Further information from another source:
Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A.
Answer:
Maternal meiosis II
Explanation:
Jill has the genotype Aa, and Jack has the genotype aa. Jack can only contribute the a, whereas Jill can contribute A or a. For the child to have 2 copies of the A allele and two copies of the a allele, that means the nondisjunction must have happened in the mother.
As for the stage of meiosis, non-disjunction in meiosis I means that homologous chromosomes fail to separate properly. This would mean that the child would inherit Aa from its mother and a from its father. This is not the case.
Non-disjunction in meiosis II means identical sister chromatids fail to separate properly, which means the child would inherit either aa from its mother, or AA from its mother, and a from its father. This could give the genotype AAa. Therefore, nondisjunction must have occurred in maternal meiosis II
<span> Wow, this is a big question, but here goes. Freckles q: 1) Since freckles is the dominant phenotype, this means that Olivia must have the recessive phenotype, which is only given by the genotype ff. FF and Ff would both give freckles because, again, freckles is dominant. So any genotype with F_ would give freckles. 2)Same answer as #1. 3) Since they only have f alleles to pass on, both the dad and the mom will pass on one f each to 100% of their children. So all will have genotype ff and no freckles.
Hairline q: Since the parents are heterozygous, this means that their genotype has one of each allele to give Hh. (Even though you said they are quadruplets, I'm going to assume they are fraternal (different DNA) b/c otherwise they'd always have the same genotype if they are identical). 1) Okay, so when you do the Punnett square for Hh x Hh, you'll get 1HH:2Hh:1hh. Since widow's peak is dominant, any genotype with H_ gives the widow's peak. So, there are 3 options from the Punnett square, so you'll have 3 of the kids with widow's peak. Theres a 3/4 chance that a child will have widow's peak from these parents. 2) Since straight hairline is recessive, you need hh to get this phenotype and there is only one option. So, there will only be one child with the straight hairline. (1/4 chance). 3) Homozygous dominant means that you have two of the dominant allele, so HH. Since there is only 1 option, there is only one child with this genotype.
Tongue q: Since rolling the tongue is a dominant trait and the parents both can't roll their tongues, they must have a homozygous recessive genotype for this to happen (remember in dominance, any genotype with a dominant allele will give the dominant phenotype), so they have tt. 1) Since again they can only pass on t alleles, the kids will all have tt, so no one can roll their tongues. 2) None of them are hybrids because there was no variety in the genotypes or anything. Both parents had tt, so they were same in genotype. 3) She will have tt because of the above stated reasons.
Dimples q: Since all four kids have dimples, the dominant phenotype, they must all have the genotype D_ (either Dd or DD). 1) Since Marcus is a hybrid, this means that he had parents that were DD x dd to give him the genotype Dd. Since he has a recessive allele d in his genotype, Olivia must have all dominant alleles to make sure that each child has at least one dominant D. So, she must have the genotype DD. 2) Since she is DD, the dominant alleles will make her have dimples.
Earlobe q: 1) Since the parents are EE both, a cross of EE x EE will give EE genotype children. So, all children have EE, this means they all have free earlobes. So the ratio is 100% free to 0% attached earlobes. 2)Homozygous means they have two of the same alleles. Since all of them have EE, 100% of them are homozygous.
PTC q: Marcus has genotype bb and Olivia has genotype Bb because she is heterozygous. 1) The cross of bb x Bb gives 1Bb:1bb, So, 1/2 can taste the paper, so 50% can taste. 2) Since Violet can't taste the paper, she must be recessive and have the genotype bb. Since both of the boys can taste, they must have the genotypes Bb. Since 1/2 is already Bb, Claudia must be bb to help create the 50% that can't taste in the kids. 3)So, 2 people out of the family can taste the paper. Even though Olivia has Bb and should be able to taste, she can't. So, only Jonas and Nathan can taste the paper.
Pheww... done.. Hope this helps! :)<span>
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:DI think it might be C. ,but i am not sure please comment if i am wrong or right also have a Great day! :D
The contents that a data in the food label is supported are
by having nutritional facts, example having the percent of carbohydrates
consumed in the food. Another thing, they often gave advice in milligrams when
having to label the nutritional fact such as having 1,300mg of calcium.
