BIOLOGY 120 Lecture 6: Biology Lecture Week 6
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Question 1
What would be the dimensions of a Punnett square for the cross Hh x hh?
4 x 4 | ||
2 x 2 | ||
2 x 1 | ||
1 x 1 |
Question 2
In dogs, the gene for fur color has two alleles. The dominant allele (G) codes for grey and the recessive allele (g) codes for black fur. The female dog is heterozygous. The male dog is homozygous recessive. Figure out the phenotypes and genotypes possible in their puppies by using a Punnett Square. (provide answers on one line with commas between - xx, xx, xx, xx)
Genotypes: , ,
Phenotypes: ,
Question 3
The fundamental Mendelian process which involves the separation of alleles located at the same spot on the chromosome would be called ___.
segregation | ||
independent assortment | ||
continuous variation | ||
discontinuous variation |
Question 4
Using the information given on dominant and recessive traits, state the phenotype for each genotype. Purple flowers are dominant to white flowers in pea plants.
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Question 5
In dogs, the gene for fur color has two alleles. The dominant allele (G) codes for grey and the recessive allele (g) codes for black fur. The female dog has black fur and the male dog is homozygous dominant. Make a Punnett square to determine the chance of getting each genotype and phenotype below.
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Question 6
Using the information given on dominant and recessive traits, state the phenotype for each genotype. Round seeds are dominant to wrinkled seeds in pea plants.
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Question 7
Using the information given on dominant and recessive traits, state the phenotype for each genotype. Hairy knuckles are dominant to non-hairy knuckles in humans.
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Question 8
An allele is ___.
one of the bases in DNA | ||
an alternate form of a gene | ||
another term for epistasis | ||
present only in males and is responsible for sex determination |
Question 9
In a cross of a round hybrid pea with a true breeding round parent (Rr x RR), what genotypic proportions would be observed in the offspring?
Half heterozygous, half homozygous dominant | ||
Half round, half wrinkled | ||
All heterozygous | ||
All round |
Question 10
What would be the dimensions of a Punnett Square depicting a dihybrid cross?
1 x 4 | ||
2 x 4 | ||
4 x 4 | ||
2 x 2 |
The TATA-binding protein (TBP) binds to the TATA box sequence in eukaryotic promoters. What is its function in transcriptional initiation?
It blocks access of RNA polymerase to the promoter, until removed by general transcription factors.
It is the subunit of prokaryotic RNA polymerase that is required to recognize promoters.
It modifies histones so nucleosomes can be removed from DNA for transcription.
It bends and partly unwinds DNA at a promoter.
The genetic code is said to be âdegenerateâ because
there are more codons than amino acids. |
there are more amino acids than codons. |
different organisms use different codons to encode the same amino acid. |
some codons specify more than one amino acid. |
Three general mechanisms appear to be responsible for the conversion of proto-oncogenes to oncogenes
overexpression, point mutations, deletions |
inversions, translocations, methylation |
familial, sporatic, phosphorylation |
None of the above account for the conversion of proto-oncogenes into oncogenes |
Transcriptional control of genes that acts by regulating the continuation of transcription is called
induction |
attenuation |
antitermination |
negative inducible control |
The genetic code is fairly consistent among all organisms. The term used to describe this consistency is
redundant |
resilient |
universal |
the central dogma |
The F, G, and H loci are linked in the order written. There are 30 cM between F and G and 30 cM between G and H. If a plant Ff Gg Hh is testcrossed, what proportion of the progeny will be ff gg hh, assuming no interference?
0.7 |
0.3 |
0.245 |
0.15 |
DNA synthesis is always from 5â to 3â because
replication must be continuous |
the strands are antiparallel |
primers lack 3â to 5â exonuclease activity |
none of the above |
The F, G, and H loci are linked in the order written. There are 30 cM between F and G and 30 cM between G and H. If a plant Ff Gg Hh is testcrossed, what proportion of the progeny will be ff gg hh, assuming no interference?
0.7 |
0.3 |
0.245 |
0.15 |
In the ZZ-ZW sex-determination system, if an AaBb female was crossed to an individual of genotype Aa Bb, what is the probability of a female offspring with the two dominant traits given by alleles A and B? Assume A and B are dominant alleles.
1/8 |
1/16 |
9/16 |
9/32 |
Which of these statements is incorrect?
Syntenic genes are located on the same chromosome. |
Independent assortment results in recombinant chromosomes. |
You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome. |
Linked genes are always syntenic. |
What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?
0.49 cM |
4.9 cM |
49 cM |
490 cM |
What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)
Human error: they should have been more careful about theirexperimental setup. |
Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits. |
Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes. |
Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean. |
When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?
9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio. |
Genotypes of the offspring can be determined based on theirphenotype. |
If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation. |
B and C |
Why do we map genes?
To understand how genes interact with each other |
Comparative genomics analysis |
To determine the genotype of an organism |
All of the above |