in the description,,,

Question

dogs have a special kind of spotted color, dominant in relation to black .in a homozygous state,a gene of black color has a lethal effect (puppies die in an embryonic state).from crossing two spotted dogs ,12 puppies were obtained

1.how many types of gametes can a spotted dog form

2, how many different phenotypes will live puppies have?

3.how many different genotypes will live puppies have?

4.how many puppies will be spotted ?

5 how many puppies can be homozygous?

Answer 1

It is not possible to answer most of these questions without making an assumption. The problem is that we have no way of knowing whether the spotted dogs were homozygous dominant or heterozygous. Even with a simplifying assumption, questions 1, 3 and 5 have multiple possible answers.

Let's assume that because 12 puppies were born from the cross (which is a large litter), both parents were not heterozygous. We might assume this because a heterozygote x heterozygote cross would produce, on average, 25% homozygous recessive embryos, which, because this is a lethal condition, would mean that there were 16 original embryos, a very large number. If this assumption is correct, then one parent could have been homozygous dominant and the other heterozygous, or both may have been homozygous dominant. From the information provided, there is no way to tell which of these is true. This means that there are multiple possible answers for some of the questions.

Because of the above, we have no way to answer question 1. A homozygous dominant spotted dog produces one type of gamete (spotted only), while a heterozygous spotted dog produces two types of gametes (50% spotted, 50% black). We don't know what genotype of spotted dog is being referred to in the question.

We also have no way of answering question 3. To see why, do two Punnet squares: one in which both parents are homozygous dominant, and one in which one parent is homozygous dominant and one is heterozygous. The crosses produce a different number of possible genotypes for the offspring.

Finally, we have no way of answering question 5. All could be homozygous, or half could be homozygous, depending upon the genotypes of the parents (see the results of the Punnet squares in the previous paragraph). If our initial assumption was wrong, it is also possible that 1/3 of the surviving puppies could be homozygous (do a Punnet square for a heterozygote x heterozygote cross and remove the homozygous recessive offspring that died as embryos).

I suspect that the question is misphrased, either in its original form or in the form transcribed here.