The pedigrees have not been provided. They have been attached.
Answer:
Pedigree 1 : Autosomal recessive
Pedigree 2 : Autosomal dominant
Pedigree 3 : Autosomal recessive
Explanation:
It has been given that the traits are autosomal i.e. they are not controlled by X and Y genes. We have to further find out if they are inherited in recessive pattern (autosomal recessive) or in dominant pattern (autosomal dominant).
Pedigree 1: Out of the three generations, the trait is only observed in one. This usually happens in recessive inheritance as two recessive alleles must be together to express the condition and this event has less chances of occurring. Also, it is visible how normal parents in first generation had a daughter with the recessive trait. This is possible if the parents were heterozygous for the trait or "carriers". They contributed one recessive allele each in the next generation due to which the daughter showed the recessive phenotype. Thus, this is most probably a recessive condition.
Pedigree 2: Out of the four generations, the trait is expressed in three. This usually happens in dominant inheritance as only one allele is needed to express the trait so both dominant homozygous and heterozygous individuals will show it. Thus, this is most probably a dominant condition.
Pedigree 3: This trait also skipped generations which is a feature of recessive traits. As with pedigree 1, the trait disappeared in pedigree and reappeared when two recessive alleles came together again. Thus, this is most probably a recessive condition.
Human structure.........................
Answer:
See the answer below
Explanation:
a) The mode of inheritance of the bar mutant is X-linked and dominant.
Assuming that barness is represented by the allele B.
Bar female =
or 
Wild type male = 
Bar female x wild type male:
x 
All of the F1 flies are bar because a single B allele is required for the trait to be expressed.
Bar male x wildtype female
x 

All the female offspring are barred while all the males are wildtype barred resulting in a 50/50 (857/905 is approximately 50/50) ratio of barred/wildtype offspring.
b) Genotypes of each of the cross above, including the parents and the F1 generations.
1.
x 
2.
x 
