Passage 6
1. C
“B” expresses Trait X, black fur, while “b” expresses Trait Y, yellow fur. Black fur (“B”) is dominant over yellow fur (“b”), so even one “B” will produce a dog with black fur. To produce yellow fur, the genotype must be “bb”.
2. C
Since dominant gene “B” expresses Trait X, which is black fur, parents with phenotypes BB and Bb would not be able to produce offspring with yellow fur. To produce a recessive phenotype, both parents must possess at least one recessive gene (“b”).
3. C
Filling out a Punnett square with “Bb” and “bb” will reveal two combinations of “Bb” and two combinations of “bb”. Only genotype “bb” can produce Trait Y, so the probability of offspring with Trait Y is ½.
4. C
Two “B” genes will produce a black puppy. One “B” and one “b” gene will produce a black puppy because “B” is dominant over “b”. Only two recessive “b” genes can produce a yellow puppy.
5. D
If both parents have only recessive genes, then their offspring can only have recessive genes. Trait X is produced by the dominant gene “B”, which is not present in either parent.
6. B
There are three combinations that will result in a black puppy and only one that will result in a yellow puppy, so the ratio is 3:1.
7. B
Only the gene combination “bb” can produce Trait Y, yellow fur. Two parents with Bb phenotypes would have a ¼ chance of producing Trait Y. Therefore, if 8 puppies were produced there would be 2 with yellow fur.