Assignment 9-3

BIBL-350 Week 4 Chapter Assignment 2 Assignment 9-1 (p. 184-185) ( 20 points ) Please complete the assignment by entering your answer(s) in the shaded areas . 9. Concordance Exercises 1. Use the concordance to answer the following questions about Acts 1:8 ( 7 points ) a. Write out the English transliterated form of the word translated “power” in Acts 1:8. dunamis ) b. How many times does this word occur in the New Testament? 83 times ) c. List the passages in Acts that translate this word as “power” 1:8; 3:12; 4:7, 33, 6:8; 8:10 & 10:38 ) d. List the passages in Acts that translate this word as “miracles” 2:22; 8:13 & 19:11 ) 2. Use the concordance to answer the following questions about Exodus 4:21. ( 4 points ) a. Write out the English transliterated form of the word translated “power” in Exodus 4:21. hand ) b. How many times does this word occur in the Old Testament? 859 times ) c. List the passages in Exodus that translate the word as “power.” 2:19, 3:20; 4:2, 4, 6, 7, 17, 20, 21; 5:21; 7:4-5, 15, 17, 19; 8:5-6, 17; 9:3, 15, 22; 10:12, 21-22; 12:11; 13:3, 9, 14, 16; 14:16, 21-22, 26-27, 29-30; 15:6, 9, 12, 20; 16:3; 17:5, 9, 11; 18:9-10; 19:13; 21:13, 20, 24; 23:1, 31; 24:11; 32:4, 11, 15; 33:22-23; 34:4, 29 & 38:21 ) 3. The NASB uses the word “examined” in 1 Corinthians 4:3, “judgment” in 1 Cor. 4:5, and “decide” in 1 Cor. 6:5 ( 4 points ) Are these the same Greek words? no Write out the English transliteration of the three Greek words translated as “examined,” “judgment,” and “decide” in these three passages. anakrino, krino, diakrino ) 4. Use the concordance to answer the following questions about the word “hope” ( 2 + 2 + 1 = 5 points ) : a. Paul uses the word “hope” in Romans 4:18. How many times total does he use this same word in his letters? 36 ) 4

Linked genes and recombinant chromosomes

Since each chromosome bears a large number of genes, we should expect that two alleles of interest might reside on the same chromosome (linked). This scenario gives a different inheritance pattern than if the genes were on different chromosomes. Consider the following example:

Sweet pea plants bear a gene for flower color (P) on the same chromosome as the gene for pollen shape (L). This diagram shows two alleles for each gene on a pair of homologous chromosomes that have replicated. Note that this plant is heterozygous for both genes, so the genotype is PpLl. But because the genes for flower color and pollen shape are linked, only two types of gametes can be produced: PL and pl.

Since these genes are linked, the Punnett square for heterozygous parents should look like this:

Cross:
PpLl x PpLl

Maternal

PL

pl

Paternal

PL

PPLL

PpLl

pl

PpLl

ppll

Note that the phenotypic ratio of the offspring is 3:1. If the genes were not linked, independent assortment would occur and the ratio would be 9:3:3:1 as you know from problems using dihybrid crosses. However, working with linked genes is not this simple due to the potential for crossing over during the first meiotic prophase. Review this animation if you do not remember how crossing over occurs. In the case of the sweet pea genes, crossing over between the flower color and pollen shape genes might look like this:

In the case shown here, the P allele on one chromatid of the P chromosome will switch with the p allele on one chromatid of the homologous chromosome. This will happen because the ends of these chromatids have swapped places (crossed over). Thus, two new linkages will be created: a chromosome with pL and a chromosome with Pl. The new chromosomes formed by crossing over are called recombinant chromosomes. Note that crossing over which occurs anywhere between the P and L loci will create new linkages.

Now let's examine an actual experiment that was performed using the above plant. Examine this cross carefully. Note that purple flower color (P) is dominant to red flower color (p) and that long pollen (L) is dominant to round pollen (l).

Experiment to determine inheritance of two traits in sweet peas

Analysis of the data from this experiment can tell us quite a bit about the inheritance of flower color and pollen shape:

This analysis of data reveals that the ratio of F2 phenotypes was neither 9:3:3:1 nor 3:1. However, if the two most common phenotypes are compared, they are close to a 3:1 ratio. This indicates that the genes are linked. Why are small numbers of other phenotypes present? This is due to crossing over between the flower color and pollen shape genes in a small number of cases to produce offspring with recombinant chromosomes. We can calculate the frequency of crossing over by dividing the number of these offspring by the total number of offspring observed:

19 + 27 / 296+19+27+85 x 100 = 10.8%

Thus we can conclude that the genes for flower color and pollen shape are linked and that the frequency of crossing over between them is 10.8%. This can also be expressed as a recombination frequency of 10.8%. Note that a large number of offspring must be observed to produce reliable ratios.

When you understand how inheritance of linked genes works, answer questions 8 and 9 .

ACTIVITY 3. POLYGENIC INHERITANCE AND EPISTASIS

There are many cases were the phenotype of a trait is determined by more than one gene. You have already learned about two such polygenic traits (human height and skin color) in the Patterns of Inheritance topic. Polygenic traits may be influenced by a number of genes and their interactions which produces a range of different phenotypes. Two examples are described below.

A cross between a white mouse and a black mouse yields progeny that are not only white and black, but also chocolate-brown.

If the cross is repeated, the ratio obtained is always 2 white mice: 1 black mouse: 1 brown mouse.  At first we may suspect incomplete dominance in one pair of alleles, but since the ratio of coat color is consistent when the same pair of mice are repeatedly mated, we should question this hypothesis. 

The unusual ratio can be explained as the result of two genes that code for enzymes in the same metabolic pathway. We can visualize this interaction by drawing a simple two-step pathway involving two enzymes, each having alleles with a dominant/recessive relationship.

Colorless compound------->Brown--------->Black
                            Enzyme  1       Enzyme 2
Alleles:                 C/c                   B/b

If a mouse possesses functional enzyme 2 (at least one B allele) that individual can convert brown pigment to black. However, since functional enzyme 1 must be present to have the brown pigment to be converted, individuals that are ccBB or ccBb or ccbb will not able to produce brown or black pigment and will be white in phenotype. So in some sense, the pathway is blocked at the point of enzyme one. The cross described above is the result of a CcBb  X ccbb mating. Diagram this cross on your worksheet to verify that the expected results are consistent with the ratios described above. Then answer question 10.

Epistasis occurs when the alleles of one gene mask the effects of a dominant allele of another gene. In the above example, epistasis occurs when cc of the gene for enzyme 1 prevents the expression of enzyme 2 even though the B gene is present (i.e. the potential black color of the hair is masked and the mouse is white).

Another example of polygenic inheritance involves genes interacting in a different fashion.   Wild mice have hair that is "mousy" brown or dull in color, quite different from the light chocolate-brown or velvet-black colored mice obtained in the cross above. This is because coat color in wild mice is the result of two genes which place color at different positions on the hair shaft.

Wild mouse

Mouse hairs (magnified)

Therefore, mice also have a third gene for coat color. The A allele for this gene gives the banding pattern on a hair, whereas aa mice lack this banding. In the cross discussed above, all offspring were aa for this coat color gene. Most of the traits that we are interested in regarding our own phenotype (such as eye color, hair color, height, and nose shape) are polygenic traits.

When you understand the above examples of polygenic inheritance and epistasis, answer questions 11 and 12.

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