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

2. Oc mutants are changes in the DNA sequence of the operator that impair the binding of the . Therefore, the lac operon associated with the Oc operator cannot be turned off. Because an operator controls only the 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 (lines 1 and 2) and therefore must represent 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 . Line 6 has c– on the same chromosome as b+ but the 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 () 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 cannot be ascertained in a that is also lacOc) but the other c– must be lacI–. 4. ß-Galactosidase Permease Part No Lactose No lactose Lactose a + + – + b + + – – c – – – – d – – – – e + + + + f + + – – g – + – + Chapter Ten 171

a. The Oc 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 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 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 . 9. The term epigenetic inheritance is used to describe heritable alterations in which the DNA sequence itself is not changed. 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 from one parent and a “normal” but inactivated allele from the other parent. 13. The inheritance of structure is thought to be responsible for the inheritance of epigenetic information. This is due to the inheritance of the code and may also include inheritance of DNA patterns. 14. Many DNA- interactions are shared by and , but the mechanisms by which 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 will include all and . b. E, G, I — all introns will be removed. c. A, C, L — the and regions will bind various transcription factors that may interact with RNA .