Ma'am Bio

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PRACTICE (25 MINUTES)
1. Divide learners into groups of four.
2. Provide copies of four sample pedigrees. (See samples in Figure 2 at the end of this document.)
3. For each pedigree, provide questions for the group to answer I. Identify the mode of inheritance
II. Write down the genotypes of specific individuals III. Compute for the probability of having an affected offspring

A. Look at the family of IV-9 and IV-10. If the trait is dominant, is it possible for them to have an affected offspring?
(Answer: NO. If the trait is dominant, then unaffected individuals are homozygous recessive. Two recessive individuals CANNOT produce a dominant offspring.)
B. If the trait is recessive, is it also possible for IV-9 and IV-10 to have an unaffected offspring?
(Answer: YES. This can happen if both parents are
heterozygous for the trait, which means they can each give a recessive allele to produce a homozygous
recessive offspring.)
C. Based on your answers for a) and b), is the trait dominant or recessive?
(Answer: RECESSIVE)
D. Give the genotypes of the following:
i. IV-9 (Answer: Dd)
ii. IV-10 (Answer: Dd)
iii. V-1 (Answer: DD or Dd)
iv. I-1 (Answer: dd)
v. I-2 (Answer: Dd)
E. If IV-9 and IV-10 were to have another child, what is the probability that they will have an affected offspring? (Answer: 1/4 or 25% following the Mendelian ratio from a hybrid cross)
A. Is this trait dominant or recessive?
(Answer: RECESSIVE. If the trait were dominant, then individuals I-3 and I-4 are both homozygous recessive, which means they CANNOT have a dominant offspring.)
B. What are the most probable genotypes of I-3 and I-4? (Answer: Dd and Dd in order for each parent to be able to contribute a recessive allele to give rise to a recessive offspring.)
C. What are the most probable genotypes of II-4 and II-5? (Answer: Dd and Dd. Same reason as b.)
D. What is the probability that II-4 and II-5 will have another normal offspring?
(Answer: 75%. A hybrid cross will produce 75% dominant offspring and 25% recessive offspring.)

A. Is the trait dominant or recessive?
(Answer: DOMINANT. If the trait were recessive, then individuals I-1 and I-2 are homozygous recessive, and they CANNOT produce a dominant affected offspring.)
B. What are the most probable genotypes of I-2 and I-3? (Answer: Dd and Dd. Each parent must be heterozygous in order to give a recessive allele to produce a recessive unaffected offspring.)
C. What is the probability that II-2 is Dd?
(Answer: 1 or 100%. II-2, together with the homozygous recessive II-1, was able to produce homozygous
recessive unaffected offspring. This can only happen if II-2 also possesses a recessive allele, which means s/he is a heterozygote.)
D. What is the probability that II-1 and II-2 will have another normal offspring?
(Answer: 1/2 or 50%. Following the Mendelian cross of dd x Dd, there is a 50% probability of producing a homozygous recessive unaffected offspring.)
A. Is the trait dominant or recessive?
(Answer: DOMINANT. If the trait were recessive, then individuals I-3 and I-4 must be homozygous recessive, and they CANNOT produce a dominant offspring.)
B. What are the genotypes of I-1 and I-2?
(Answer: dd and dd. Since the trait is dominant, it follows that unaffected individuals are homozygous recessive.)
C. What is the probability that I-1 and I-2 will have an affected offspring?
(Answer: 0. Homozygous recessive individuals CANNOT produce an offspring with a dominant trait.)
D. What are the genotypes of I-3 and I-4?
(Answer: Dd and Dd. Each parent must have a recessive allele in order to produce a homozygous recessive offspring.) E. What is the probability that II-6 is Dd?
(Answer: 2/3. II-6’s parents are both heterozygotes. Following the Mendelian cross of Dd x Dd, the probabilities of
occurrence of phenotypes in this cross are 25% (1/4) DD, 50% (2/4) Dd, and 25% (1/4) dd, giving a ratio of 1:2:1. Since II-6 is already affected, then his phenotype is dominant. Therefore, the probability of II-6 being affected is 0. So instead of a ratio of 1:2:1, the ratio to be considered should now be just 1:2 (DD:Dd). The probability of II-6 being Dd should now be 2/3.)
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