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Genetics Problems of All Types

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Attached are some genetics problems that i have some trouble solving. Please solve these problems (showing work as to how you arrived at the solutions with the answers).

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Solution Summary

The solution assists with questions regarding genetics of all types. The solution is detailed and provides a step-by-step answer.

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Genetics Problems of All Types!

ABO Blood Types
Drosophila: Fruit Flies
Autosomal/Sex-Linkage
Monohybrids
Dihybrids
F1 and F2 genotypes and phenotypes
Incomplete Dominance

It's all here!

Please see the attached Word document for solution guide.
Question 4.4
An IAi genotype is an A blood type because the "A" allele is present. If an individual with this genotype mates with another individual with this genotype, you can get an ii genotype, an "O" blood type.

IA i
IA IAIA IAi
i IAi ii

If an IAi genotype ("A" blood type) mates with an IBi genotype ("B" blood type) like this, we can still produce the ii ("O" blood type) offspring.

IA i
IB IBIA IBi
i IAi ii

An AB child ( IBIA) cannot be produced from a cross between an "A" and a "O" blood type. Why not? Because there's no IB allele in any of the parents.

Likewise, an O child cannot be produced if one of the parents is AB. Why not? There needs to be an "i" allele from each parent to make an ii offspring. However, an AB parent is IAIB - clearly, there's no i allele there to give to the child.

An "A" blood type offspring can be produced from an cross between an AB and a B. See below:

IA IB
IB IBIA IBIB
i IAi IBi

Question 4.9
A red-flowered snapdragon is CRCR while a white one is CWCW.

CR CR
CW CWCR CWCR
CW CWCR CWCR
All the F1 end up being CWCR which is the heterozygote. These are all pink.

Now, let's cross these F1 pink flowered snapdragons and see what we get.
CW CR
CW CWCW CWCR
CR CWCR CRCR

¼ of the F2 are white (CWCW). ¼ of the F2 are red (CRCR). ½ of the F2 are pink (CWCR).

Now, let's cross the F1 to the red parent: CWCR x CRCR
CW CR
CR CWCR CRCR
CR CWCR CRCR
½ are pink (CWCR) and ½ are red (CRCR).

You should be able to do the cross between the F1 and the white parent: CWCR x CWCW. Just follow the patterns above and you should be fine!

Question 4.13
A short-haired cream guinea pig must be L? CyCw.
A long-haired white guinea pig must be ll CwCw.

Now, we know that a long-haired cream baby is produced. Therefore, the baby must be ll CyCw. In order to get this offspring, the genotype of the short-haired cream guinea pig parent L? CyCw must be Ll CyCw. How do we know? Because a long allele (l) must come from each parent in order to have a long haired child. Remember, long hair is recessive and is only expressed in the homozygote.
Therefore, we want to cross this child with the short-haired cream parent.
These are the genotypes of the parents in this cross: ll CyCw x Ll CyCw
I'll set it up for you. You can fill in the Punnett square and compute the ratios of the ...

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