4. cats have a gene that codes for the color of their fur. suppose that a certain breed of cat can have black, gray, or white fur. black fur is dominant, white fur is recessive, and gray fur is intermediate (i. e., cats with gray fur possess one allele for black fur and one for white fur). if a gray cat and a white cat have kittens, what phenotypes could the kittens exhibit •
a) black, gray, or white fur •
b) only gray fur •
c) black or white fur •
d) gray or white fur

D) gray or white fur

Explanation:

Suppose the gene for fur color has alleles B and b. Here, B imparts black fur while b gives white fur. Since B is incompletely dominant over b, the heterozygous genotype "Bb" would give intermediate gray fur. Hence, the cross between gray (Bb) and white cat (bb) would give progeny in 1 gray: 1 white ratio.

step 1: initiation

initiation is the beginning of transcription. it occurs when the enzyme rna polymerase binds to a region of a gene called the promoter. this signals the dna to unwind so the enzyme can ‘‘read’’ the bases in one of the dna strands. the enzyme is now ready to make a strand of mrna with a complementary sequence of bases.

step 2: elongation

elongation is the addition of nucleotides to the mrna strand. rna polymerase reads the unwound dna strand and builds the mrna molecule, using complementary base pairs. there is a brief time during this process when the newly formed rna is bound to the unwound dna. during this process, an adenine (a) in the dna binds to an uracil (u) in the rna.

step 3: termination

termination is the ending of transcription, and occurs when rna polymerase crosses a stop (termination) sequence in the gene. the mrna strand is complete, and it detaches from dna.