answer

watching

views

9 Oct 2018

Plant breeders studying genes affecting leaf shape in the plant Arabidopsis identified six independent recessive mutations that resulted in mutant serrated leaves rather than wild type smooth leaves. They began to perform complementation tests, but some of the tests could not be completed due to a fire in the greenhouse. Their results are indicated in the table on the right:

mutations

1

2

3

4

5

6

1

â

+

â

*

+

2

â

â

3

â

â

4

â

**

5

â

+

6

â

The asterisks in the table represent two complementation tests that were not performed. What would be expected?

* would not show complementation and ** would show complementation

both * and ** would not show complementation

both * and ** would show complementation

* would show complementation and ** would not show complementation

answer

watching

947

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Deanna HettingerLv2

11 Oct 2018

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Related questions

Cystic Fibrosis (CF) is an autosomal recessive genetic disease that is characterized by the secretion of thick, abnormal mucous in the lungs. The mutated gene that leads these symptoms is the CF transmembrane conductance regulator (CFTR gene). The encoded protein is a chloride channel that moves chloride ions out of epithelial cells. In a homozygous individual, the lack of this chloride channel function in a mutant slows the flow of water into extracellular spaces. This lack of water export leads to the heavy mucus in the lungs of CF patients. Although children with CF would die in childhood, current treatments provide life expectancies into the 30âs.

CF is the most frequently inherited genetic disorder among American children. It affects 1 of every 2,500 children; 30,000 currently have the disease. Although its frequency is higher in European populations, it affects all human populations. Heterozygous carrier frequencies in the US are estimated to be 1 out of 29 Caucasians, 1 out of 46 Hispanics, 1 out of 61 African Americans and 1 out of 90 Asians. There are genetic screens for carriers of CF but they are not routinely performed despite this high frequency of carriers. At least all states in the US proscribe testing of newborns to identify children with CF early so that treatment can be started. Newborn tests for CF: http://www.cff.org/AboutCF/Testing/NewbornScreening/ScreeningforCF/ . If CF is such a common genetic disease, and such a simple Mendelian trait, why is genetic screening of parents not done more frequently? Is it just economic, a pain to do, or is it more complex such that the results of genetic testing are not just positive or negative?

Answer the following questions:
1. If two parents are heterozygous carriers of CF, show a pedigree with four children showing the incidence of carriers and afflicted children. If each unafflicted child (carrier or not) mates with another person, who has a 1 out of 30 chance of being a carrier of CF, what is the expected incidence CF disease in the third generation (frequency of CF in any one of grandchildren of the original couple)?

2. Pleiotropy of Cystic Fibrosis. Watch the two videos on the left at:

http://www.cftrscience.com/MOD-animation
or, if you can not view these, read the short NIH Genetics Home Reference for Cystic Fibrosis http://ghr.nlm.nih.gov/condition/cystic-fibrosis

Connect the description with one of the following videos, or related ones (there are many) you see:
video (10 minutes) of daily life of a child with Cystic Fibrosis http://www.youtube.com/watch?v=WUsuGKzdDg8
Lauraâs (20 year old) video diary of life with CF http://www.youtube.com/watch?v=Pgrf8qJXhpc&feature=related

In your own words, describe the range of CF phenotypes. List the other organs besides lungs that are affected and how does it affects those organs.

3.The DNA sequence of the CFTR gene is 170,000 base pairs in length and encodes the CFTR protein that is 1480 amino acids long. Open the accompanying text file (in Canvas/Files/Discussion folder) âCFTR gene DNA sequenceâ. You will see an alignment of the NIH human genome sequence of the CFTR gene â called âReferenceâ rather than âwild-typeâ- against a gene sequence containing a common mutation in CFTR. Look at the alignment and determine how the DNA sequence differs between them. What is the position number where they differ? How many and what nucleotides do they differ in? A â-â mark in one sequence in the alignment means that that base sequence is missing the corresponding bases in the other sequence. The sequence is a long one but the differences are in bolded bright red text. Matching sequence is marked with a | linking the two sequences. Look near the sequence AAATATCATC if you use FIND tool in Word.

