Which test is used to provide information about how well the observed values fit the expected values?
CHI-SQUARE TEST
41. (Problem 22b) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria, and her mother has normal metabolism. If Sally's parents have another child, what is the probability that this child will have alkaptonuria? A. 100% B. 25% D. 50% E. 0% C. 75%
D. 50%
42. (Problem 22c) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria, and her mother has normal metabolism. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria? B. 25% A. 100% C. 75% D. 50% E. 0%
D. 50%
In a monohybrid cross: A. only a single character is studied. B. parents differ from each other in only one trait. C. only one parent is hybrid; the other is pure-breeding. D. A and B. E. All of the above.
D. A and B.
33. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type A Female parent= blood type A Kittens= 10 kittens with blood type A A. Male I(A)I(A) × Female I(A)i(B) B. Male I(A)i(B) × Female I(A)I(A) E. No possible parental genotype is listed. C. Male I(A)I(A) × Female I(A)I(A) D. All of the above are possible genotypes for the parents.
D. All of the above are possible genotypes for the parents.
22. (Problem 14b) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the orange parent. B. Orange (Rr) C. Cream (rr) D. Both A and B E. Both B and C A. Orange (RR)
D. Both A and B
7. (Problem 6) The additive rule of probability B. indicates that the probability of a single mutually exclusive event can be determined by adding the probabilities of the two or more different ways in which this single event could take place. C. indicates that the probability of two events occurring together is the product of their probabilities of occurring independently. D. Both A and B are correct descriptions of the additive rule. E. Both A and C are are correct descriptions of the additive rule. A. allows for predicting the probability of a single event that can occur in two or more ways.
D. Both A and B are correct descriptions of the additive rule.
28. (Problem 17c) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W.E. Castle and J.C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W.E. Castle and J.C. Phillips, 1909, Science 30:312-313). What, if anything, does this experiment indicate about the validity of the pangenesis and the germ-plasm theories discussed in Chapter 1? A. The production of black guinea pig offspring suggests that the allele for black coat color was passed along to the offspring from the transplanted ovary, thus supporting the germ-plasm theory. D. Both A and B describe outcomes that indicate the validity of the germ-plasm theory but not the pangenesis theory. E. The experiment indicates nothing regarding the validity of either the pangenesis or germ-plasm theory. B. Because no white guinea pigs were produced, no white coat alleles traveled to the ovary and into the gametes of the white female, thus indicating pangenesis did not occur. C. The male white guinea pig likely had a greater contribution to the phenotypes of the offspring, thus invalidating the pangenesis theory.
D. Both A and B describe outcomes that indicate the validity of the germ-plasm theory but not the pangenesis theory.
43. (Problem 23) Suppose that you are raising Mongolian gerbils. You notice that some of your gerbils have white spots, whereas others have solid coats. What type of cross could you carry out to determine whether white spots are due to a recessive allele? C. Cross a gerbil with white spots to another gerbil with white spots. If this cross produces three-fourths gerbils with solid coats and one-fourth with white spots, then white spots are due to a recessive allele. B. Cross a gerbil with white spots to another gerbil with white spots. If this cross produces one-half gerbils with solid coats and one-half with white spots, then white spots are due to a recessive allele. D. Cross a gerbil with white spots to another gerbil with a solid coat. If this cross produces either all solid or one-half gerbils with solid coats and one-half with white spots, then white spots are due to a recessive allele. E. Cross a gerbil with white spots to another gerbil with a solid coat. If this cross produces only gerbils with white spots, then white spots are due to a recessive allele. A. Cross a gerbil with white spots to another gerbil with white spots. If this cross produces only gerbils with solid coats, then white spots are due to a recessive allele.
D. Cross a gerbil with white spots to another gerbil with a solid coat. If this cross produces either all solid or one-half gerbils with solid coats and one-half with white spots, then white spots are due to a recessive allele.
17. (Problem 12, part 1) The inheritance of red hair was discussed in the introduction to this chapter. At times in the past, red hair in humans was thought to be a recessive trait and, at other times it was thought to be a dominant trait. What features of heritance would red hair be expected to exhibit as a recessive trait? E. No correct answer is listed. B. Red hair would appear only in children who had at least one parent with red hair. C. Two red hair parents could have some offspring with non-red hair. A. Non-red hair would often appear in children whose parents had red hair. D. In matings between two red hair parents, all of the offspring would be expected to have red hair.
D. In matings between two red hair parents, all of the offspring would be expected to have red hair.
32. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type A Female parent= blood type A Kittens= 7 kittens with blood type A, 2 with blood type B A. Male I(A)I(A) × Female i(B)i(B) B. Male I(A)i(B) × Female i(B)i(B) C. Male i(B)i(B) × Female I(A)I(A) D. Male I(A)i(B) × Female I(A)i(B) E. No possible parental genotype is listed.
