OBJECT To study the prepared pedigree charts of genetic traits such as rolling of tongue, blood groups, widows peak, colour blindness etc.

                            EXPERIMENT 09

OBJECT 

To study the prepared pedigree charts of genetic traits such as rolling of tongue, blood groups, widows peak, colour blindness etc. 

Requirements :-

Prepared pedigree charts on the genetic traits. 

PROCEDURE

Observe the given pedigree chart and write comment on it.

Problem 1

Rolling of tongue appears in the progeny due to recessive gene. Find out the possible genotype of the family members in the following pedigree.

Solution/Comments

The trait is present in the father parent due to presence of two recessive genes (1-2aa). The trait can appear in the progeny only when it becomes homozygous recessive. Since, only one of the progeny carries the trait, the mother parent must be heterozygous (test cross-Aa x aa = 50% heterozygous, 50% recessive), i.e., I-1 Aa II-1 is aa. Il-2, 11-3 and II-4 are heterozygous (test cross) and,therefore, Aa The cross between II-1 and her husband also produces one homozygous recessive (III-2 = aa). This is possible only if the outsider is heterozygous (Aa).Naturally III-1 is also heterozygous (Aa).II-3 is heterozygous (Aa). Her husband can be either heterozygous (Aa x Aa AA, 2Aa, aa) or homozygous dominant (Aa x AA 2AA, 2Aa). Since none of the progeny is with recessive rolling tongue the possibility is that the new entrant in the pedigree is homozygous dominant (AA). III-3, III-4, III-5 are either AA or Aa. 

Problem 2

In the pedigree given below, indicate whether the shaded symbols belong to dominant or recessive trait. Also give genotype of the whole pedigree.

Solution/Comments:-

Since the shaded symbol appear in all the offspring father must be homozygous dominant while the mother homozygous recessive (AA x aa = all Aa) because in all other cases this possibility is absent(Aa x aa = 2Aa + 2aa ; aa x AA = all Aa ; aa x Aa = 2aA + 2aa). All the member of II generation  will therefore be heterozygous ( Aa) this is further confirmed  by marriage of II-1 with homozygous recessive ( Aa x aa = aa + Aa) bears children of both the parental types. Marriage of II-3 with the homozygous recessive can produce both recessive and heterozygotes.