Genetics Practice Problems Solutions

Here are the answers and explanations for the genetics review problems from class today. There will be five problems on the test.

1. Colorblindness is a sex-linked, recessive trait. Remember that sex-linked traits are on the X chromosome, which means that men only have one copy of the gene. In this example, we are told that the man has colored vision, so his genotype must be X^CY. The woman also has colored vision, so she could be X^CX^C or X^CX^c. Since one of her sons is colorblind, she must be X^CX^c, since sons inherit their X chromosome from their mothers.

2. This is a relatively simple dihybrid cross. If we use B for brown eyes and b for blue eyes; and H for brown hair and h for blonde hair, the genotype of the man is BbHH, and the woman is bbhh. Doing the cross, their children have a 50% chance of being BbHh and a 50% chance of being bbHh. Answering the question posed on the sheet, there is 0 chance their children will have blue eyes and blonde hair.

3. This is an example of a dihybrid cross where one of the genes displays incomplete dominance - red and white flowers give pink. We are crossing two F₁ plants, so the phenotype of both parents is TtRr. For this type of dihybrid cross with complete dominance, we would normally expect the 9:3:3:1 ratio. However, with incomplete dominance there is a new phenotypic class, since the heterozygous individuals are distinct from the homozygous dominant individuals. The expected phenotypic ratios are then 3 tall, red-flowered; 6 tall, pink-flowered; 3 tall, white-flowered; 1 dwarf, red-flowered; 2 dwarf, pink-flowered; and 1 dwarf, white-flowered.

4. A dihybrid cross with a lethal allele. If an individual is homozygous recessive for the l allele, they will not survive, and are not counted in the phenotypic ratios for the answer. The parental genotypes are LlBb and Llbb. After throwing out the individuals with the lethal gene combination, the phenotypic ratio in the offspring is 1 normal-legged, brown; 1 normal-legged,white; 2 deformed-legged, brown; and 2 deformed-legged, white.

5. Gene linkage. You absolutely must know how to analyze these types of data, and tell the difference between parental and recombinant phenotypes. The data that are presented are from a testcross on the F₁ generation. The genotypes for this cross are CcShsh crossed with ccshsh (remember, a test cross is always performed with a homozygous recessive individual). The phenotypes of the parents are colored, full seeds and colorless, shrunken seeds.

If we assume that these genes are going to follow the Mendelian laws of inheritance, we predict that the offspring would have equal numbers of the four possible phenotypic classes: colored, full; colored, shrunken; colorless, full; and colorless, shrunken. HOWEVER, that is NOT what the data show. Two of the phenotypic classes, colored, full seeds and colorless, shrunken seeds are MUCH more common than the other two. These two common phenotypes are called the PARENTAL phenotypes, since they resemble the parents of the cross. The other two phenotypes, which are much less common, are called RECOMBINANT phenotypes, since these gene combinations do not exist in the parental generation.

To calculate the map distance, we need to calculate the recombination frequency, which is simply the percentage of offspring that show recombinant phenotypes. For this problem, the answer is (515 + 489) / 8368 = 12%. This means that the genes are 12 map units apart.

6. This is an example of epistasis - one gene is influencing the expression of a second gene at a second location. In this case, dogs that are homozygous recessive for the e gene will be yellow, regardless of what alleles are at the location that determines pigment color (B for black and b for chocolate). The phenotypes of the parents are BbEe. Doing the cross results in a 9 black to 3 chocolate to 4 yellow labs.

7.

...........7..........3...................15..................5..........
-----/-----------/----------/---------------------------/-------------/---
.....b..........d...........a...........................c.............e

8. This is a simple incomplete dominance cross. The heterozygous individuals have green flowers. A cross of two green flowers gives results of 1 blue, 2 green and 1 yellow.

9. The genotype of the woman must be ii, since that is the only possibility for type O blood. Her baby, with type A blood, must have at least one i allele from the mother. Therefore, the babies genotype must be I^Ai. The I^A allele must come from the father. The only man with an I^A allele to contribute is man #2.

10. Pedigree A is an autosomal recessive trait. Pedigree B is a sex-linked trait because many more males exhibit the trait than females (7 vs 2). Pedigree C is a dominant autosomal trait. To differentiate recessive and dominant traits, there are a few things to look for. First is that recessive traits tend to skip generations. Look at generations I and III in pedigree A. For a dominant trait, at least one parent must exhibit the trait in order for it to be passed on to the offspring.

11. I treated this example like a regular dihybrid cross, but in this case the dominant allele will change based on the sex of the individual. The two parental genotypes are BbXX and BbXY. For female offspring, there will be 3 with hair for every 1 bald; and for males there will be 1 with hair for every 3 bald.