Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
<h2>Ovulation </h2>
Explanation:
In women, luteinizing hormone carries out different roles in the two halves of the menstrual cycle
- In week one to two of the cycle, luteinizing hormone is required to stimulate the ovarian follicles in the ovary to produce the female sex hormone, oestradiol
- Around day 14 of the cycle, a surge in luteinizing hormone levels causes the ovarian follicle to tear and release a mature oocyte (egg) from the ovary, a process called ovulation
- For the remainder of the cycle (weeks three to four), the remnants of the ovarian follicle forms a corpus luteum
- Luteinizing hormone stimulates the corpus luteum to produce progesterone, which is required to support the early stages of pregnancy, if fertilization occurs
Answer:
because he trusted no one
Explanation:
The answer is prime mover
Explanation:
The genes that determine color and pattern are linked.
- Solid pattern allele (L) is dominant over lined pattern (l)
- Blue color allele (B) is dominant over aqua color (b)
<h3><u>Initial cross</u></h3>
Homozygous lined aqua beetle has the genotype <em>lb/lb</em>
It can produce only 1 type of gamete: <em>lb</em>
X
Homozygous solid blue beetle has the genotype<em> </em><em>LB/LB</em>
It can produce only 1 type of gamete:<em> </em><em>LB</em>
The F1 is formed by the union of the parents' gametes, so it will have the genotype <em>lb/LB</em>
<h3><u>F1 test cross</u></h3>
Test crosses involve a homozygous recessive individual.
<u>The homozygous recessive individual</u><u><em> lb/lb</em></u><u> can only produce 1 type of gamete:</u>
<u>The F1 individual </u><em><u>lb/LB</u></em><u> can produce 4 types of gametes:</u>
- lb - parental
- LB - parental
- lB - recombinant
- Lb - recombinant
