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# I II III - PowerPoint PPT Presentation

In humans, deafness ( d ) and blindness ( b , due to the disease retinitis pigmentosum) are determined by recessive alleles at X-chromosome loci that are 18 map units apart. Consider the pedigree shown below.

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Presentation Transcript
In humans, deafness (d) and blindness (b, due to the disease retinitis pigmentosum) are determined by recessive alleles at X-chromosome loci that are 18 map units apart. Consider the pedigree shown below.
• a. What are the possible genotypes for individual III-3? What is the probability for each possible genotype?
• b. What is the probability that individual III-3 can have a son who is both blind and deaf?
• c. What are the possible genotypes for individual III-5? What is the probability for each possible genotype?
• d. What is the probability that individual III-5 can have a son who is both blind and deaf?

I

II

III

Problem 5

Problem Set 1Winter 2011

1 2 3 4 5 6 7

XbD

XBd

1. Determine the Genotype for II-1

Let B= normal vision, b= blindness 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 blind children. (It is neither possible nor necessary to determine whether the chromosome from her mother is a parental or recombinant combination.)

X

XbD

XBd

XbDY

Considering only female offspring,

II-2 passes his X chromosome.

XbD

0.41

XbD

0.41 XbDXbD

0.41

XBd

0.41 XBdXbD

0.09 XbdXbD

0.09

Xbd

0.09 XBDXbD

0.09

XBD

II-1 can produce four different gametes.

Two are parentals: XbD and XBd. Two are recombinants: Xbd and XBD. Since the genes are 18 map units apart, the frequencies of recombinants must add to 18% and the parentals must add to 82%.

Probability this can happen

Total possible outcomes

3. Determine the possible genotypes for III-3

XbD

0.41

XbD

0.41 XbDXbD

0.41

XBd

0.41 XBdXbD

0.09 XbdXbD

0.09

Xbd

0.09 XBDXbD

0.09

XBD

III-3 is blind 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: 0.82 XbDXbD or 0.18XbdXbD

0.18 Xbd

XbD

0.82 XbD

I

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 18% probability. Or it can be part of a parental chromosome, XbD,which should occur with 82% probability.

Includes 0.41 Xbd as parental and 0.09Xbd as recombinant.
• There are three events that must occur for III-3 to have a blind and deaf son:
• 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

Total possible outcomes

5. Determine the possible genotypes for III-5

XbD

0.41

XbD

0.41 XbDXbD

0.41

XBd

0.41 XBdXbD

0.09 XbdXbD

0.09

Xbd

0.09 XBDXbD

0.09

XBD

III-5 has normal 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: 0.82 XBdXbD or 0.18XBDXbD

0.82 XBd

XbD

0.18 XBD

XbD

I

II

III

XBdY

XbD

XBd

XbDY

III-5 is female, so she must have received the XbD chromosome from her father. She has normal vision, so she must receive an XB from her mother. This XB can be part of a recombinant chromosome, XBD, which should occur with 18% probability. Or it can be part of a parental chromosome, XBd, which should occur with 82% probability.

• There are three events that must occur for III-5 to have a blind and deaf son:
• 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: 0.41 XBd and 0.41 XbDRecombinant: 0.09 Xbd and 0.09 XBD