Concept explainers
A heterozygous individual has a _______ for a trait being studied.
- a. pair of identical alleles
- b. pair of nonidentical alleles
- c. haploid condition, in genetic terms
Concept introduction: The pair of chromosomes where one of the pairs is of the maternal source and the other is of the paternal source that has genes in the same loci with variable alleles of the gene is called homologous chromosomes. The allele is the modification of the gene. The alleles are of two types, namely dominant and recessive.
Answer to Problem 1SQ
Correct answer: A heterozygous individual has a pair of nonidentical alleles for a trait being studied. Therefore, option b. is correct.
Explanation of Solution
Reasons for the correct statement:
Homozygous individuals have the same alleles on the gene loci either dominant homozygous or recessive homozygous. Heterozygous individuals have one dominant allele and one recessive allele. Thus, two different types of alleles of the same gene are present in the gene loci that are not identical to each other.
Option b. is given as “a pair of nonidentical alleles”
As heterozygous individuals possess different alleles of the same gene for the trait expressed, option b. is correct.
Reasons for the incorrect statements:
Option a. is given as “a pair of identical alleles”.
Homozygous individuals only have the same allele on both homologous chromosomes that is, the pair of identical alleles. Therefore, option a. is incorrect.
Option c. is given as “haploid condition, in genetic terms”.
Haploid condition refers to half the number of chromosomes. A heterozygous individual has two sets of chromosomes. Therefore, option c. is incorrect.
Hence, options a. and c. are incorrect.
Want to see more full solutions like this?
Chapter 13 Solutions
Biology: The Unity and Diversity of Life (MindTap Course List)
- A trait that is present in a male child but not in either of his parents is characteristic of inheritance. a. autosomal dominant b. autosomal recessive c. X-linked recessive d. It is impossible to answer this question without more informationarrow_forwardA mother who is blood type a B has a child who is a B also. A potential father‘s blood type O. A geneticist concludes that. A. He cannot be the father B. He is very likely to be the father C. No conclusion can be drawnarrow_forwardTay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forward
- Chromosomes undergo division during meiosis. Such chromosomal behaviors are consistent with Mendel's laws of a. Unit Characters b. None of these c. Dominance and Recessiveness d. Independent Assortment e. Secondary Nondisjunction Chromosomes undergo division during meiosis. Such chromosomal behaviors are consistent with Mendel's laws of a. Unit Characters b. None of these c. Dominance and Recessiveness d. Independent Assortment e. Secondary Nondisjunctionarrow_forwardMendel’s crossing of round-seeded pea plants with wrinkled-seeded pea plants resulted in progeny that all had round seeds. This indicates that the wrinkled-seed trait is Select one: a. rare b. abnormal c. dominant d. codominant e. recessivearrow_forwardpleasee answer item no. 4, 5, and 6arrow_forward
- A. heterozygous blue-eyed man is marrying a homozygous brown-ayed woman. Determine the ratio of the following from the offspring. B. Genotype C. Phenotypearrow_forwardAnswer all the questions about the problemarrow_forwardShaded individuals in this pedigree express a particular trait. Based on this pedigree, what is the mode of inheritance of the trait? O A. autosomal dominant O B. autosomal recessive OC sex-linked dominant D. sex-linked recessivearrow_forward
- Two parents without cystic fibrosis have a child with cystic fibrosis. The allele for cystic fibrosis is therefore _________ to the non-disease allele Group of answer choices A. codominant B. recessive C. dominant D. epistatic E. incompletely dominantarrow_forwardWebbed fingers is inherited as an X-linked disease An unaffected male marries an affected female. a. Draw a Punnett square of the possible offspring. b. List the phenotypes of the possible children c. Draw a pedigree that displays the inheritance in you Punnett squarearrow_forwardAlkaptonuria is a metabolic disorder in which affected people produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, buther brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. Q. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?arrow_forward
- Biology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning