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# Genetics problem HELP?

Topic: Different ways of writing a ratio
June 21, 2019 / By Meade
Question: Someone tried to answer it yesterday but I'm still not sure A creature has colorful shells and stores sucrose in its body. Two purebred strains of it are obtained. The first is dark brown with all dark brown progeny and the second is green with all green progeny. He crossed them to produce heterozygotes. When he crossed the heterozygotes, the 32 offspring were: 18 dark brown, 6 red, 6 green, 2 yellow 1) How many pairs of genes (not alleles) control shell color? 2) Write genotypes of parents and draw Punnett square for cross between heterozygotes. Ok, I think it is 2 pairs but not sure... Also, I am not sure what the parents would be and what each genotype in the square represents....big help needed.

## Best Answers: Genetics problem HELP?

Lauryn | 5 days ago
Ok, here's the answer: 1) 2 pairs GgBb x GgBb if you cross in all possible ways will produce 16 different phenotypes (which happens to be half of the 32 total offspring in your example): These 9 will produce the brown (brown requires at least one dominant gene in each of the pairs of genes) GGBB (1) GgBB (2) GGBb (2) GgBb (4) These 3 will produce either green or red (red or green requires at least one dominant gene in ONE of the pairs only) GGbb (1) Ggbb (2) These 3 will also produce either green or red (red of green requires at least one dominant gene in the OTHER pair of genes) ggBB (1) ggBb (2) This one will produce yellow (yellow, as the rarest, requires all recessive genes in both gene pairs) ggbb (1) Only include these 16 possible offspring in your punnett square. So, in your example, there were 32 kids, all you have to do is multiply each amount displayed here in this example by two to get 32 kids -- but remember, even with 32 kids or 64 kids or 128 kids, what matters is the ratio of different combinations of genes (there are only 16 possible combos). So, on to the parents. This is where it gets interesting. In order to produce heterozygous offspring in our example (GgBb), the parents could be 6 possible genotype combinations: GGBB (brown) and Ggbb (green) or GgBB (brown) and GGbb (green) or GgBB (brown) and Ggbb (green) or GGBb (brown) and Ggbb (green) or GgBb (brown) and Ggbb (green) or GgBb (brown) and GGbb (green) Your teacher is trying to trick you with that first sentence that says "all progeny were green" and "all progeny were brown." We have no clue who the original parents were mating with in order to produce these 100% green and 100% brown progeny. We also do not have any idea HOW MANY heterozygous offspring (GgBb) these parents had when they mated. Therefore, as long as it is POSSIBLE to produce heterozygotes at all, then that is a possible genotype. Hope that helped. (ps the person above me is wrong because they said the parents were BBgg (brown) and bbGG (green) then went on to say that a kids with BBgg genotype were red)
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Ok, here's the answer: 1) 2 pairs GgBb x GgBb if you cross in all possible ways will produce 16 different phenotypes (which happens to be half of the 32 total offspring in your example): These 9 will produce the brown (brown requires at least one dominant gene in each of the pairs of genes) GGBB (1) GgBB (2) GGBb (2) GgBb (4) These 3 will produce either green or red (red or green requires at least one dominant gene in ONE of the pairs only) GGbb (1) Ggbb (2) These 3 will also produce either green or red (red of green requires at least one dominant gene in the OTHER pair of genes) ggBB (1) ggBb (2) This one will produce yellow (yellow, as the rarest, requires all recessive genes in both gene pairs) ggbb (1) Only include these 16 possible offspring in your punnett square. So, in your example, there were 32 kids, all you have to do is multiply each amount displayed here in this example by two to get 32 kids -- but remember, even with 32 kids or 64 kids or 128 kids, what matters is the ratio of different combinations of genes (there are only 16 possible combos). So, on to the parents. This is where it gets interesting. In order to produce heterozygous offspring in our example (GgBb), the parents could be 6 possible genotype combinations: GGBB (brown) and Ggbb (green) or GgBB (brown) and GGbb (green) or GgBB (brown) and Ggbb (green) or GGBb (brown) and Ggbb (green) or GgBb (brown) and Ggbb (green) or GgBb (brown) and GGbb (green) Your teacher is trying to trick you with that first sentence that says "all progeny were green" and "all progeny were brown." We have no clue who the original parents were mating with in order to produce these 100% green and 100% brown progeny. We also do not have any idea HOW MANY heterozygous offspring (GgBb) these parents had when they mated. Therefore, as long as it is POSSIBLE to produce heterozygotes at all, then that is a possible genotype. Hope that helped. (ps the person above me is wrong because they said the parents were BBgg (brown) and bbGG (green) then went on to say that a kids with BBgg genotype were red)