4957 Shares

Genetics problem HELP?

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 Answer

Best Answers: Genetics problem HELP?

Lauryn 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)
👍 278 | 👎 5
Did you like the answer? Genetics problem HELP? Share with your friends

We found more questions related to the topic: Different ways of writing a ratio


Lauryn Originally Answered: Genetics problem HELP?
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)
Lauryn Originally Answered: Genetics problem HELP?
The results are coming out like a typical dihybrid cross, in a 9:3:3:1 ratio. If you assume that there are two genes controlling shell color, and the original parental generations are BBgg or bbGG, then the original F1 generation would be all BbGg. Then, these individuals would mate, and their offspring (the F2 generation) would be (reading across on a Punnett square from left to right on the first row: BBGG (brown), BBGg (brown), BbGG (brown), BbGg (brown), on the second row: BBGg (brown), BBgg (red), BbGg (brown), Bbgg (red), on the third row: BbGG (brown), BbGg (brown), bbGG (green), bbGg (green), and on the bottom row: BbGg (brown), Bbgg (red), bbGg (green) and bbgg (yellow).

Jobeth Jobeth
The results are coming out like a typical dihybrid cross, in a 9:3:3:1 ratio. If you assume that there are two genes controlling shell color, and the original parental generations are BBgg or bbGG, then the original F1 generation would be all BbGg. Then, these individuals would mate, and their offspring (the F2 generation) would be (reading across on a Punnett square from left to right on the first row: BBGG (brown), BBGg (brown), BbGG (brown), BbGg (brown), on the second row: BBGg (brown), BBgg (red), BbGg (brown), Bbgg (red), on the third row: BbGG (brown), BbGg (brown), bbGG (green), bbGg (green), and on the bottom row: BbGg (brown), Bbgg (red), bbGg (green) and bbgg (yellow).
👍 120 | 👎 4

Jobeth Originally Answered: Biology/genetics questions?
- the enzyme helicase is responsible for breaking hydrogen bonds ligase for ligation and polymerase for synthesis - phosphate and deoxyribose play structural role , they should be stronger , the basses just store the genatic information
Jobeth Originally Answered: Biology/genetics questions?
the daddy has one X chromosome which has the mutant gene. the feminine has inherited this X chromosome from her father and one customary from her mom. one million. in case you're informed THAT THE OFFSPRING IS A BOY: To have a colorblind son, the husband's genotype would not rather make a distinction because of the fact the daddy will supply the Y chromosome. Then, there's a 50% risk that the son will inherit from her mom the mutant X chromosome, and 50% risk that the son will inherit from her mom the conventional X chromosome. Then looking as an entire: 50% risk to have a common son, and 50% risk to have a colorblind son. 2. in case you're informed THAT THE OFFSPRING IS a woman: in case you have been questioning approximately if that they had a daughter: right here, the husband's genotype concerns, because of the fact the daddy will supply the daughter the mutant X chromosome, and as quickly as back, 50% risk of inheriting a mutant X chromosome from the mummy, and 50% risk of inheriting the conventional X chromosome from the mummy. Then looking as an entire: in the event that they have a daughter, there will be 50% risk that the daughter is a service of the trait, and 50% risk that the daughter has the ailment. observing the vast image, in case you do not comprehend the intercourse of the offspring: 25% risk of a common boy 25% risk of a colorblind boy 25% risk of a service lady 25% risk of a colorblind lady

If you have your own answer to the question different ways of writing a ratio, then you can write your own version, using the form below for an extended answer.