In another file, you will an alignment of the reference translated into a protein sequence, compared to the mutant allele translated into protein sequence. The differences are again in red whereas the areas where the proteins match are in black. Look at the alignment. How do these two proteins differ? Again, find the position number and describe how they differ at that location. A â-â mark in the alignment means that amino acid is missing when compared to the other sequence.

4. Given that the CFTR gene can be analyzed for a mutation that causes CF, couples interested in being tested as carriers face several choices given the results.
What could be a consequence of a false positive result (couple is identified as carriers but they would not produce any children with symptoms)?

What could be a consequence of a false negative result (a test indicates the couple is not both carriers but they would produce some children with CF)?

5. The sensitivity of the test can vary with ethnic group. The following is a chart from the NCBI Genetic testing site. A sensitivity less than 100% would be likely to produce false negative results from an inability to determine that someone is carrier.

CF Carrier Test Sensitivity by Ethnic Group http://www.ncbi.nlm.nih.gov/projects/GeneTests/static/concepts/primer/ptscarr5.shtml
Ethnic Group ¹	Population Frequency of CF	Sensitivity of Carrier Detection ²
Caucasian	1 in 29	90%
Ashkenazi Jewish	1 in 29	97%
Native American	1 in 32	94%
Latino	1 in 46	57%
African-American	1 in 60	75%
Asian	1 in 90	30%

Why might different ethnic populations differ in the ability of the genetic test to identify carriers of CF?

1. A species of flowering plant includes plants with blooms that range from bright golden yellow to very pale yellow; no flowers carrying the yellow pigment allele appear white. The allele for yellow flower color:

a. must be a suppressor of another allele.
b. demonstrates incomplete penetrance.
c. demonstrates variable expressivity.
d. must be a modifier of another allele.
e. demonstrates variable penetrance.

2. Coat color in mice is determined by two alleles acting at a single locus: B is dominant and results in mice with black coats, and b is recessive and results in mice with white coats. In a litter of 12 mice, nine are black (B/â) and three are white (b/b). What is the penetrance of the B coat color allele?

a. zero
b. 25 percent
c. 50 percent
d. 75 percent
e. 100 percent

3. Semisterility in nearly 50 percent of the progeny they produce is often an indication that:

a. the parents are translocation heterozygotes.
b. both parents carry pericentric inversions.
c. both parents carry polytene chromosomes.
d. one of the parents carries a chromosomal deletion that removes a telomere.
e. one of the parents carries a chromosomal deletion that removes a chromosome's centromere.

4. A researcher is studying melanoma cells derived from a patient's skin tumor. He determines that all melanoma cells are 2n + 1, but the chromosome number in all normal skin cells is 2n. The 2n + 1 melanoma cells likely arose from:

a. mitotic nondisjunction.
b. meiotic nondisjunction.
c. mitotic recombination.
d. parthenogenesis.
e. colchicine treatment.

5. Why won't a balancer chromosome in Drosophila produce viable crossover products when paired with a wild-type chromosome?

a. Balancer chromosomes contain multiple inversions.
b. Balancer chromosomes contain multiple deletions.
c. Balancer chromosomes have no centromere.
d. Balancer chromosomes have no telomeres.
e. None of the answers is correct.

6. A doctor notes that a patient's genome contains 100 potentially harmful recessive mutations; however, the patient is heterozygous for all of the loci considered and thus does not demonstrate any abnormal phenotypes. The 100 recessive mutations the doctor discovered represent the patient's:

a. bivalents.
b. monosomy.
c.genetic load.
d. amphidiploidy.
e.nondisjunction.

7.The term describing the relationship between the number of copies of a gene and how much product is made from the gene is:

gene-dosage effect.
gene balance.
dosage compensation.
transcription.
aneuploidy.

