Genetics HW #4

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table. Parent genotypes: Cc Dd × cc dd Progeny genotypes: Cc Dd - 222 Cc dd - 280 cc Dd - 280 cc dd - 218 total - 1000 Given this data, use Table 5.2 to find the most accurate range within which the p value falls.

0.001 < p < 0.01

In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy. If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

0.015

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: px sp cn - 1461 px sp cn+ - 3497 px sp+ cn - 1 px sp+ cn+ - 11 px+ sp cn - 9 px+ sp cn+ - 0 px+ sp+ cn - 3482 px+ sp+ cn+ - 1539 total - 10,000 What is the coefficient of coincidence in this region?

0.16

In Drosophila, the genes y, f, and v are all X-linked. Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are testcrossed. The F2 are distributed as follows: y f v - 3210 y f v+ - 72 y f+ v - 1024 y f+ v+ - 678 y+ f v - 690 y+ f v+ - 1044 y+ f+ v - 60 y+ f+ v+ - 3222 total - 10,000 What is the coefficient of coincidence in this region?

0.4

Genes A and B are on different (nonhomologous) chromosomes. If an individual's genotype is Aa Bb, what is the genotypic ratio of gametes produced?

1 A B : 1 A b : 1 a B : 1 a b

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table. Parent genotypes: Cc Dd × cc dd Progeny genotypes: Cc Dd - 222 Cc dd - 280 cc Dd - 280 cc dd - 218 total - 1000 The chi-square value is the sum for all progeny classes of (observed-expected)2/expected. Using the chi-square test for goodness of fit, calculate the chi-square value to test the null hypothesis that genes C and D are unlinked. What is the chi-square value?

14.4

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred. The F2 are distributed as follows: sn car - 55 sn car+ - 45 sn+ car - 45 sn+ car+ - 55 total - 200 What is the χ2 value for a chi-square test for goodness of fit of the null hypothesis?

2.0

Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross L M / l m × l m / l m would be L m / l m?

25%

A dihybrid testcross is made between genes H and I. Four categories of offspring are produced: H I, H i, h I, and h i. How many degrees of freedom would there be in a chi-square test for goodness of fit of the null hypothesis that the H and I genes are unlinked?

3

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table. Parent genotypes: Cc Dd × cc dd Progeny genotypes: Cc Dd - 222 Cc dd - 280 cc Dd - 280 cc dd - 218 total - 1000 Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

3

DNA polymerases use their ________ activity to remove a mismatched base.

3' → 5' exonuclease

A region of DNA has six copies of a trinucleotide repeat. During one round of replication, the template strand slips as shown in the diagram. How many repeats will the DNA made using the newly synthesized strand as a template have after the next round of replication? (SEE DIAGRAM IN Q#42)

4

The map of a chromosome interval is: A——10 m.u.——B——40 m.u.——C From the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny?

40

In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y+ car+ / y car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny from such a female would be recombinant for y and car?

45%

Suppose the map for a particular human chromosome interval is: a——1 m.u.——b——1 m.u.——c——1 m.u.——d——1 m.u.——e——1 m.u.——f In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a-f interval?

5%

Assume that the mutation rate for a given gene is 5 × 10−6 mutations per generation. For that gene, how many mutations would be expected if 10 million sperm are examined?

50

In Drosophila, the genes b, c, and sp are linked and arranged as shown below: b——30 m.u.——c——20 m.u.——sp This region exhibits 90% interference. How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?

6

A region of DNA has six copies of a trinucleotide repeat. During one round of replication, the newly synthesized strand slips as shown in the diagram. If the loop is not repaired (cut out), how many repeats will the DNA made using as a template the newly synthesized strand have after the next round of replication? (SEE DIAGRAM in Q#41)

8

In a frameshift mutation all of the amino acids before the shift are changed.

False

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred. The F2 are distributed as follows: sn car - 55 sn car+ - 45 sn+ car - 45 sn+ car+ - 55 total - 200 If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

Genes sn and car are not linked.

In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman?

H r / h R

Genes A and B are close together on the same chromosome. As shown in the diagrams of meiotic cells from an A B / a b individual, crossing-over occurs between A and B in some meioses, but most of the time crossing-over occurs somewhere else on the chromosome. In total, what gametes will be produced by an A B / a b individual?

More A B and a b; and fewer A b and a B

Genes A and B are close together on the same chromosome. As shown in the diagrams of meiotic cells from an A B / a b individual, crossing-over occurs between A and B in some meioses, but most of the time crossing-over occurs somewhere else on the chromosome. If the arrangement of alleles is altered, so the individual's genotype is A b / a B, what gametes will be produced overall?

More A b and a B; and fewer A B and a b

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table. Parent genotypes: Cc Dd × cc dd Progeny genotypes: Cc Dd - 222 Cc dd - 280 cc Dd - 280 cc dd - 218 total - 1000 What is a reasonable conclusion based on the chi-square analysis?

