Education.com
Try
Brainzy
Try
Plus

Patterns of Inheritance Practice Problems

based on 2 ratings
By — McGraw-Hill Professional
Updated on Aug 19, 2011

Review the following concepts if needed:

Patterns of Inheritance Practice Problems

Practice 1

Black coat of guinea pigs is a dominant trait; white is the alternative recessive trait. When a pure black guinea pig is crossed to a white one, what fraction of the black F2 is expected to be heterozygous?

Solution 1

As shown in Example 2.9, the F2 genotypic ratio is 1 BB : 2 Bb : 1 bb. Considering only the black F2, we expect 1 BB : 2 Bb or two out of every three black pigs to be heterozygous; the fraction is 2/3.

Practice 2

A dominant gene b + is responsible for the wild-type body color of Drosophila; its recessive allele b produces black body color. A testcross of a wild-type female gave 52 black and 58 wild type in the F1. If the wild-type F1 females are crossed to their black F1 brothers, what genotypic and phenotypic ratios would be expected in the F2? Diagram the results using the appropriate genetic symbols.

Solution 2

The dash in the genotype of the parental female indicates that one allele is unknown. Since the recessive black phenotype appears in the F1 in approximately a 1 : 1 ratio, we know that the female parent must be heterozygous b + b. Furthermore, we know that the wild-type F1 progeny must also be heterozygous. The wild-type F1 females are then crossed with their black brothers:

The expected F2 ratio is therefore the same as that observed in the F1, namely, 1 wild type : 1 black.

Practice 3

If a black female guinea pig is testcrossed and produces two offspring in each of three litters, all of which are black, what is her probable genotype? With what degree of confidence may her genotype be specified?

Solution 3

The female parent could be homozygous BB or heterozygous Bb and still be phenotypically black; hence, the symbol B–. If she is heterozygous, each offspring from this testcross has a 50% chance of being black. The probability of six offspring being produced, all of which are black, is (1/2)6 = 1/64 = 0.0156 = 1:56%. In other words, we expect such results to occur by chance less than 2% of the time. Since it is chance that operates in the union of gametes, she might actually be heterozygous and thus far only her B gametes have been the "lucky ones" to unite with the b gametes from the white parent. Since no white offspring have appeared in six of these chance unions, we may be approximately 98% confident (1 – 0:0156 = 0:9844 or 98.44%), on the basis of chance, that she is of homozygous genotype (BB). It is possible, however, for her very next testcross offspring to be white, in which case we would then become certain that her genotype was heterozygous Bb and not BB.

View Full Article
Add your own comment