lu DIlA OleCures would be found in this bivalent? 2points) Below is the karyotyp

Question

lu DIlA OleCures would be found in this bivalent? 2points) Below is the karyotype of an organism with three homologous chromosomes. The location of 5 genes on the chromosomes is shown. The individual has the genotype LINnQqAARr Draw a picture of an oocyte of this organism showing how the chromosomes would look at the a. b. Show the steps of Meiosis I and Meiosis Il that would need to happen in order for this oocyte to c. Indicate which steps are part of meiosis I and which are part of meiosis II. Be sure to show d. Clearly indicate where any crossing over occurs and where independent assortment occurs end of S phase. produce a gamete with the genotype LnqAr. enough detail so that I know that you understand the purpose of meiosis I and meiosis

Explanation / Answer

Ans for Q.6

Blood groups A and B hshow dominance over blood groups O.

Here as the child has blood group O-ve, mother must be heterozygous for the A+ and O- alleles. Father is O-ve so he is homozygous for O-allele.

The genotype and phenotypes of future generations can be predicted using punnet's square method.

So the couple as 50%chance of having A+ and O-child phenotypically. Genotypic ratio is 1:1 means half of the children will be heterozygous dominent with A+phenotype and half will be homozygous recessive for O-.

Answer for Q. 10:

According to Hardy Weinberg law,

E+e =1

Hence as given frequently of e is 1 in 10000

Therefore, E+1/10000=1

i.e. E= 1- 0.00001= 0.9999

Hence frequency of E in the population is 0.9999 and e is 0.0001

For heterozygous frequencies,

The law states that E 2 + 2Ee+ e 2 =1

as per previous ccalculations,

(0.9999)2+2 (Ee)+(0.0001)2.  = 1

Ee=[ 1- (0.9999)2 - 10-8] ÷2

Ee= 1-0.9998 /2( as 10-8 is a very small no.)

Ee frequncy is 0.0001

So frequency of E= 0.9999, e= 0.0001 and Ee= 0.0001

Mother/ Father O-ve O-ve A+ A+O- A+O- O- O-O- O-O-

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