1. The I gene determines the synthesis of a repressor molecule, which blocks expression of the lac operon and which is inactivated by the inducer. The presence of the repressor I+ will be dominant to the absence of a repressor I–. Is mutants are unresponsive to an inducer. For this reason, the gene product cannot be stopped from interacting with the operator and blocking the lac operon. Therefore, Is is dominant to I+.

2. Oc mutants are changes in the DNA sequence of the operator that impair the binding of the lac repressor. Therefore, the lac operon associated with the Oc operator cannot be turned off. Because an operator controls only the genes on the same DNA strand, it is cis (on the same strand) and dominant (cannot be turned off). 3. a. You are told that a, b, and c represent lacI, lacO, and lacZ, but you do not know which is which. Both a– and c– have constitutive phenotypes (lines 1 and 2) and therefore must represent mutations in either the operator (lacO) or the repressor (lac I). b– (line 3) shows no ß-gal activity and by elimination must represent the lacZ gene. Mutations in the operator will be cis-dominant and will cause constitutive expression of the lacZ gene only if it’s on the same chromosome. Line 6 has c– on the same chromosome as b+ but the phenotype is still inducible (owing to c+ in trans). Line 7 has a– on the same chromosome as b+ and is constitutive even though the other chromosome is a+. Therefore a is lacO, c is lacI, and b is lacZ. b. Another way of labeling mutants of the operator is to denote that they lead to a constitutive phenotype; lacO– (or a–) can also be written as lacOc. There are also mutations of the repressor that fail to bind inducer (allolactose) as opposed to fail to bind DNA. These two classes have quite different phenotypes and are distinguished by lacIs (fails to bind allolactose and leads to a dominant uninducible phenotype in the presence of a wild-type operator) and lacI– (fails to bind DNA and is recessive). It is possible that line 3, line 4, and line 7 have lacIs mutations (because dominance cannot be ascertained in a cell that is also lacOc) but the other c– alleles must be lacI–. 4. ß-Galactosidase Permease Part No lactose Lactose No lactose Lactose a + + – + b + + – – c – – – – d – – – – e + + + + f + + – – g – + – + Chapter Ten 171

a. The Oc mutation leads to the constitutive synthesis of ß-galactosidase because it is cis to a lacZ+ gene, but the permease is inducible because the lacY+ gene is cis to a wild-type operator.

b. The lacP– mutation prevents transcription so only the genes cis to lacP+ will be transcribed. These genes are also cis to Oc so the lacZ+ gene is transcribed constitutively.

c. The lacIs is a trans-dominant mutation and prevents transcription from either operon. d. Same as part c. e. There is no functional repressor made (and one operator is mutant as well). f. Same as part b. g. Both operators are wild type and the one functional copy of lacI will direct the synthesis of enough repressor to control both operons. 9. The term epigenetic inheritance is used to describe heritable alterations in which the DNA sequence itself is not changed. Paramutation and parental imprinting are two such examples. 11. Imprinted genes are functionally hemizygous. Maternally imprinted genes are inactive when inherited from the mother, and paternally imprinted genes are inactive when inherited from the father. A mutation in one of these genes is dominant when an offspring inherits a mutant allele from one parent and a “normal” but inactivated allele from the other parent. 13. The inheritance of chromatin structure is thought to be responsible for the inheritance of epigenetic information. This is due to the inheritance of the histone code and may also include inheritance of DNA methylation patterns. 14. Many DNA-protein interactions are shared by prokaryotes and eukaryotes, but the mechanisms by which proteins bound to DNA at great distances from the start of transcription affect that transcription is unique to eukaryotes. Also, mechanisms of gene regulation based on chromatin structure are distinctly eukaryotic. 29. a. D through J — the primary transcript will include all exons and introns. b. E, G, I — all introns will be removed. c. A, C, L — the promoter and enhancer regions will bind various transcription factors that may interact with RNA polymerase.