BIOLOGY 130 CHAPTER 12 STUDY QUESTIONS

BIOLOGY 130 CHAPTER 12 STUDY QUESTIONS Question 1 of 10 1.0/ 1.0 Points Determine what statements describe Huntington’s disease? Select all that apply. A. The disease is caused by a delayed action gene B. The diseases is autosomal dominant C. The disease is sex­linked so it is mostly seen in men D. Offspring have a 50% chance of inheriting the gene E. t is recessive lethal Question 2 of 10 1.0/ 1.0 Points What statement describes X­linked traits? Select all that apply. A. Women can be carriers because they can be heterozygous for the trait B. X­linked traits are more common in men C. Males are never carriers since they only receive one X chromosome D. Males inherit X­linked traits from their fathers E. Women only have to have one affected gene to show the trait Question 3 of 10 1.0/ 1.0 Points If genes are completely linked, all the offspring will be . A.dominant B.hybrids C.parental D.recessive E.recombinants Question 4 of 10 1.0/ 1.0 Points A and Z are independent events. The probability that A occurs is ¼. The probability that Z occurs is ¼. What is the probability that both will occur? A.1/4 B.1/8 C.1/3 D.1/2 E.1/16 Question 5 of 10 1.0/ 1.0 Points In a monohybrid cross from heterozygous parents, of the offspring would be what genotypes? Select all that apply. A. Homozygous recessive B. Homozygous dominant C. Heterozygotes D. Homozygous recessives E. Homozygous dominants F. Heterozygote Question 6 of 10 1.0/ 1.0 Points Sickle cell anemia is a recessive trait. A cross between two heterozygotes would produce what phenotypic ratio? A.1:2:1 B.3:1 C.2:2 D.1:3 E.4:0 Question 7 of 10 1.0/ 1.0 Points Low pitched male voices (RR) and a high pitched male voice (rr). Heterozygotes have a baritone voice. This is an example of . A.incomplete dominance B.codominance C.Mendelian genetics (dominance and recessive D.multiple alleles Question 8 of 10 1.0/ 1.0 Points If a cross for widow’s peak was conducted between a homozygous dominant and a heterozygote, it is expected that all organisms of the offspring will have widow’s peak. Even though all have the widow’s peak, they can . A. be phenotypically different B. have more than one allele from a parent for the trait C.be genotypically different D.have the exact same alleles Question 9 of 10 1.0/ 1.0 Points The allele for widows peak (H) is dominant for the allele for no widows peak (h). At a different gene locus, the allele for hitchhikers thumb (D) is dominant to the allele for non­hitchhikers thumb (d). A man is heterozygote for the traits and marries a woman who has no widows peak and is heterozygote for hitchhikers thumb. What is the man’s genotype? A.HHDD B.HhDd C.HHdd