Category: Answers

Chapter 3

3.4. b: 0, c: ½

3.8. Panel A is anaphase I of meiosis, because the homologous chromosomes are paired. Panel B is anaphase II of meiosis, because the chromosome number has been reduced by half. Panel C is anaphase of mitosis, because the homologous chromosomes are not paired.

3.10: 5/6 expected to survive.

3.12: A) [0.5 x 0.5] + [0.5 x 0.5] + [0.5 x 0.5] = 0.75. B) 0.5 x 0.5 = 0.25

3.17: Note first that there is no chance of the X-linked color-blindness allele being passed to II-4, because he inherits his Y chromosome from his father (I-2) and his X chromosome from his normal mother (I-1). Because we are told that I-2 is the only possible source of the mutant gene, this means that there is no possibility that the male III-3 carries the mutant gene. (Moreover, III-3 is not color blind, and so his X chromosome must be nonmutant.) On the other hand, the female II-2 must be a carrier for the color-blindness allele because she inherits an X chromosome from her father (male I-2), whom we know carries the allele for color blindness. Given that II-2 is heterozygous, there is a probability of 1/2 that she transmits the allele to her daughter (III-2) and another probability of 1/2 that III-2 transmits the allele to the male IV-1. Hence, the total probability that IV-1 is color blind is 1 x 1/2 x 1/2 = 1/4.

3.19: There are 1000 progeny altogether, and the expected numbers are 562.5, 187.5, 187.5, and 62.5, respectively, from top to bottom. This chi-square test has 3 degrees of freedom (four classes of data); the final chi-square value is 3.6. The P value is about 0.33. The data give no reason to reject the hypothesis of 9:3:3:1 segregation.

 

Chapter 5

5.12: The mom has a Robertsonian translocation joining the long arm of Chr. 21 with the long arm of another acrocentric chromosome. The child has 46 chromosomes. This karyotype differs from the usual situation for trisomy 21 because the extra chromosome 21 is not a free chromosome.

 

Additional Question not in your textbook:

#1. An insect from a strain that is true-breeding for white eyes was crossed to an insect from a strain that is true-breeding for red eyes. All the F₁s had white eyes. The F₂s consisted of:

243 white eyed, 58 red eyed, 21 cream eyed.

Provide the genotypes for the parents, F₁s and F₂s of the cross.

Hint: epistasis!

 

How to solve this problem? First, start by analyzing the F₂ phenotypic ratios:

Total progeny = 322

White = 243/322 ≈ 12/16

Red = 58/322 ≈ 3/16

Cream = 21/322 ≈ 1/16

The 16 in the denominator indicates that the F₁ is segregating for two gene pairs, and thus that the F₁ is a dihybrid. The F₁ can arbitrarily be given the genotype of Aa Bb. The F₂s can now be analyzed with respect to a modified 9:3:3:1 dihybrid ratio:

9 A_ B_

3 A_ bb

3 aa B_

1 aa bb

Thus,

P        aa BB red     x       AA bb white

F1      Aa Bb white

F2      A_ _ _        white

aa B_ red

aa bb  cream

 

ANSWERS: BI311 Homework Assignment #2

Chapter 2

2.2:   2/3. Because the seed is round it cannot be ww. Ww seeds give 2 bands in the gel. Of the round seeds, 1/3 are WW and 2/3 are Ww.

2.4.    Note that you assume ‘rare’ alleles to exist in heterozygous state.

a) parents must be Aa x aa so the risk of having an Aa child is 1/2

b) parents must be Aa x Aa so the risk of having an aa child is 1/4

c) parents must be Aa x aa so the risk of having an aa child is 1/2

 

2.8.

a) Individual II-2 is aa

b) Both individuals I-1 and I-2 must be Aa

c) Individuals II-1 and II-3 are either AA or Aa

d) Because II-3 is phenotypically normal, the odds that they are heterozygous carrier is 2/3

 

2.16:   Due to Mendelian segregation, the expected numbers are: a) 0; b) 100; c) 200; d) 100; e) 0; f) 0

 

2.21.

a) Recessive allele because the affected individual had unaffected parents. Also, the parents of the affected individual are related.

b) The double line indicates a mating between relatives.

c) III-1 and III-2 are first cousins.

d) I-1 and I-2 could be (AA) or (Aa). At least one individual must be (Aa) or both could be (Aa). Not possible for both to be (AA). II-1 and II-4 are most likely (AA) because they a allele is a rare trait. Individuals that marry into a family are assumed to be wildtype unless there is evidence to the contrary.   III-1 and III-2 are both (Aa) because they have an affected offspring (IV-4).

