* doesn't allow me to use the words h0m0_zyg0us and h3ter0_zyg0us because apparently, it's inappropiate. smh.
Practice #4
If you want to produce the greatest percentage of speckled chickens, one parent has to be h0m0_zyg0us for one trait and the other must be h0m0_zyg0us for the other trait. Doing so will result in 100% of the offspring being speckled as they will receive one allele from dad and one allele from mom and since it's co-dominant, both alleles will show in the offspring.
Practice #3
a) In order for them to have a h0m0_zyg0us recessive (brown shell), both parents must contain the recessive allele which means that both parents are h3ter0_zyg0us (Hh) for both traits. This would also result in three offspring having a green shell and one offspring having a brown shell.
b)
Group S: we do not know the genotype of the first parent therefore there will be two possible answers. Having a h0m0_zyg0us dominant (HH) and h0m0_zyg0us recessive (hh) parent will lead to all the offspring having a green shell as all of them will contain the dominant allele. Having a h3ter0_zyg0us (Hh) parent and a h0m0_zyg0us recessive (hh) parent will result in half the offspring having a green shell and the other half having a brown shell.
Group T: since both parents are h0m0_zyg0us recessive, all offspring will be brown as they will only be able to receive a recessive allele from both parents.
Practice #2
The answer is similar to Practice #3, Question B, Group S. A) will have two red roses and two white roses while B) will have all red roses.
Practice #1
a) Brown fur is the dominant phenotype.
b) The male will be h0m0_zyg0us recessive (bb) as it has black fur while the female will be h0m0_zyg0us dominant (BB) as it has brown fur.
