A black guinea pig is mated to a white guinea pig, and they produce 12 black offspring Then, the same white guinea pig is mated to a different black guinea pig, and they produce 7 black and 5 white offspring. What is the best explanation for the difference between these two matings? In your answer, you must include genotypes for the parents and offspring of each cross.

Human sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?

You decide to breed a pair of guinea pigs, one black and one white. In guinea pigs, black fur is caused by a dominant allele (B) and white is due to homozygosity for a recessive allele (b) at the same locus. Your guinea pigs have seven offspring, four black and three white. What are the genotypes of the parents? Why is there a 1:1 ratio in this cross?

You want to determine whether genes a and b are linked. What cross would you use and why? How would this cross tell you if they are linked?

Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?

Imagine that you are performing a cross involving seed color in garden pea plants. What traits would you expect to observe in the F1 offspring if you cross true-breeding parents with green seeds and yellow seeds? Yellow seed color is dominant over green. a. only yellow-green seeds b. only yellow seeds c. 1:1 yellow seeds:green seeds d. 1:3 green seeds:yellow seeds

Imagine that you are performing a cross involving seed texture in garden pea plants. You cross true-breeding round and wrinkled parents to obtain F1 offspring. Which of the following experimental results in terms of numbers of plants are closest to what you expect in the F2 progeny? a. 8lOroundseeds b. 8lOwrinkledseeds c. 405:395 round seeds:wrinkled seeds d. 610:190 round seeds:wrinkled seeds

As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny

The following pedigree shows the pattern of inheritance of red-green color blindness in a family. Females are shown as circles and males as squares; the squares or circles of individuals affected by the trait are filled in black. What is the chance that a son of the third-generation female indicated by the arrow will be color blind if the father is not color blind? If he is color blind?

Go back to Problem 1, and assume you now study a third gene having alleles C and c. For each genotype listed, what type(s) of gametes can be produced? a. AABBCC b. AaBBcc c. AaBBCc d. AaBbCc