8. The gene for upright ears in rabbits has two alleles. The E allele is dominant and results in upright ears, and the recessive e allele results in droopy ears. Of 100 rabbits surveyed, 87 had a genotype for upright ears (E/â). However, only 72 of these 87 rabbits had upright ears; the remaining 15 rabbits had completely droopy ears. This suggests that the allele for upright ears:

a. has incomplete penetrance.
b. has complete penetrance.
c. demonstrates variable expressivity.
d. is lethal.
e. is incompletely dominant.

9. A researcher has selfed an autotetraploid plant with the genotype D/D/d/d, and in the resulting progeny she observes a 35:1 phenotypic ratio. What can she conclude from this observation?

a. The progeny from this cross do not demonstrate a standard Mendelian phenotypic ratio.
b. In the 35:1 phenotypic ratio, the 1 represents the phenotype of the d/d/d/d progeny.
c. During meiosis in the D/D/d/d parent plant, it is likely that only bivalents, no quadrivalents or trivalents, formed.
d. It is likely that in this plant species, gametes associate with one another randomly.
e. All of the answers are correct.

10. A researcher is studying fur color in a new species of bats. He notices that fur color, which is due to the presence of a dominant allele (G), ranges from light gray to very dark gray. In addition, some bats that carry a G allele have no fur pigmentation whatsoever; their coats appear white. This is an example of:

a. only variable expressivity.
b. only incomplete penetrance.
c. both variable expressivity and incomplete penetrance.
d. recessive allele lethality.
e. pleiotropy.

amethystdinosaur303

**Please answer ALL 10 QUESTION**

a. must be a suppressor of another allele.

b. demonstrates incomplete penetrance.

c. demonstrates variable expressivity.

d. must be a modifier of another allele.

e. demonstrates variable penetrance.

a. zero

b. 25 percent

c. 50 percent

d. 75 percent

e. 100 percent

3. Semisterility in nearly 50 percent of the progeny they produce is often an indication that:

a. the parents are translocation heterozygotes.

b. both parents carry pericentric inversions.

c. both parents carry polytene chromosomes.

d. one of the parents carries a chromosomal deletion that removes a telomere.

e. one of the parents carries a chromosomal deletion that removes a chromosome's centromere.

a. mitotic nondisjunction.

b. meiotic nondisjunction.

c. mitotic recombination.

d. parthenogenesis.

e. colchicine treatment.

5. Why won't a balancer chromosome in Drosophila produce viable crossover products when paired with a wild-type chromosome?

a. Balancer chromosomes contain multiple inversions.

b. Balancer chromosomes contain multiple deletions.

c. Balancer chromosomes have no centromere.

d. Balancer chromosomes have no telomeres.

e. None of the answers is correct.

a. bivalents.

b. monosomy.

c.genetic load.

d. amphidiploidy.

e.nondisjunction.

7.The term describing the relationship between the number of copies of a gene and how much product is made from the gene is:

a. gene-dosage effect. gene balance.

b. dosage compensation.

c. transcription.

d. aneuploidy.

a. has incomplete penetrance.

b. has complete penetrance.

c. demonstrates variable expressivity.

d. is lethal.

e. is incompletely dominant.

9. A researcher has selfed an autotetraploid plant with the genotype D/D/d/d, and in the resulting progeny she observes a 35:1 phenotypic ratio. What can she conclude from this observation?

a. The progeny from this cross do not demonstrate a standard Mendelian phenotypic ratio.

b. In the 35:1 phenotypic ratio, the 1 represents the phenotype of the d/d/d/d progeny.

c. During meiosis in the D/D/d/d parent plant, it is likely that only bivalents, no quadrivalents or trivalents, formed.

d. It is likely that in this plant species, gametes associate with one another randomly.

e. All of the answers are correct.

a. only variable expressivity.

b. only incomplete penetrance.

c. both variable expressivity and incomplete penetrance.

d. recessive allele lethality.

e. pleiotropy.