Unit 2 Review Questions
(Student textbook pages 281–5)
11. G1: rapid growth and cell activity; S: DNA synthesis and replication; G2: cell prepares for division
12. A karyotype is a picture of the set of chromosomes that an individual has. It is used to determine sex, diagnose monosomy and trisomy disorders or chromosome abnormalities.
13. same size, position of centromere, and banding patterns; they are not identical because each chromosome could carry diff erent alleles for the same genes
14. Haploid cells contain half the set of chromosomes. They are also called gametes. In humans they are found in the testes and ovaries. Diploid cells contain a full set of chromosomes, and are called somatic cells. They are found in all tissues except the germ tissues.
16. a. 2 pairs
b. 2 chromosomes
18. Dominant and recessive refer to alleles. Dominant alleles make proteins and are always expressed in phenotype. Recessive alleles do not make proteins and are only expressed when the individual has the homozygous recessive genotype.
20. Autosomal recessive inheritance refers to the inheritance of a recessive trait that is found on one of the autosomes (non-sex chromosomes). Cystic fibrosis is an autosomal recessive disorder. A person will only have CF if they are homozygous recessive for the trait.
25. Because males have one X chromosome, if they have the recessive allele it will always be expressed.
33. a. Round is dominant since all F1 off spring are round.
b. The data is very close to the Mendelian ratio. I would expect 75 percent of the off spring to be round and 75.8 percent were round. The percent may not be the same due to the small number of off spring observed.
34. Yes, if both parents are heterozygous for hairline
36. a. Both parents are PpTt.
b. PT, Pt, pT, pt
37. a. Dominant. If shaded was recessive, all off spring would have to be recessive.
b. If A = affected, I-1, I-2 are Aa, II-1 is AA or Aa (since both parents are heterozygous and the disorder-causing allele is dominant), and II-2 and II-3 are aa.
39. Baby 1 belongs to Mr. and Mrs. Jones. Baby 2 belongs to the Guttierez family. Mr. and Mrs. Guttierez cannot have a type O child as neither carries the recessive allele.
40. If the father is not hemophiliac there is no chance of a daughter having the disease. There is a 25 percent chance that the son is a carrier since the mother is a carrier.
48. Perform a test cross by crossing the unknown parent with a homozygous recessive parent (tt). One short off spring tells us that the parent is heterozygous:
t TT Tt
t Tt tt
No short off spring tells us the parent might not be heterozygous:
t TT Tt
t Tt Tt
58. a. As certain traits were bred into the breed, natural variation was decreased. Perhaps by increasing the jaw strength of a dog, you also decrease the hip stability because the genes are linked. Some breeders breed to close relatives, which increases the chance for disorders to be exhibited.
b. Breed with dogs from different areas of the country or world. Make sure that you are not breeding close relatives.
63. Knowing your genetic profile may enable you to seek treatment, change your lifestyle to reduce risk, or make an informed decision on the risk of having children. However, this knowledge could reduced your freedom, create discrimination (in hiring, getting insurance, or finding a life partner), and increase stress. These implications show that society needs to consider how the information can and should be used.
Unit 2 Self-Assessment Questions
(Student textbook pages 286–7)
12. The cells that result from meiosis merge during fertilization to produce the first cell (zygote) of an organism. This cell will carry all errors that occurred during meiosis and is the foundation for the entire organism. Mitosis begins and continues for the life of the organism. Only the cells that divide from a cell that has a mistake in mitosis will carry that error.
22. Th e Rousseaus are parents of Baby 1; Baby 2 belongs to the Sakic family.
23. a. Let X m = Duchenne muscular dystrophy allele and let X M = normal allele; female carrier X M X m ,
normal male X M Y, aff ected male X m Y
c. No. The father must have the recessive allele on his X chromosome for him to father a daughter with the disorder (she must be homozygous for the allele). The father is normal and therefore has the dominant allele.