Mid-Term Review - Unit 3

Sickle-shaped red blood cells result from a mutation in the gene that codes for hemoglobin. This mutation results in Sickle-cell anemia. A Partial sequence of bases form a normal hemoglobin gene and a sequence that results in Sickle-cell anemia are shown.

Normal Hemoglobin: T - G - A - G - G - T - C - T - C - C - T - C

Sickle-cell Hemoglobin: T- G - A - G - G - T - C - A - C - C - T - C

What type of a mutation is depicted in this sequence?

• Substitution (point mutation)

• Insertion

• Deletion

• Frameshift

Tags

NGSS.HS-LS3-2

In the P generation, a homozygous tall pea plant is crossed with a short pea plant. The probability that an F2 plant will be tall is...

• 25%

• 50%

• 75%

• 100%

Tags

NGSS.HS-LS3-3

Chickens have different types of feathers. Frizzled feathers curl toward a chickens head. Assume that feather type is determined by a single gene and that the allele for frizzled feathers is dominant over the allele for straight feathers. In a cross between two chickens with straight feathers, what percentage of offspring can be expected to have frizzled feathers?

• 0%

• 25%

• 50%

• 100%

Tags

NGSS.HS-LS3-3

A punnet square shows all of the following EXCEPT...

• all possible results of a genetic cross

• the genotypes of the offspring

• the alleles in the gametes of each parent

• the actual results of a genetic cross

Tags

NGSS.HS-LS3-1

Which of the following can be determined by examining this karyotype?

• The presence of an additional chromosome.

• The presence of a homozygous genotype.

• The person's genome

• The person's phenotype

Tags

NGSS.HS-LS3-1

The table below lists the traits for fruit color where allele R exhibits incomplete dominance over R'. Heterozygous fruit have orange phenotypes. What percentage of offspring are expected to have an orange phenotype if the parent plants are orange (RR') and yellow (R'R')?

• 25%

• 50%

• 75%

• 100%

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-3

In guinea pigs, black coat color (B) is dominant over white coat color (b) and short hair (S) is dominant over long hair (S). Based upon the dihybrid cross shown below, what are the genotype and phenotype for letter X?

• Bbss; brown coat, long hair

• BbSs; black coat, short hair

• BBss; black coat, long hair

• Bbss; black coat, long hair

Tags

NGSS.HS-LS3-1

NGSS.HS-LS3-2

The table shows a DNA sequence and three typs of mutations that can change the DNA sequence. Which muation will cause a transition to stop?

• Mutations 1 and 3 only

• Mutation 1 only

• mutation 2 only

• mutations 1,2,and 3

Tags

NGSS.HS-LS1-1

NGSS.HS-LS3-2

The structure labeled A in Figure 10-3 is called the

• Centriole

• Sister chromatids

• Centromere

• Spindle

Tags

NGSS.HS-LS1-4

The structure labeled B in Figure 10-3 is called the

• Centriole

• Sister Chromatids

• Centromere

• Spindle

Tags

NGSS.HS-LS1-4