Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
Answer:
nope that not possible because O has no antigens which can donate any blood group but A donates to same blood group bcoz of same antigen.
Answer:
Parietal cells in the gastric glands in the stomach.
Explanation:
The main constituent of gastric acid is hydrochloric acid produced and that is roduced by parietal cells in the gastric glands in the stomach.
Crossing over in germinal cells during gametogenesis is one of the primary causes of this variance. In humans, gametes are generated during meiosis, which reduces the chromosome number to 23.
One cell divides twice during meiosis to create four daughter cells. These four daughter cells are haploid, meaning they have half the amount of chromosomes as the parent cell. Meiosis generates sex cells or gametes? (eggs in females and sperm in males).
During meiosis, homologous chromosomes are paired, resulting in gene recombination.
Non-sister chromatids exchange genetic material, resulting in novel allele combinations.
Learn more about to meiosis visit here;
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