I II III
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I II III 2 3 4 5 6 Normal Vision Normal Hearing Blindness Normal Hearing Normal Vision Deafness Blindness Deafness 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? Problem 5 Problem Set 1 Fall 2009
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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. - PowerPoint PPT Presentation
Transcript of I II III
I II III
1 2 3 4 5 6 7
Normal VisionNormal Hearing
Blindness Normal Hearing
Normal VisionDeafness
BlindnessDeafness
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?
Problem 5Problem Set 1
Fall 2009
1. Determine the Genotype for II-1
I II III
Let B= normal vision, b= blindness D= normal hearing, d= deafness
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.)
XBdXbD
2. Determine the Female Offspring Possible for II-1 and II-2
XBdXbD XbDYX
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%.
XbD
XbD
XBd
Xbd
XBD
Parentals add to 82%
Considering only female offspring,II-2 passes his X chromosome.
Recombinantsadd to 18%
0.41 XbDXbD
0.41 XBdXbD
0.09 XbdXbD
0.09 XBDXbD
0.41
0.41
0.09
0.09
3. Determine the possible genotypes for III-3
0.820.090.41/0.41
III-3 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.
XbD
XbD
XBd
Xbd
XBD
Parentals add to 82%
Recombinantsadd to 18%
0.41 XbDXbD
0.41 XBdXbD
0.09 XbdXbD
0.09 XBDXbD
0.41
0.41
0.09
0.09
Probability this can happen
Total possible outcomes
Genotypes for III-3: 0.82 XBdXbD or 0.18 XBDXbD
0.180.090.09/0.41
3A. An easier way to determine the possible genotypes for III-3
I II III
XBdY
XBdXbD XbDY
0.18 XBD0.82 XBd XbDXbD
III-3 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.
4. Determine the probability that III-3 can have a blind and deaf son.
There are three events that must occur for III-3 to have a blind and deaf son:
1. She must have the XBdXbD genotype.2. 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.
3. The father must pass a Y chromosome.
0.0369 0.5 x 0.09 x 0.82Y)bdP(X
Y)passes P(dad x )bdX passes P(mom x )bDXBdX is P(momY)bdP(X
Possible gametes for XBdXbD
Parental: 0.41 XBd and 0.41 XbD
Recombinant: 0.09 Xbd and 0.09 XBD
Xbd is one of two recombinant gametes with a frequency of 0.09
5. Determine the possible genotypes for III-5
III-5 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.
XbD
XbD
XBd
Xbd
XBD
Parentals add to 82%
Recombinantsadd to 18%
0.41 XbDXbD
0.41 XBdXbD
0.09 XbdXbD
0.09 XBDXbD
0.41
0.41
0.09
0.09
0.820.090.41/0.41
0.180.090.09/0.41
Probability this can happen
Total possible outcomes
Genotypes for III-5: 0.82 XbDXbD or 0.18 XbdXbD
5A. An easier way to determine the possible genotypes for III-5
III-5 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.
I II III
XBdY
XBdXbD XbDY
XbD
XbD
0.18 Xbd
0.82 XbD
6. Determine the probability that III-5 can have a blind and deaf son.
There are three events that must occur for III-5 to have a blind and deaf son:
1. She must have the XbdXbD genotype.2. She must pass the Xbd chromosome to her offspring. The father
must pass a Y chromosome.
0.045 0.5 x 0.5 x 0.18Y)bdP(X
Y)passes P(dad x )bdX passes P(mom x )bDXbdX is P(momY)bdP(X
Includes 0.41 Xbd as parental and 0.09 Xbd as recombinant.
