Genetic problems please????????????????

Question

Genetic problems please????????????????
16. In watermelons, solid green (G) color is dominant over striped pattern (g), and short shape (S) is dominant over long shape (s). What are the genotype and the phenotype of the offspring for a striped homozygous short watermelon and a green, short watermelon that’s heterozygous for both traits?
Answer is ggS_ and GgSs
So my question how do we know if the first one is ggSs or ggSS. They did not even tell us?

how in this world am i going to find out if dominant gens are homo or herto pleaseeeee help me

Answer 1

I believe ecolink has it correct. The question is vague, but it appears to say that the parent ggS_ is homozygous for striped *and* short. The question clearly says striped is homozygous, but it is unclear if the adjective is to be applied to short also. If you assume that ggS_ is meant to be homozygous and since striped is g and short is S, then the genotype must be ggSS. However, if the question is meant to be intentionally vague, then you have to do the crosses with the underscore designating the unknown allele. This is more in line with real experiments where you are trying to find out the genotype of the parent. So, you have to look at your phenotypes in the offspring from the cross to determine the parental genotype. If they gave you some of the offspring genotypes or phenotypes, then you can use those to determine the parental genotype. Good luck!

Answer 2

The problem says that the first parent is striped, homozygous short. Striped can only be gg; homozygous short is SS; so first parent is ggSS. You already have the second parent, GgSs.

Answer 3

1) genotype = ggSS to have striped it has to be homozygous gg. and since it says homozygous short...then you have ss not SS

2) GgSs short and heterozygous for both

Answer 4

options-blowing question. Leah is partly maximum options-blowing and on route notwithstanding the bigger challenge arises at the same time as purely one determine is a service. In that situation a field attempt will not in any respect paintings and the breeder has basically accomplished extra to damage their breed through passing on a mutation 50% of the time without threat of detecting it. that is very real of universal sires who're used very many times and/ or through use of man made insemination the position their genes are disseminated in an somewhat severe frequency between something of the inhabitants. "field" attempting out is likewise each and every from time to time referred to as "attempt" matings. that is an somewhat negative and unacceptable technique for figuring out inherited issues even before there have been any DNA tests accessible. Breeders will not in any respect get rid of inherited ailments with attempt matings for the easy reason that there are this style of large volume of mutations and... you won't be able to continuously change into attentive to those that are distributors. also, if a service to service attempt mating takes position then the prospect of manufacturing an affected animal is 25% so the prospect nonetheless exists that an affected isn't created from each and every mating, now to not teach through that factor it really is too previous due... so why blindly mate without understanding even if a mutation exists with the prospect of propagating extra of a similar challenge? that is a stupid mind-set noticeably now at the same time as there are this style of large volume of DNA tests and well being monitors accessible. the challenge is not basically with the situation "what if both father and mother are distributors of a recessive ailment" yet extra so "what if purely one determine is a service." it really is at the same time as the challenge maintains to flow undetected.