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Deafness and red-green colorblindness in humans are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below. What are the possible genotypes for individual III-3 ? What is the probability for each possible genotype?

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- Deafness and red-green colorblindness in humans are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.
- What are the possible genotypes for individual III-3? What is the probability for each possible genotype?
- What is the probability that individual III-3 can have a colorblind son with deafness?
- What are the possible genotypes for individual III-5? What is the probability for each possible genotype?
- What is the probability that individual III-5 can have a colorblind son with deafness?

I

II

III

Problem 5

Problem Set 1Fall 2007

1 2 3 4 5 6 7

X are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.bD

XBd

1. Determine the Genotype for II-1Let B=color vision, b=colorblindness D= normal hearing, d=deafness

I

II

III

XBdY

II-1 must receive XBd from her father.She must receive an XD from her mother since she is not deaf.The X from her mother must also have Xb, since II-1 has colorblind children. (It is not possible or necessary to determine whether the chromosome from her mother is a parental or recombinant combination.)

Parentals are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.add to 68%

Recombinantsadd to 32%

2. Determine the Female Offspring Possible for II-1 and II-2X

XbD

XBd

XbDY

Considering only female offspring,

II-2 passes his X chromosome.

XbD

0.34

XbD

0.34 XbDXbD

0.34

XBd

0.34 XBdXbD

0.16 XbdXbD

0.16

Xbd

0.16 XBDXbD

0.16

XBD

II-1 can produce four different gametes.

Two are parentals: XbD and XBd. Two are recombinants: Xbd and XBD. Since the genes are 32 map units apart, the frequencies of recombinants must add to 32% and the parentals must add to 68%.

Probability this can happen are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.

Total possible outcomes

Parentals add to 68%

Recombinantsadd to 32%

3. Determine the possible genotypes for III-3XbD

0.34

XbD

0.34 XbDXbD

0.34

XBd

0.34 XBdXbD

0.16 XbdXbD

0.16

Xbd

0.16 XBDXbD

0.16

XBD

III-3 is colorblind so she can have only two of the four possible genotypes shown in the Punnett square. Use the probability rule of the chance that something can happen out of the total possible outcomes to normalize these probabilities so that the sum equals 1.

Genotypes for III-3: 0.68 XbDXbD or 0.32XbdXbD

0.68 X are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.bD

0.32 Xbd

XbD

3A. An easier way to determine the possible genotypes for III-3I

II

III

XBdY

XbD

XBd

XbDY

XbD

III-3 is female, so she must have received the XbD chromosome from her father. She is colorblind, so she must receive an Xb from her mother. This Xb can be part of a recombinant chromosome, Xbd, which should occur with 32% probability. Or it can be part of a parental chromosome, XbD,which should occur with 68% probability.

Includes 0.34 X are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.bd as parental and 0.16Xbd as recombinant.

4. Determine the probability that III-3 can have a colorblind son with deafness.- There are three events that must occur for III-3 to have a colorblind son with deafness:
- She must have the XbdXbD genotype.
- She must pass the Xbd chromosome to her offspring.
- The father must pass a Y chromosome.

Probability this can happen are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.

Total possible outcomes

Parentals add to 68%

Recombinantsadd to 32%

5. Determine the possible genotypes for III-5XbD

0.34

XbD

0.34 XbDXbD

0.34

XBd

0.34 XBdXbD

0.16 XbdXbD

0.16

Xbd

0.16 XBDXbD

0.16

XBD

III-5 has color vision and normal hearing so she can have only two of the four possible genotypes shown in the Punnett square. Use the probability rule of the chance that something can happen out of the total possible outcomes to normalize these probabilities so that the sum equals 1.

Genotypes for III-5: 0.68 XBdXbD or 0.32XBDXbD

0.32 X are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.BD

XbD

0.68 XBd

5A. An easier way to determine the possible genotypes for III-5I

II

III

XBdY

XbD

XBd

XbDY

XbD

III-5 is female, so she must have received the XbD chromosome from her father. She is not colorblind, so she must receive an XB from her mother. This XB can be part of a recombinant chromosome, XBD, which should occur with 32% probability. Or it can be part of a parental chromosome, XBd,which should occur with 68% probability.

X are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.bd is one of two recombinant gametes with a frequency of 0.16

6. Determine the probability that III-5 can have a colorblind son with deafness.- There are three events that must occur for III-5 to have a colorblind son with deafness:
- She must have the XBdXbD genotype.
- She must pass the Xbd chromosome to her offspring. This requires a recombination event and is one of four types of gametes she can produce.
- The father must pass a Y chromosome.

Possible gametes for XBdXbDParental: 0.34 XBd and 0.34 XbDRecombinant: 0.16 Xbd and 0.16 XBD

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