Mendelian Genetics (Unit 5 Standard B)

In buffalo, being boring (B) is dominant to being busy (b). A true-breeding, busy male is crossed with a true-breeding, boring female.  What are the possible genotypes of their offspring?

• Only Bb

• BB and bb

• BB, Bb, and bb

• Only BB

Tags

NGSS.HS-LS3-1

In corn, smooth kernels (K) are dominant to wrinkled kernels (k).  A farmer crosses two corn plants that are heterozygous for kernel texture.  What are the odds that the offspring will have smooth kernels?

• 0/4

• 1/4

• 2/4

• 3/4

• 4/4

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-2

In a flowering plant, tall (T) is dominant to short (t), and blue flowers (B) is dominant to white flowers (b). A tall plant with white flowers (Ttbb) is crossed with a short plant with blue flowers (ttBb). What is the chance that the offspring will be short with white flowers?

• 3/16

• 1/4

• 1/2

• 1/8

• 3/4

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-2

NGSS.HS-LS3-3

The pedigree below tracks the presence of attached earlobes through a family's generation. Having attached earlobes is an autosomal recessive trait.  What is the genotype of individual I-1?

• ee

• Ee

• EE

• XeY

• XEY

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-2

The Russian blue is a rare breed of cat that is susceptible to developing cataracts on the eyes. Scientists hypothesize that this condition is inherited as a result of a mutation. Figure 1 shows a pedigree obtained in a study of cats owned by members of the Russian Blue Club in Sweden.  Based on the inheritance pattern shown in Figure 1, which of the following best predicts the nature of the original mutation?

• A recessive mutation on the X chromosome

• A recessive mutation on a somatic chromosome

• A dominant mutation on the X chromosome

• A dominant mutation on a somatic chromosome

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-2

In fruit flies, purple eyes and ebony body are traits that display autosomal recessive patterns of inheritance. In a genetics experiment, students cross wild-type flies with flies that have purple eyes and ebony bodies. The students observe that all the flies in the  F1  generation have normal eyes and a normal body color. The students then allow the  F1  flies to mate and produce an  F2  generation. The students record observations about the flies in the  F2  generation and use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the students’ chi-square goodness-of-fit test is presented in Table 1.  The students choose a significance level of  p=0.01 . Which of the following statements best completes the next step of the chi-square goodness-of-fit test? Use the AP biology formulas sheet to view the equation and chi square table.

• The calculated chi-square value is 2.11, and the critical value is 7.82.

• The calculated chi-square value is 2.11, and the critical value is 11.35.

• The calculated chi-square value is 10.48, and the critical value is 7.82.

• The calculated chi-square value is 10.48, and the critical value is 11.35.

Tags

NGSS.HS-LS3-3

In a strain of tomato plants, short plant height and small fruit size are traits that display autosomal recessive patterns of inheritance. To In a strain of tomato plants, short plant height and small fruit size are traits that display autosomal recessive patterns of inheritance. To investigate whether the traits segregate independently, researchers cross a pure-breeding line of tall tomato plants that have large fruits with a pure-breeding line of short tomato plants that have small fruits. The researchers observe that all the plants in the  F1  generation are tall and have large fruits. The researchers cross the  F1  plants with one another to generate an  F2  generation. The researchers record observations for the  F2  generation and will use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the chi-square goodness-of-fit test is shown in Table 1.  The researchers choose a significance level of  p=0.05 . Which of the following best completes the chi-square goodness-of-fit test? Use the AP biology formula sheet.

• The calculated chi-square value is 9.24, and the critical value is 7.82. The null hypothesis of independent assortment can be rejected.

• The calculated chi-square value is 9.24, and the critical value is 9.49. The null hypothesis of independent assortment cannot be rejected.

• The calculated chi-square value is 13.13, and the critical value is 7.82. The null hypothesis of independent assortment can be rejected.

• The calculated chi-square value is 13.13, and the critical value is 9.49. The null hypothesis of independent assortment cannot be rejected.

Tags

NGSS.HS-LS4-3