One can say with a high degree of confidence that genes C and D are linked.

Genes A and B are so close together on the same chromosome that crossing-over occurs between A and B in only 1 in a million meioses. An AA BB individual mates with an aa bb individual to produce an F1 that is A B / a b. If 100 progeny are observed, what genotypic ratio is expected in the gametes produced by the F1?

Only parental 1 A B : 1 a b

What is the consequence to a bacterial cell of a mutation that inactivates the enzyme responsible for methylating the A in the DNA sequence 5′ GATC?

Parental and new DNA strands cannot be distinguished during mismatch repair.

A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide. How is this repair system able to determine which mismatched base is incorrect?

Some bases of the parental strand are methylated, while none of the new strand are.

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table. Parent genotypes: Cc Dd × cc dd Progeny genotypes: Cc Dd - 222 Cc dd - 280 cc Dd - 280 cc dd - 218 total - 1000 If only 100 progeny had been counted and the same proportions of progeny genotypes observed, how would the p value and the conclusion drawn about linkage change?

The p value would increase, and the likelihood of linkage decreases.

The addition of a single base pair to a gene's DNA sequence can cause a frameshift mutation.

True

In a tautomeric shift,

hydrogen atoms move to form a base with altered hydrogen properties.

Alkylating agents, such as ethylmethane sulfate (EMS), are mutagens because they

add ethyl or methyl groups to bases.

After a tautomeric shift in adenine,

adenine bonds with cytosine.

A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap) and grey body (gy). The F1 mice were wild-type for both traits. When the F1 were interbred, the F2 were distributed as follows: Females: all wild type (200) Males: wild type (91), apricot (11), grey (9), apricot & grey (89) Which of the following statements is correct?

ap and gy are X-linked and 10 map units apart

A nucleotide deletion during DNA replication

causes the amino acids encoded after the deletion to be incorrect.

What structure is responsible for maintaining the connection between homologous chromosomes until anaphase of meiosis I?

cohesin complexes that are located distal to the crossover and hold sister chromatids together

UV light and other ionizing radiations damage DNA molecules by

creating thymine dimers between adjacent thymines in the DNA chain.

Another name for a chromosome is a _______, because it contains alleles that are often inherited together.

linkage group

When DNA is damaged by UV light and is not repaired,

neither DNA replication nor transcription can occur and the organism will probably die.

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred. The F2 are distributed as follows: sn car - 55 sn car+ - 45 sn+ car - 45 sn+ car+ - 55 total - 200 What is the p value from this test? (Pick the most accurate choice.)

p > 0.5

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: px sp cn - 1461 px sp cn+ - 3497 px sp+ cn - 1 px sp+ cn+ - 11 px+ sp cn - 9 px+ sp cn+ - 0 px+ sp+ cn - 3482 px+ sp+ cn+ - 1539 total - 10,000 Which of the three genes is in the middle?

px

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: px sp cn - 1461 px sp cn+ - 3497 px sp+ cn - 1 px sp+ cn+ - 11 px+ sp cn - 9 px+ sp cn+ - 0 px+ sp+ cn - 3482 px+ sp+ cn+ - 1539 total - 10,000 What is the genotype of the females that gave rise to these progeny?

px sp cn+ / px+ sp+ cn

The first step in base excision repair is

recognizing and excising the incorrect base from a nucleotide.

Proofreading by DNA polymerase involves the removal of

several bases on the newly-synthesized strand of DNA.

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: px sp cn - 1461 px sp cn+ - 3497 px sp+ cn - 1 px sp+ cn+ - 11 px+ sp cn - 9 px+ sp cn+ - 0 px+ sp+ cn - 3482 px+ sp+ cn+ - 1539 total - 10,000 Which of the following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

sp——0.21 m.u.——px——30.01 m.u.——cn

In light repair,

the covalent bonds between the thymine dimers are broken.

The diploid garden pea plant has 14 chromosomes. The haploid fungus Neurospora crassa has 7 chromosomes. Neither organism has separate male and female individuals. Therefore, the number of linkage groups in these two organisms is

the garden pea has 7 linkage groups, and Neurospora has 7.

Exposure to UV light from the sun or tanning beds causes

thymine dimers.

Replacing an adenine nucleotide with a guanine is an example of a

transition.

Replacing a thymine nucleotide with a guanine is an example of a

transversion.

In Drosophila, the genes y, f, and v are all X-linked. Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are testcrossed. The F2 are distributed as follows: y f v - 3210 y f v+ - 72 y f+ v - 1024 y f+ v+ - 678 y+ f v - 690 y+ f v+ - 1044 y+ f+ v - 60 y+ f+ v+ - 3222 total - 10,000 Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?

y——15 m.u.——v——22 m.u.——f