 

Chapter 3

3.1. After the centromeres have split, each former sister chromatid is counted as a chromosome in its own right, so at anaphase there are 92 chromosomes.

 

Additional Questions not in your textbook:

#1. On the basis of Mendel’s observations, predict the results from the following crosses with peas:

a) true-breeding tall peas x true-breeding dwarf peas

all tall

b) the offspring of (a) self-fertilized

¾ tall, ¼ dwarf

c) offspring of (a) crossed with the original tall parent

all tall

d) the offspring of (a) crossed with original dwarf parent

½ tall, ½ dwarf

 

#2. In pigeons, a dominant allele C causes a checkered pattern in feathers, the recessive allele c produces a plain pattern. Feather color is controlled by an independently assorting gene; where the dominant B allele causes red feathers and the recessive allele b produces brown feathers. You cross a true-breeding checkered red pigeon with a true-breeding plain brown pigeon.

a) What is the phenotype of the F1 offspring?

checkered and red feathers

b) If these F1 offspring are crossed with each other, what phenotypes will appear in the F2 generation, and in what proportions?

9 checkered red : 3 checkered brown : 3 plain red : 1 plain brown

 

 

Correct answers in bold.

1. (5 points) Which of the following is NOT a property associated with DNA?

a. double helix

b. sugar-phosphate backbone

c. parallel strands (5’ to 3’)

d. nucleic acids

 

2. (5 points) The process by which a polypeptide (or, protein) is produced from a template RNA molecule is referred to as:

a. replication

b. translation

c. transcription

d. reverse transcription

 

3. You have been given a tube containing a DNA sample. The DNA is linear and 1,000 base pairs (bp) in length. To further study this DNA, you decide to use the restriction enzyme PvuII and agarose gel electrophoresis. The diagram in the box below shows how PvuII digests DNA molecules, with the arrowheads indicating strand cleavages sites.

Part A. (5 points) Which of the following terms best describes the ends of DNA molecules that result after digestion with PvuII?

a. Blunt-ended

b. Sticky-ended

c. Terminated

d. Replicated

Part B. (5 points) After performing the restriction digestion of your DNA sample with PvuII and running an agarose gel, you observe three bands on the gel: 50 bp, 450 bp, and 500 bp. How many recognition sites for PvuII are present in your original linear DNA sample? Write the answer in below.

Two

Chapter 1

1.10: 5’-ATGAC-3’

1.12: 5’-TGTCGTATTTGCAAG-3’

1.18: A) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly. B) This mutation alters the start codon into a non-start codon. Thus, translation will not start with the first AUG; translation would start with the next AUG down the mRNA, or not at all if it is too far down the molecule. C) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly (no change). D) The Val is changed into an Ala. E) Met-Ser-Thr-Ala-Val-Leu.

1.19: There are 64 possible codons and only 20 amino acids; hence, many amino acids are specified by two or more codons. A mutation that changes a codon for a particular amino acid into a synonymous codon for the same amino acid does not change the amino acid sequence.

1.20: a) Y, Z and W missing, X in excess; b) Z and W missing, Y in excess; c) W missing, Z in excess

 

Chapter 6

6.1: Each fragment terminates with a single-stranded overhang whose base sequence is complementary to that of any other fragment produced by the same restriction enzyme.

6.10: The DNA in this species is replicated by a totally conservative mechanism, where the parental strands stay together and the daughter strands stay together.

6.14:

——————(HindIII)———–(PstI)———–(HindIII)————-

1.5 kb                                 0.5 kb             0.8 kb                    0.3kb