D. Male I(A)i(B) × Female I(A)i(B)
30. (Problem 18) In cats, blood type A results from an allele I(A) that is dominant over an allele i(B) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter. Male parent= blood type B Female parent= blood type B Kittens= 6 with blood type B B. Male I(A)i(B) × Female i(B)i(B) C. Male i(B)i(B) × Female I(A)i(B) E. No possible parental genotype is listed. A. Male I(A)I(A) × Female i(B)i(B) D. Male i(B)i(B) × Female i(B)i(B)
D. Male i(B)i(B) × Female i(B)i(B)
21. (Problem 14a) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. The F1 are intercrossed to produce the F2. What are genotypes of the parents, the F1, and the F2? C. Parents (RR × Rr); F1 (Rr); F2 (RR, Rr, and rr) D. Parents (RR × rr); F1 (Rr); F2 (RR, Rr, and rr) A. Parents (Rr × Rr); F1 (Rr); F2 (RR, Rr, and rr) E. Parents (RR × rr); F1 (Rr); F2 (RR and Rr) B. Parents (RR × RR); F1 (RR); F2 (RR)
D. Parents (RR × rr); F1 (Rr); F2 (RR, Rr, and rr)
15. (Problem 11, part 1) How is the chi-square goodness-of-fit test used to analyze genetic crosses? E. All of the above are ways in which the goodness-of-fit chi-square test is used to analyze genetic crosses. B. To determine whether the genetic cross has been carried out correctly C. To evaluate whether the correct genetic explanation for the results observed has been determined A. To determine the likelihood that the phenotypic ratio observed is accurate D. To evaluate the role of chance in causing deviations between observed and the expected numbers of offspring produced in a genetic cross
D. To evaluate the role of chance in causing deviations between observed and the expected numbers of offspring produced in a genetic cross
38. (Problem 21) In sheep, lustrous fleece results from an allele (L) that is dominant over an allele (l) for normal fleece. A ewe (adult female) with lustrous fleece is mated with a ram (adult male) with normal fleece. The ewe then gives birth to a single lamb with normal fleece. What are the likely genotypes of the two parents? B. ewe (ll); ram (LL) C. ewe (Ll); ram (Ll) E. The genotypes of the parents cannot be determined from the information provided. D. ewe (Ll); ram (ll) A. ewe (LL); ram (ll)
D. ewe (Ll); ram (ll)
10. (Problem 8, part 1) The chromosome theory of heredity states that D. genes are located on chromosomes. E. chromosomes are made of DNA and protein. B. genes are inherited from a common ancestor. C. chromosomes segregate during mitosis. A. DNA is the genetic material.
D. genes are located on chromosomes.
The dihybrid cross: A. was used by Mendel to develop the law of segregation. B. uses two parents, both of which are heterozygous. C. shows a 3:1 phenotypic ratio in the next generation. D. is a cross that involves two traits. E. None of the above.
D. is a cross that involves two traits.
Gregor Mendel used peas in his experiments for all of the following reasons EXCEPT: A. he could grow the plants all year in Moravia. B. he had a number of strains with obviously different traits. C. he could count a large number of experimental results. D. he could control the fertilization. E. Actually, all of these were important reasons to use peas.
E. Actually, all of these were important reasons to use peas.
1. (Problem 1) Why was Mendel's approach to the study of heredity so successful? B. The seven characteristics he chose to study were important because they exhibited only a few distinct phenotypes and did not show a range of variation. C. By looking at each trait separately and counting the numbers of the different phenotypes, he was able to detect mathematical ratios of progeny phenotypes. D. He adopted an experimental approach and applied the scientific method. From his observations, he proposed hypotheses that he was then able to test empirically. A. He chose to work with a plant that was easy to cultivate, grew relatively rapidly, and produced many offspring whose phenotype was easy to determine. E. All of the above
E. All of the above
(Problem 13, part 1) Which of the following is a characteristic of an organism that would make it suitable for studies of the principles of inheritance? A. Easy to grow and maintain B. Grows rapidly, producing many generations in a short period of time C. Produces large numbers of offspring D. Has distinctive phenotypes that are easy to recognize E. All of the above are characteristics of an organism that would make it well suited for genetic studies.
