1. From a mating between two normal winged drosophila, 27 dumpy winged and 79 normal winged offspring were produced.      i. What is the nature of the dumpy winged gene?      ii. What...

1. From a mating between two normal winged drosophila, 27 dumpy winged and 79 normal winged offspring were produced.
      i. What is the nature of the dumpy winged gene?
      ii. What were the parental genotypes?

 2. In a cross between a dumpy winged fly and one of the parents in the previous question, how many normal winged flies would be expected among 120 offspring?

3. Albinism, the inability to synthesize chlorophyll, is a recessive character in a number of green plant species.   If a tobacco plant known to be heterozygous for albinism is self-pollinated and 600 of its seeds are subsequently germinated, how many seedlings would be expected to be albino? And how many would be expected to have the parental genotype?

4. A man and his wife can both taste phenylthiourea.   They have four children, two of whom are unable to taste.   What are the parental genotypes?

Asked on by tchrome

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mvcdc | Student, Graduate | (Level 2) Associate Educator

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1. Mating two normal winged drosophila resulted to 27 dumpy- (n) and 79 normal-winged (N) offsprings. Approximately, we have 3/4 N and 1/4 n. This means that normal-winged is a dominant phenotype -dumpy-wingedness is recessive. This offspring ratio can only arise if both parents are heterozygous normal - their genotype is Nn. (Note Nn x Nn results to NN, Nn, Nn, nn - 3/4 normal, 1/4 dumpy)

2. Crossing a dumpy winged and a drosophila from number 1 means getting the offspring from the cross: Nn x nn. This results to Nn, Nn, nn, nn. Since normal is dominant, we have 50% normal and 50% dumpy. Hence, from 120 offsprings, we expect approximately 60 offsprings will be normal.

3. A heterozygous albino tobacco plant would have a phenotype of Nn (N being normal, n being albino). The cross we're concerned with is Nn x Nn. The results from this cross are NN, Nn, Nn, nn. Only 1/4 or 25% are albino, and 50% would have the parental genotype (Nn). (Note that 75% would have the parental phenotype - normal). Hence, from 600 offsprings we expect that approximately 150 will be albino and 300 will have the parental genotype, Nn. 

4. The ability to taste phenylthiourea is a dominant trait. (Let P be the ability to taste it, p for not having the ability). We know that 50% of the children have the pp genotype. This can only happen if the couple are both heterozygous dominant (Pp). (Note: if either the mom or the dad is homozygous dominant PP, the offsprings would be PP, PP, Pp and pp - hence only one would not be able to have the ability to taste). [You can read up on the third source if you're interested on this] 

Note that we are assuming Mendellian Inheritance. You can also use the punnet square to try to make sense of the cross and the 'counting' of the offpsrings. All these are referenced below.

Sources:

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tchrome's profile pic

tchrome | (Level 2) eNoter

Posted on

In the answer when you said, 'We know that 50% of the children have the pp genotype. This can only happen if the couple are both heterozygous dominant (Pp).'

With taster being dorminant, I dnt think there would be 50% taster and 50% non-taster. Crossing Pp x Pp results in 75% taster and 25% non-taster then PP x Pp would result in all offspring being tasters.

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