E. All of the above are characteristics of an organism that would make it well suited for genetic studies.
4. (Problem 4) How are Mendel's principles different from the concept of blending inheritance discussed in Chapter 1? A. According to Mendel's principles, alleles (or genetic factors) are discrete units that remain separated within an individual organism. B. According to the concept of blending inheritance, offspring result from a blending of their parents' genetic material. C. According to the concept of blending inheritance, once blended, the genetic material from each parent is no longer discrete units that remain separated within an individual organism. E. All of the above are differences between Mendel's principles and the concept of blending inheritance. D. According to the concept of blending inheritance, genetic material once blended cannot be separated in future generations.
E. All of the above are differences between Mendel's principles and the concept of blending inheritance.
The events of meiosis explain Mendel's law of segregation in that: A. meiosis is a reductional division. B. meiosis produces cells that are haploid. C. fertilization produces a new diploid cell from two haploid gametes. D. B and C. E. All of the above.
E. All of the above.
A. allows for predicting the probability of two or more independent events occurring together. B. indicates that the probability of a single mutually exclusive event can be determined by adding the probabilities of the two or more different ways in which this single event could take place C. indicates that the probability of two events occurring together is the product of their probabilities of occurring independently. E. Both A and C are correct descriptions of the product rule. D. Both A and B are correct descriptions of the product rule.
E. Both A and C are correct descriptions of the product rule.
23. (Problem 14c) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. Give the genotypes and phenotypes of a backcross between the F1 and the cream parent. B. Orange (Rr) C. Cream (rr) E. Both B and C A. Orange (RR) D. Both A and B
E. Both B and C
14. (Problem 10, part 2) In which phase of mitosis are the principles of segregation and independent assortment at work? A. Prophase C. Anaphase E. These principles do not apply to mitosis. B. Metaphase D. Telophase
E. These principles do not apply to mitosis.
A monohybrid cross is performed and many offspring of two expected phenotypes are observed. However, the observed numbers of offspring are different than the expected numbers of offspring. If a chi-square value for this cross is calculated to be 0.05, what is the probability that the difference between the observed and expected numbers of offspring is due to chance? -Less than 0.01 -Less than 0.05 -About 0.975 -Greater than 0.5
Greater than 0.5
Which of the following characteristics of an organism would make it the most difficult for studying heredity? -Production of many offspring -A variety of heritable phenotypes that can be easy to differentiate -Rapid growth to maturity -Large size, making them expensive to maintain
Large size, making them expensive to maintain
Gregor Mendel received broad scientific training during his early university studies. Which of the following subjects do you think provided Mendel with the necessary knowledge to analyze phenotypic ratios from his experiments?
Mathematics
Why did Gregor Mendel end his genetics experiments?
Mendel became too busy after he was elected abbot of his monastery.
Which genetic principle states that each individual possesses two alleles coding for a trait and that these alleles separate when gametes are formed?
Principle of segregation
If one parent had the genotype RrYy, what would be the genotype of the other parent in a dihybrid test cross?
Rryy
How could two genetically identical oak trees have different heights when tree height is a heritable trait? -The height of the trees may differ if one of them received more sunlight, water, and fertilizer than the other. -The seeds that gave rise to these two trees were generated from two different sources of pollen. -Genes do not contribute to the limits on the height of oak trees. -Mutations stunted the growth of one of the trees. -Independent assortment of alleles for height during meiosis can produce offspring with varying degrees of height at maturity.
The height of the trees may differ if one of them received more sunlight, water, and fertilizer than the other.
Where did Gregor Mendel acquire knowledge of the scientific method that he applied to his study of inheritance?
The university of vienna
Mendel's approach to the study of heredity was effective because Please choose the correct answer from the following choices, and then select the submit answer button. he adopted an experimental approach. he conducted studies on an appropriate model organism. he kept careful records. All of the above
all of the above
A _____ can be used to determine the phenotypes and expected proportions with complex crosses
branch diagram
A cross between parents that differ in two characteristics is a
dihybrid cross
An individual possessing two different alleles is a
heterozygote
A specific place on a chromosome occupied by an allele is called a
locus
Mendel developed his theory of inheritance by conducting crosses with which organism?
peas
The genetic principle that states that alleles at one locus do not affect the segregation of alleles at another locus is called the
principle of independent assortment
________ is the likelihood of the occurrence of a particular event, usually expressed as a proportion or ratio
probability
The principle of independent assortment is an extension of the principle of
segregation
When some factor other than chance is responsible for differences between observed and expected value
significant
A chi-square value is determined by calculating the _____ of all the squared differences between observed and expected and divide by the expected values.
sum
A chi-square test is calculated for the observed and expected numbers of progeny from a genetic cross. The probability associated with the chi-square value is 0.023. This is the probability that -the cross was carried out correctly. -chance produced the observed numbers of progeny. -the observed values are expected in this type of genetic cross. -the difference between observed and expected numbers is due to chance.
the difference between observed and expected numbers is due to chance.
